It was two years ago today...

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Originally Posted by regnis

JB--As soon as you say one is an ace, the other cards in that suit are disqualified. If it is the ace of spades, there can't be 2 aces of spades, so the other spades are out. That only leaves the 6 hearts. It is the same as the dice.

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Ooh... I just saw that Regnis also found the error in jbjb's question.

I love this. regnis is a genius attorney.

Alan, the Two Dice Puzzle thread at WoV ended with you and Shack supposed to meet up for a little gambling game. He was offering you a 9 to 1 payoff every time both dice were 2's, but only got even money if they weren't. You wanted to film the whole thing and put it on youtube. What happened? Where's the youtube video?

We never met up. The couple of times I tried were holiday weekends and it was not convenient and then it just slipped away.

The side bet really was a "side bet" and had nothing to do with the actual question. Frankly, I was hoping to win the side bet at a real game of craps because I tend to throw outside numbers with my cross-sixes set which would actually give me an edge. I tend to throw a lot of hard-4s.

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Originally Posted by Alan Mendelson

I tend to throw a lot of hard-4s.

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But wasn't a cup supposed to be used? BTW, with variance being what it is you had a pretty good chance to win with only 10 or 20 rolls.

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Originally Posted by mickeycrimm

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Originally Posted by Alan Mendelson

I tend to throw a lot of hard-4s.

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But wasn't a cup supposed to be used? BTW, with variance being what it is you had a pretty good chance to win with only 10 or 20 rolls.

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I don't remember and in the end I was going to buy lunch for him anyway. It was a lunch bet but we would be keeping score with the points. It's ancient history, or do you want to make a big deal out of it because some of you are starting to realize that the answer is really 1/6 ??

Ya know, since these guys "think" it's 1 in 6, they must surely "think" that the probability of the "other" die being either 1, 3, 4, 5, or 6 is 1 in 3 since there are twice as many of those numbers left. :-)

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Originally Posted by jbjb

Ya know, since these guys "think" it's 1 in 6, they must surely "think" that the probability of the "other" die being either 1, 3, 4, 5, or 6 is 1 in 3 since there are twice as many of those numbers left. :-)

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Nothing to say about what Regnis said about your cards/math?

You DON'T KNOW what ace is there so again, you CANNOT eliminate the other cards. You're attorney is overruled!!

But sure, IF I said it "was the ace of spades", then, and only then, YES it would be 1 in 6.

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Originally Posted by jbjb

You DON'T KNOW what ace is there so again, you CANNOT eliminate the other cards. You're attorney is overruled!!

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OMG. Read your own question. Try to understand your own error.

There is no error. You're ASSUMING the Ace of spades is there. You'll be wrong half the time.

When I look at the two cards drawn and at least one is an ace, the other card will either be the last ace or ANY OF THE OTHER TEN CARDS. Therefore, AGAIN for the millionth time, the answer is 1 in 11!

Okay jbjb, go back and reread your original post again. You said there were two stacks of cards and you chose one card from the stack of spades. When you chose one card from the stack of spades you were left with six cards in the stack of hearts. That's 1/6 to make a pair. It doesn't matter if the spade was the ace or not. Any spade gives you 1/6 hearts to make a pair.

Here are the conditions in your question:

"Using playing cards. I take the ace through 6 of spades and ace through 6 of hearts. In their own piles. Mix them up separately and pull one from each pile and look at both SIMULTANEOUSLY."

If you pick from just one pile there are only six cards. 1/6

Adios.

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Originally Posted by Alan Mendelson

Okay jbjb, go back and reread your original post again. You said there were two stacks of cards and you chose one card from the stack of spades. When you chose one card from the stack of spades you were left with six cards in the stack of hearts. That's 1/6 to make a pair. It doesn't matter if the spade was the ace or not. Any spade gives you 1/6 hearts to make a pair.

Here are the conditions in your question:

"Using playing cards. I take the ace through 6 of spades and ace through 6 of hearts. In their own piles. Mix them up separately and pull one from each pile and look at both SIMULTANEOUSLY."

If you pick from just one pile there are only six cards. 1/6 Adios.

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Alan, read the quote carefully. He said "pull one from each pile." How did you get from that, that he pulled just one card and it was from the spades pile?"

jbjb's example is actually pretty good. It points to people thinking about dice having six sides when they should be thinking about dice having six numbers.

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Originally Posted by regnis

Eddie took a shot at my question about the peeker, albeit a failed shot. I'm waiting for someone to explain how the same dice can have different odds.

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regnis, do we agree that the peeker has MORE information about the 2 dice than the non-peeker?

When asked, "What is the probability that both dice are showing a 2?" the peeker knows which dice (i.e. the red dice or the green dice) is a 2 and can eliminate 5 possibilities that the non-peeker cannot.

1. If the red dice is a 2, there is a 1/6 chance the green dice is a 2. Agreed?
2. If the green dice is a 2 there is a 1/6 chance the red dice is a 2. Agreed?

The peeker only has to contend with statement 1 or statement 2. NOT BOTH. So, if the question is posed to him, 1 die is revealed and the other "spinning" he could answer 1/6.

The non-peeker has to consider both statements doesn't he?

You have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."
What is the probability that both dice are showing a 2?



I cannot see anything in the original question that would allow the non-peeker to eliminate any of the 11 possibilities of 2 dice where at least 1 of the dice is a 2.

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Originally Posted by a2a3dseddie

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Originally Posted by regnis

Eddie took a shot at my question about the peeker, albeit a failed shot. I'm waiting for someone to explain how the same dice can have different odds.

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regnis, do we agree that the peeker has MORE information about the 2 dice than the non-peeker?

When asked, "What is the probability that both dice are showing a 2?" the peeker knows which dice (i.e. the red dice or the green dice) is a 2 and can eliminate 5 possibilities that the non-peeker cannot.

1. If the red dice is a 2, there is a 1/6 chance the green dice is a 2. Agreed?
2. If the green dice is a 2 there is a 1/6 chance the red dice is a 2. Agreed?

The peeker only has to contend with statement 1 or statement 2. NOT BOTH. So, if the question is posed to him, 1 die is revealed and the other "spinning" he could answer 1/6.

The non-peeker has to consider both statements doesn't he?

You have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."
What is the probability that both dice are showing a 2?



I cannot see anything in the original question that would allow the non-peeker to eliminate any of the 11 possibilities of 2 dice where at least 1 of the dice is a 2.

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Eddie-I appreciate that you and I can disagree on this without animosity, name calling etc. But here again is my take.
The dice don't know or care which one is a 2. And the odds of an event don't change by virtue of being the peeker or the other guy. Let's not concern ourselves with red or green or left or right. As Alan has stated for 2 years now, if one die is stated to have a 2 showing, the other possible numbers on that die are out. And just because the peeker knows which die has the 2 and the other guy doesn't, can't change the odds of the other die being a 2. The odds are what they are.

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Originally Posted by RS__

It comes down to this -- what happens if neither die is a 2?

Or in other words, does he always truthfully say, "At least one die is an X."? In this case, assuming he picks a random die to express the value (either the one closer to him, first one he saw, etc.), then the answer would be 1/6.

But if he can only say this when one of the dice is a 2...and doesn't say anything if neither die is a 2, then it's 1/11.


Can we agree on that, Alan?

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No comments??

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Originally Posted by regnis

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Originally Posted by a2a3dseddie

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Originally Posted by regnis

Eddie took a shot at my question about the peeker, albeit a failed shot. I'm waiting for someone to explain how the same dice can have different odds.

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regnis, do we agree that the peeker has MORE information about the 2 dice than the non-peeker?

When asked, "What is the probability that both dice are showing a 2?" the peeker knows which dice (i.e. the red dice or the green dice) is a 2 and can eliminate 5 possibilities that the non-peeker cannot.

1. If the red dice is a 2, there is a 1/6 chance the green dice is a 2. Agreed?
2. If the green dice is a 2 there is a 1/6 chance the red dice is a 2. Agreed?

The peeker only has to contend with statement 1 or statement 2. NOT BOTH. So, if the question is posed to him, 1 die is revealed and the other "spinning" he could answer 1/6.

The non-peeker has to consider both statements doesn't he?

You have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."
What is the probability that both dice are showing a 2?



I cannot see anything in the original question that would allow the non-peeker to eliminate any of the 11 possibilities of 2 dice where at least 1 of the dice is a 2.

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Eddie-I appreciate that you and I can disagree on this without animosity, name calling etc. But here again is my take.
The dice don't know or care which one is a 2. And the odds of an event don't change by virtue of being the peeker or the other guy. Let's not concern ourselves with red or green or left or right. As Alan has stated for 2 years now, if one die is stated to have a 2 showing, the other possible numbers on that die are out. And just because the peeker knows which die has the 2 and the other guy doesn't, can't change the odds of the other die being a 2. The odds are what they are.

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My thoughts exactly.

I agree 100% with everything you've written here. But the peeker is at a different point in the resolution of the dice throw. That's the difference.

I posted a craps analogy earlier that will help demonstrate this.

You have to place a bet on hard 4.

Would you rather place the bet before any of the 2 dice are thrown or place the bet after we first set one of the dice to a 2 and just roll the other die? Are the odds the same? Of course they aren't and it's not just because once you set one of the dice to a 2 that an easy 4 will not be possible.

On the Wizard of Vegas site, here's what someone posted as a bet:

1) shake dice in the cup and then turn upside down on the table
2) reveal both dice
3) pay 8:1 if dice show 2-2
4) get paid 1:1 if dice show 2-1, 2-3, 2-4, 2-5, or 2-6
5) no action if dice show any other combination

Are these terms agreeable to you regnis or Alan?

I just booked my flight to Las Vegas and will be there from June 21 to June 28. I would be willing to meet either of you or anyone else believing it's 1/6.

Alan suggested betting stakes be lunch. It doesn't matter to me.

I suggest we each start with $25 in quarters. We play until one person has no quarters left and we use the $50 to get lunch somewhere.

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Originally Posted by regnis

Eddie-I appreciate that you and I can disagree on this without animosity, name calling etc. But here again is my take.
The dice don't know or care which one is a 2. And the odds of an event don't change by virtue of being the peeker or the other guy. Let's not concern ourselves with red or green or left or right. As Alan has stated for 2 years now, if one die is stated to have a 2 showing, the other possible numbers on that die are out. And just because the peeker knows which die has the 2 and the other guy doesn't, can't change the odds of the other die being a 2. The odds are what they are.

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I hope this finally makes sense to those of you who maintain the answer is 1/11 to this particular question.

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Originally Posted by Alan Mendelson

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Originally Posted by regnis

Eddie-I appreciate that you and I can disagree on this without animosity, name calling etc. But here again is my take.
The dice don't know or care which one is a 2. And the odds of an event don't change by virtue of being the peeker or the other guy. Let's not concern ourselves with red or green or left or right. As Alan has stated for 2 years now, if one die is stated to have a 2 showing, the other possible numbers on that die are out. And just because the peeker knows which die has the 2 and the other guy doesn't, can't change the odds of the other die being a 2. The odds are what they are.

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I hope this finally makes sense to those of you who maintain the answer is 1/11 to this particular question.

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No it DOESN'T eliminate the other numbers, PERIOD!

If your so confident, take the 8:1 bet for $100 per roll. Win $800 when 2-2 shows and lose $100 when 2-1, 2-3, 2-4, 2-5 or 2-6 shows. Guarantee you'll go broke.

jbjb stop embarrassing yourself.

Take a die and show it with a 2 on top. The other five faces on that die are now excluded.

Take the second die and examine it. It has six faces. One of those six faces will match the two.

Now, let's take your stack of suited cards.

Select one of the six spades. By choosing one spade you eliminate the other spades from consideration. In fact, you said that in your own writing.

Now, you have six hearts. One of those hearts will make a pair with the spade you selected.

If you don't understand this, you've inhaled too much smoke in casinos.

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Originally Posted by Alan Mendelson

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Originally Posted by arcimede$

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Originally Posted by Alan Mendelson

Absolutely false statements.

Yes, a "peeker" can see either one die or both dice and say truthfully and correctly "at least one die is a 2." Do you know how to speak the English language Arc?

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Silly nonsense. If the die he sees when he sees only one die is not a 2 and the die he does not see is a 2, then he is not answering the question "correctly". Face-palm.

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Arc, now you're going to start playing word games? Your response was to regnis who was talking about one die being a 2. I was challenging your use of the English language and what "at least one" means. Of course if the peeker sees two 6s he cannot truthfully say at least one die is a 2 which sets up the original question.

Take your face palm and slap yourself.

Regnis: too bad you don't have this gang in court. They would be pleading for a deal. They don't know what hit them.

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As usual when Alan is made to look foolish he tries to change the subject. You supported Regnis statement so please explain to everyone how the peeker could know "at least one die is a 2" when he doesn't see both of the die.

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Originally Posted by jbjb

Using playing cards. I take the ace through 6 of spades and ace through 6 of hearts. In their own piles. Mix them up separately and pull one from each pile and look at both SIMULTANEOUSLY. I say "at least one is an ace, what is the probability the other is an ace?" Any of the 11 cards could be the "other one." Same 1 in 11 as the dice problem and people problem I mentioned earlier

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Originally Posted by Alan Mendelson

Now, let's take your stack of suited cards.

Select one of the six spades. By choosing one spade you eliminate the other spades from consideration. In fact, you said that in your own writing.

Now, you have six hearts. One of those hearts will make a pair with the spade you selected.

If you don't understand this, you've inhaled too much smoke in casinos.

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I don't see where jbjb said this.

Like the dice problem, you don't know which card is the ace. Since you don't know, all of the 11 combinations where one of the cards is an ace are in play.

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Originally Posted by arcimede$

As usual when Alan is made to look foolish he tries to change the subject. You supported Regnis statement so please explain to everyone how the peeker could know "at least one die is a 2" when he doesn't see both of the die.

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Are you fucking insane? He only has to see ONE die to make the statement "at least one die is a two." You don't have to see both dice. And in the question it is stated that the "peeker" truthfully says one die is a 2.

So what the fuck is your problem?

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Originally Posted by a2a3dseddie

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Originally Posted by jbjb

Using playing cards. I take the ace through 6 of spades and ace through 6 of hearts. In their own piles. Mix them up separately and pull one from each pile and look at both SIMULTANEOUSLY. I say "at least one is an ace, what is the probability the other is an ace?" Any of the 11 cards could be the "other one." Same 1 in 11 as the dice problem and people problem I mentioned earlier

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Originally Posted by Alan Mendelson

Now, let's take your stack of suited cards.

Select one of the six spades. By choosing one spade you eliminate the other spades from consideration. In fact, you said that in your own writing.

Now, you have six hearts. One of those hearts will make a pair with the spade you selected.

If you don't understand this, you've inhaled too much smoke in casinos.

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I don't see where jbjb said this.

Like the dice problem, you don't know which card is the ace. Since you don't know, all of the 11 combinations where one of the cards is an ace are in play.

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This is where jbjb said it. Again, we have a reading comprehension problem with the 1/11 gang:

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Originally Posted by jbjb

Using playing cards. I take the ace through 6 of spades and ace through 6 of hearts. In their own piles. Mix them up separately and pull one from each pile and look at both SIMULTANEOUSLY.

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Does jbjb now want to tell us that he didn't intend to have separate piles, spades and hearts?

It's all about the English language folks.

Alan, had you been there when Galileo announced that the earth circled the sun instead of the sun circling the earth you would have told him he was full of shit.

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Originally Posted by Alan Mendelson

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Originally Posted by arcimede$

As usual when Alan is made to look foolish he tries to change the subject. You supported Regnis statement so please explain to everyone how the peeker could know "at least one die is a 2" when he doesn't see both of the die.

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Are you fucking insane? He only has to see ONE die to make the statement "at least one die is a two." You don't have to see both dice. And in the question it is stated that the "peeker" truthfully says one die is a 2.

So what the fuck is your problem?

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Alan, perhaps this is where our interpretations of the problem differ from yours.

I guess you are looking at this from a 1 roll and 1 roll only point of view.

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Originally Posted by Alan Mendelson

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Originally Posted by a2a3dseddie

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Originally Posted by Alan Mendelson

Now, let's take your stack of suited cards.

Select one of the six spades. By choosing one spade you eliminate the other spades from consideration. In fact, you said that in your own writing.

Now, you have six hearts. One of those hearts will make a pair with the spade you selected.

If you don't understand this, you've inhaled too much smoke in casinos.

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I don't see where jbjb said this.

Like the dice problem, you don't know which card is the ace. Since you don't know, all of the 11 combinations where one of the cards is an ace are in play.

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This is where jbjb said it. Again, we have a reading comprehension problem with the 1/11 gang:

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Originally Posted by jbjb

Using playing cards. I take the ace through 6 of spades and ace through 6 of hearts. In their own piles. Mix them up separately and pull one from each pile and look at both SIMULTANEOUSLY.

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Does jbjb now want to tell us that he didn't intend to have separate piles, spades and hearts?

It's all about the English language folks.

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Alan, when jbjb says:

"at least one is an ace, what is the probability the other is an ace?"

Couldn't he be holding:

As 2h
As 3h
As 4h
As 5h
As 6h

Ah 2s
Ah 3s
Ah 4s
Ah 5s
Ah 6s

Or

Ah As
?

This is where, if I may remind, I pointed out that peeking at the dice as written is a sequential event, not simultaneous.

Further, as written, there is only one roll at a time. Again, sequential.

The writer presented the question as this being one roll. He did that explicitly to create confusion.

I'm not sure the writer intended to create confusion. I think the math guys got confused because the writer set it up as a very simple question. Our 1/11 gang can't see the simple answer. No... they had to go and take the complicated route.

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Originally Posted by Alan Mendelson

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Originally Posted by arcimede$

As usual when Alan is made to look foolish he tries to change the subject. You supported Regnis statement so please explain to everyone how the peeker could know "at least one die is a 2" when he doesn't see both of the die.

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Are you fucking insane? He only has to see ONE die to make the statement "at least one die is a two." You don't have to see both dice. And in the question it is stated that the "peeker" truthfully says one die is a 2.

So what the fuck is your problem?

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Why didn't you answer my question, Alan. Instead you more or less admitted you know you are wrong by swearing. If he sees only one die and it is a 6, he cannot KNOW whether the other die is a 2. Therefore, he can not "truthfully" say whether or not one die is a 2. He HAS to see both die to "truthfully" answer the question.

Face-palm.

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Originally Posted by Alan Mendelson

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Originally Posted by arcimede$

As usual when Alan is made to look foolish he tries to change the subject. You supported Regnis statement so please explain to everyone how the peeker could know "at least one die is a 2" when he doesn't see both of the die.

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Are you fucking insane? He only has to see ONE die to make the statement "at least one die is a two." You don't have to see both dice. And in the question it is stated that the "peeker" truthfully says one die is a 2.
So what the fuck is your problem?

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Well, the problem is if the peeker looks at the first die and it is a 1,3,4,5, or 6, which most of the time it will be one of those numbers, he has to look at the other die to see if it is a 2. Duh!

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Originally Posted by mickeycrimm

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Originally Posted by Alan Mendelson

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Originally Posted by arcimede$

As usual when Alan is made to look foolish he tries to change the subject. You supported Regnis statement so please explain to everyone how the peeker could know "at least one die is a 2" when he doesn't see both of the die.

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Are you fucking insane? He only has to see ONE die to make the statement "at least one die is a two." You don't have to see both dice. And in the question it is stated that the "peeker" truthfully says one die is a 2.
So what the fuck is your problem?

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Once again, of course one die has to be a 2 in order to make the statement "at least one die is a 2."

Arc is again trying to twist words. We went through this before.

This is Arc's routine. Ive seen it for years through his battles with Singer.

Well, the problem is if the peeker looks at the first die and it is a 1,3,4,5, or 6, which most of the time it will be one of those numbers, he has to look at the other die to see if it is a 2. Duh!

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Once again, of course one die has to be a 2 in order to make the statement "at least one die is a 2."

Arc is again trying to twist words. We went through this before.

This is Arc's routine. Ive seen it for years through his battles with Singer.

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Originally Posted by arcimede$

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Originally Posted by Alan Mendelson

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Originally Posted by arcimede$

As usual when Alan is made to look foolish he tries to change the subject. You supported Regnis statement so please explain to everyone how the peeker could know "at least one die is a 2" when he doesn't see both of the die.

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Are you fucking insane? He only has to see ONE die to make the statement "at least one die is a two." You don't have to see both dice. And in the question it is stated that the "peeker" truthfully says one die is a 2.

So what the fuck is your problem?

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Why didn't you answer my question, Alan. Instead you more or less admitted you know you are wrong by swearing. If he sees only one die and it is a 6, he cannot KNOW whether the other die is a 2. Therefore, he can not "truthfully" say whether or not one die is a 2. He HAS to see both die to "truthfully" answer the question.

Face-palm.

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We fucking went through this before. Of course the peeker has to see a 2 in order to make the statement "at least one die is a two."

You made this ridiculous comment the first time responding to regnis who discussed seeing a 2.

Then you tried to twist your words.

Quit it. You pulled this same shit in your battles with Singer over the years and I'm not going to fall for it.

Fuck off.

And you cheated on your taxes.

I asked this before and no one gave a response:

Had neither of the dice been a 2, what would the peeker have done? Would he have said "At least one of the dice is a 1" (or 3 or 4 or 5 or 6)? Or would he have said nothing and rolled again and again, until at least one of the dice is a 2?

This is probably the most integral part to answering the question.

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Originally Posted by RS__

Had neither of the dice been a 2, what would the peeker have done? Would he have said "At least one of the dice is a 1" (or 3 or 4 or 5 or 6)? Or would he have said nothing and rolled again and again, until at least one of the dice is a 2?

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Don't we have enough problems with the original question? Now you want to ask all of these other questions?

The original question said the "peeker" truthfully said at least one of the two dice is a 2. That has led to two years of debate, side bets, and alternative questions that (in one case) the author didn't even understand what he himself wrote.

And you want to know what would the peeker do if there wasn't a 2?

Better idea: ask Arc why he didn't declare his non-W2G wins on his tax return? Ask jbjb if it is okay to draw more than one spade from his pile of six spades? Ask everyone else what 2-1 equals?

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Originally Posted by Alan Mendelson

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Originally Posted by arcimede$

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Originally Posted by Alan Mendelson

Are you fucking insane? He only has to see ONE die to make the statement "at least one die is a two." You don't have to see both dice. And in the question it is stated that the "peeker" truthfully says one die is a 2.

So what the fuck is your problem?

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Why didn't you answer my question, Alan. Instead you more or less admitted you know you are wrong by swearing. If he sees only one die and it is a 6, he cannot KNOW whether the other die is a 2. Therefore, he can not "truthfully" say whether or not one die is a 2. He HAS to see both die to "truthfully" answer the question.

Face-palm.

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We fucking went through this before. Of course the peeker has to see a 2 in order to make the statement "at least one die is a two."

You made this ridiculous comment the first time responding to regnis who discussed seeing a 2.

Then you tried to twist your words.

Quit it. You pulled this same shit in your battles with Singer over the years and I'm not going to fall for it.

Fuck off.

And you cheated on your taxes.

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No one is twisting words. You are simply denying the obvious fact that the peeker has to see both die making Regnis statement nonsense and showing you will say anything to protect your ego.

Now you are bringing up my taxes which you obviously don't understand very well either (is anyone surprised?)? So, I guess we can add a complete lack of integrity to your growing list of faults.

So, once again .... How can the peeker answer the question "truthfully" if he only sees one die? If you contnue to avoid the question that is just as good as admitting he can't answer truthfully and therefore looking at one die is NOT an option.

Now, quit your incessant whining and just go away. Isn't that your usual mode of operation?

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Originally Posted by Alan Mendelson

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Originally Posted by RS__

Had neither of the dice been a 2, what would the peeker have done? Would he have said "At least one of the dice is a 1" (or 3 or 4 or 5 or 6)? Or would he have said nothing and rolled again and again, until at least one of the dice is a 2?

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Don't we have enough problems with the original question? Now you want to ask all of these other questions?

The original question said the "peeker" truthfully said at least one of the two dice is a 2. That has led to two years of debate, side bets, and alternative questions that (in one case) the author didn't even understand what he himself wrote.

And you want to know what would the peeker do if there wasn't a 2?

Better idea: ask Arc why he didn't declare his non-W2G wins on his tax return? Ask jbjb if it is okay to draw more than one spade from his pile of six spades? Ask everyone else what 2-1 equals?

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Because if you interpreted it the "he re-rolls until he gets at least one die as 2" then you'd have one answer. And if you interpreted it as "he would just say 'at least one die is a ____' [some other value]", then you'd have a different answer.

So I'd like to know what your interpretation is. If you've interpreted it differently, then under that interpretation you'd be correct.

Quote:

---

Originally Posted by arcimede$

No one is twisting words. You are simply denying the obvious fact that the peeker has to see both die making Regnis statement nonsense and showing you will say anything to protect your ego.

---

I see we have a real problem here with the English language. So I will answer this... slowly.

The peeker only needs to see one die with a 2 to say, truthfully, "at least one die is a two." It is not necessary to see both dice. The peeker only needs to see one die.

If the peeker sees a 2 and a 6, he can truthfully say "at least one die is a two."
If the peeker sees a 2 and never stops to look at the second die, he can truthfully say "at least one die is a two."

The term "at least" means, according to Dictionary.com: "at the lowest estimate or figure" and they give this example: "The repairs will cost at least $100."

All the peeker has to do is see one die.

Now, why are you even raising this? We know from the original question that truthfully one of the dice is showing a 2. Challenging the presence of a 2 serves what purpose?

There are two dice. There is a two on one of those two dice. That die with a two cannot change. Deal with that reality.

I can't wait to see what regnis says.

Quote:

---

Originally Posted by RS__

Quote:

---

Originally Posted by Alan Mendelson

Quote:

---

Originally Posted by RS__

Had neither of the dice been a 2, what would the peeker have done? Would he have said "At least one of the dice is a 1" (or 3 or 4 or 5 or 6)? Or would he have said nothing and rolled again and again, until at least one of the dice is a 2?

---

Don't we have enough problems with the original question? Now you want to ask all of these other questions?

The original question said the "peeker" truthfully said at least one of the two dice is a 2. That has led to two years of debate, side bets, and alternative questions that (in one case) the author didn't even understand what he himself wrote.

And you want to know what would the peeker do if there wasn't a 2?

Better idea: ask Arc why he didn't declare his non-W2G wins on his tax return? Ask jbjb if it is okay to draw more than one spade from his pile of six spades? Ask everyone else what 2-1 equals?

---

Because if you interpreted it the "he re-rolls until he gets at least one die as 2" then you'd have one answer. And if you interpreted it as "he would just say 'at least one die is a ____' [some other value]", then you'd have a different answer.

So I'd like to know what your interpretation is. If you've interpreted it differently, then under that interpretation you'd be correct.

---

My interpretation is simply this:

A peeker looks and sees at least one die is a 2. That die will not change. It will always be a 2.

Now you tell me: if one die is always a two, and there are two dice, what are the odds that the other die is also showing a 2? Is it still 1/11 for a six-sided die?

LOL

Quote:

---

Originally Posted by jbjb

If your so confident, take the 8:1 bet for $100 per roll. Win $800 when 2-2 shows and lose $100 when 2-1, 2-3, 2-4, 2-5 or 2-6 shows. Guarantee you'll go broke.

---

I'll take this bet for one roll. I think money well spent if I lose. How many rolls can I bet on?

Quote:

---

Originally Posted by coach belly

Quote:

---

Originally Posted by jbjb

If your so confident, take the 8:1 bet for $100 per roll. Win $800 when 2-2 shows and lose $100 when 2-1, 2-3, 2-4, 2-5 or 2-6 shows. Guarantee you'll go broke.

---

I'll take this bet for one roll. I think money well spent if I lose. How many rolls can I bet on?

---

Coach, you would go broke with this.

When rolling two dice simultaneously, 2-2 shows up 1/36 times.
There are 11 combinations showing at least one 2, and only one combination that shows 2-2. So you are facing 10 losing decisions for each winning decision.

Instead, ask jbjb if he would accept this bet:

Set a die on the number 2 and roll a second die. If the second die also shows a 2 you will get paid $800. If the second die shows any other number you will pay him $100.

Update: I just rolled a single die ten times: I got two 2s for a win of $1600. I got other numbers on the 8 other rolls for a loss of $800. Net win: $800.

Quote:

---

Originally Posted by Alan Mendelson

Coach, you would go broke with this.

Instead, ask jbjb if he would accept this bet:

Set a die on the number 2 and roll a second die. If the second die also shows a 2 you will get paid $800. If the second die shows any other number you will pay him $100.

Update: I just rolled a single die ten times: I got two 2s for a win of $1600. I got other numbers on the 8 other rolls for a loss of $800. Net win: $800.

---

I won't go broke. I'm gonna hit and run...quit while ahead.

Are you trying to scare jbjb away from my bet with your bogus results?

Coach I will be there to record the betting and rolls of the dice.

Quote:

---

Originally Posted by Alan Mendelson

Coach I will be there to record the betting and rolls of the dice.

---

I was counting on that.

Quote:

---

Originally Posted by Alan Mendelson

My interpretation is simply this:

A peeker looks and sees at least one die is a 2. That die will not change. It will always be a 2.

Now you tell me: if one die is always a two, and there are two dice, what are the odds that the other die is also showing a 2? Is it still 1/11 for a six-sided die?

LOL

---

I'm trying to give you the benefit of the doubt. I don't think you realize that, though.

I'm not sure what's so funny. If you mean you set one die to a '2', and roll the other one, then it's pretty obvious there's a 1/6 chance it lands on '2'. Why would you even ask that question?

Quote:

---

Originally Posted by Alan Mendelson

We fucking went through this before. Of course the peeker has to see a 2 in order to make the statement "at least one die is a two."

You made this ridiculous comment the first time responding to regnis who discussed seeing a 2.

Then you tried to twist your words.

Quit it. You pulled this same shit in your battles with Singer over the years and I'm not going to fall for it.

Fuck off.

And you cheated on your taxes.

---

Quote:

---

Originally Posted by Alan Mendelson

Are you fucking insane? He only has to see ONE die to make the statement "at least one die is a two." You don't have to see both dice. And in the question it is stated that the "peeker" truthfully says one die is a 2.

So what the fuck is your problem?

---

Quote:

---

Originally Posted by Alan Mendelson

You guys are having a real problem with reality, aren't you?

There are only two dice. You can't use the same die twice to figure the answer to the question.

If I tell you that at least one of the two dice is showing a 2, then the answer must be with the remaining die.

It doesn't matter which of the two dice we say is showing a two. Try to get this to sink in: It doesn't matter which of the two dice we say is showing a two.

Why (please tell me) do you think it matters with a two dice problem?

If we have two dice and one of them is known to show a 2, then only one die remains. (Hint: 2-1=1) And on that remaining die (it doesn't matter which one it is) there are only 6 faces. Of those six faces, only one face is a 2.

What the fuck is the difference which die it is in a two fucking dice problem?

---

Quote:

---

Originally Posted by Alan Mendelson

This is not a question about rolling dice and seeing results.

This is a simple question: You have two dice one of which is a 2. What are the odds that the other die is also showing a two.

You can moan and groan all you want about conditional bullshit and look at graphs all you want showing the number of dice combinations displaying a 2...

But the simple fucking answer is 1/6.

Now fuck off.

---

Quote:

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Originally Posted by regnis

Name calling and swearing etc. a clear sign of ignorance. Just sayin.

---

+1

Quote:

---

Originally Posted by RS__

Quote:

---

Originally Posted by Alan Mendelson

My interpretation is simply this:

A peeker looks and sees at least one die is a 2. That die will not change. It will always be a 2.

Now you tell me: if one die is always a two, and there are two dice, what are the odds that the other die is also showing a 2? Is it still 1/11 for a six-sided die?

LOL

---

I'm trying to give you the benefit of the doubt. I don't think you realize that, though.

I'm not sure what's so funny. If you mean you set one die to a '2', and roll the other one, then it's pretty obvious there's a 1/6 chance it lands on '2'. Why would you even ask that question?

---

So there is no misunderstanding: when two dice are rolled and you are told that one die has landed on the 2, there is a 1/6 chance that the other die is also a 2.

When you roll two dice, there are 11 combinations on those two dice that include a 2, and one of those 11 combinations is 2-2. That's 1/11.

Do you realize what the difference is? The difference is when you consider TWO dice or you consider only ONE die. When someone tells you that one die has landed on a 2, then there are only six possibilities to consider for the second die.

That's what Ive been telling you for two years. It's what regnis has been telling you. And redietz even acknowledges the difference in the two situations because as he put it, telling someone that a 2 has been rolled is sequential and not simultaneous.

Frankly, as I've said all along, it's a matter of reading comprehension and understanding the words of the problem. Your math isn't wrong, but you are applying the math incorrectly.

Quote:

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Originally Posted by Alan Mendelson

Coach I will be there to record the betting and rolls of the dice.

---

Alan's true nature is starting to manifest itself. Alan, you were chosen to receive arci and Singer's tax returns because of your "impeccable integrity." You've now shot that all to hell with your attack on arci over his tax returns. You gave everyone your word that you would not disclose any information other than the gambling win/loss. If you seen something to cause you to think he may have cheated his taxes, this is information you never should have divulged here. We now know that you ABSOLUTELY CAN'T BE TRUSTED. Furthermore, you never attacked Singer in such a way over his taxes when he openly admitted never paying any.

If you will screw arci like this then you will screw any of us on anything. I wonder how many people you screwed as a journalist. Integrity, my ass. You have none.

PS: You never received your buttboy Singer's tax returns, did you?

Quote:

---

Originally Posted by mickeycrimm

This is a an excellent question to debate here, except Rob, per the usual, jumping in and flaming the thread with his childish outbursts.

---

I made this comment early on in this thread. Now Alan is flaming the thread with childish outbursts. Singer and Mendelson are two peas in a pod.

Quote:

---

Originally Posted by RS__

I asked this before and no one gave a response:

Had neither of the dice been a 2, what would the peeker have done? Would he have said "At least one of the dice is a 1" (or 3 or 4 or 5 or 6)? Or would he have said nothing and rolled again and again, until at least one of the dice is a 2?

This is probably the most integral part to answering the question.

---

No kidding.

Wow...this is a strange reaction. Why is the wormm so nervous about Alan's involvement?

BTW...when I read "record" I took it to mean capture on video...you can do that, right Alan?

Quote:

---

Originally Posted by mickeycrimm

Quote:

---

Originally Posted by Alan Mendelson

Coach I will be there to record the betting and rolls of the dice.

---

Alan's true nature is starting to manifest itself. Alan, you were chosen to receive arci and Singer's tax returns because of your "impeccable integrity." You've now shot that all to hell with your attack on arci over his tax returns. You gave everyone your word that you would not disclose any information other than the gambling win/loss. If you seen something to cause you to think he may have cheated his taxes, this is information you never should have divulged here. We now know that you ABSOLUTELY CAN'T BE TRUSTED. Furthermore, you never attacked Singer in such a way over his taxes when he openly admitted never paying any.

If you will screw arci like this then you will screw any of us on anything. I wonder how many people you screwed as a journalist. Integrity, my ass. You have none.

---

Quote:

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Originally Posted by arcimede$

Quote:

---

Originally Posted by Alan Mendelson

Quote:

---

Originally Posted by arcimede$

Why didn't you answer my question, Alan. Instead you more or less admitted you know you are wrong by swearing. If he sees only one die and it is a 6, he cannot KNOW whether the other die is a 2. Therefore, he can not "truthfully" say whether or not one die is a 2. He HAS to see both die to "truthfully" answer the question.

Face-palm.

---

We fucking went through this before. Of course the peeker has to see a 2 in order to make the statement "at least one die is a two."

You made this ridiculous comment the first time responding to regnis who discussed seeing a 2.

Then you tried to twist your words.

Quit it. You pulled this same shit in your battles with Singer over the years and I'm not going to fall for it.

Fuck off.

And you cheated on your taxes.

---

No one is twisting words. You are simply denying the obvious fact that the peeker has to see both die making Regnis statement nonsense and showing you will say anything to protect your ego.

Now you are bringing up my taxes which you obviously don't understand very well either (is anyone surprised?)? So, I guess we can add a complete lack of integrity to your growing list of faults.

So, once again .... How can the peeker answer the question "truthfully" if he only sees one die? If you contnue to avoid the question that is just as good as admitting he can't answer truthfully and therefore looking at one die is NOT an option.

Now, quit your incessant whining and just go away. Isn't that your usual mode of operation?

---

Why does the peeker need to see both dice? That is not explicitly stated.

You have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."
What is the probability that both dice are showing a 2?

Does the original question and scenario sound more like

Quote:

---

Originally Posted by Alan Mendelson

When rolling two dice simultaneously, 2-2 shows up 1/36 times.
There are 11 combinations showing at least one 2, and only one combination that shows 2-2. So you are facing 10 losing decisions for each winning decision.

---

or

Quote:

---

Originally Posted by Alan Mendelson

Set a die on the number 2 and roll a second die. If the second die also shows a 2 you will get paid $800. If the second die shows any other number you will pay him $100.

---

?

The original question asks:

What is the probability that both dice are showing a 2?

There is a subtle difference between the original question and "If one dice is a 2 what is probability the other dice is a 2?"

The original question is asking for the probability of specific combination of a roll of 2 dice where we know what the value of 1 of the dice. Since we don't know which dice has landed on the 2, we need to consider all combinations that satisfy the peeker's only information given to us:

"At least one of the dice is a 2."

"There are 11 combinations showing at least one 2, and only one combination that shows 2-2"

So, while:

"when two dice are rolled and you are told that one die has landed on the 2, there is a 1/6 chance that the other die is also a 2."

Is true if we know which dice is a 2, that's not what was asked.

There is always a 1/6 chance any number could appear on a die.

But the original question is asking for a specific combination of 2 2.

These 1 in 6 fools need to go back to high school and it shows a college degree is worthless.

"At least one die" is NOT the same as "a certain die."

You guys keep making this ridiculous statement:

Quote:

---

Originally Posted by a2a3dseddie

Since we don't know which dice has landed on the 2, we need to consider all combinations

---

If you have two dice, and you know one die is a 2, how many possible combinations could there be? There are only two dice. One is a 2.

It doesn't matter which of the two dice is showing the 2.

Maybe on a chart showing the combinations of dice it matters, but when you are dealing with two physical dice it doesn't matter which is showing a two.

Take your minds off your charts and graphs and put two dice in front of you. Set one of the dice as a 2. What's the answer to the problem? It's one out of six.

Now, set the other die to a 2. What's the answer now? It's still 1/6.

Set both dice to a 2. What's the answer now? It's still 1/6.

We are told one of the two dice is a 2. That means ignore your charts and graphs showing all the 11 combinations with a 2. Ignore them. Put them aside. They don't count anymore. If you don't believe me ask regnis. Ask your mailman. Ask your six year old kid.

Now, regarding Arc. He made claims here about all his winnings and when he submitted his tax returns to prove his winnings the only thing he declared were a couple of $4,000 royals. He admitted here on the forum that he did not include his non W2G wins. He admitted it, mickey.

Arc, there is no requirement to see both dice. Stop your argument.

Quote:

---

Originally Posted by jbjb

These 1 in 6 fools need to go back to high school and it shows a college degree is worthless.

"At least one die" is NOT the same as "a certain die."

---

Okay so if I am a fool, what's the different in a two dice problem? Educate us oh great wise one. And also explain what you meant by your question with your pile of six spades and your pile of six hearts.

Simply enough, when two dice are rolled SIMULTANEOUSLY and AT LEAST ONE IS A 2, the other will be a 2, wait for it..... 1 in 11. When you SET A DIE TO 2 and roll a SINGLE DIE, "that" die will be a 2, wait for it.....1 in 6 times.

Quote:

---

Originally Posted by jbjb

Simply enough, when two dice are rolled SIMULTANEOUSLY and AT LEAST ONE IS A 2, the other will be a 2, wait for it..... 1 in 11. When you SET A DIE TO 2 and roll a SINGLE DIE, "that" die will be a 2, wait for it.....1 in 6 times.

---

Okay. And when you are told that ONE of the dice is a 2, what are the odds that the second die is also a 2? Are you telling me that after you know that one die is a two, that the chances for the second die to also be a two are 1/11??? How can that be when the second die only has six sides?

This thread may go down in history as the dumbest of all time. Alan must have been in a mood to start an argument. Maybe it it could be studied by a group of sociologists/psychiatrists, it would keep them busy for years.

Quote:

---

Originally Posted by quahaug

This thread may go down in history as the dumbest of all time. Alan must have been in a mood to start an argument. Maybe it it could be studied by a group of sociologists/psychiatrists, it would keep them busy for years.

---

It's a conspiracy launched by me with the cooperation of regnis and a few others.

Maybe it's to show just how much logic it takes to understand two physical dice.

Quote:

---

Originally Posted by Alan Mendelson

And when you are told that ONE of the dice is a 2, what are the odds that the second die is also a 2?

---

That is not the original question. This is.

What is the probability that both dice are showing a 2?


You said it yourself:

Quote:

---

Originally Posted by Alan Mendelson

When you roll two dice, there are 11 combinations on those two dice that include a 2, and one of those 11 combinations is 2-2. That's 1/11.

---

Quote:

---

Originally Posted by Alan Mendelson

It's a conspiracy launched by me with the cooperation of regnis and a few others.

Maybe it's to show just how much logic it takes to understand two physical dice.

---

I'm surprised that nobody has run a computer simulation for this.

I have read here about million+ hand VP simulations, using all different strategies.

This dice problem seems easier to simulate, but nobody in the 11/1 crowd has offered one.

Pages and pages of debate on at least 2 different forums, but what does the computer simulation show?

You are told TRUTHFULLY that at least one of the two dice in the cup is showing a 2. So again, what are the odds that the second die is also showing a 2? It is still 1/11 for a six sided die? Or is it 1/6?

Respond to the question and its conditions.

The condition is one die is known to be a two. It is not the same as throwing two dice simultaneously and knowing that there are 11 combinations showing at least one two. Repeat: it is not the same as throwing two dice simultaneously and knowing that there are 11 combinations showing at least one two.

This is a reading comprehension issue for those of you who think the answer is 1/11.

Quote:

---

Originally Posted by coach belly

Quote:

---

Originally Posted by Alan Mendelson

It's a conspiracy launched by me with the cooperation of regnis and a few others.

Maybe it's to show just how much logic it takes to understand two physical dice.

---

I'm surprised that nobody has run a computer simulation for this.

I have read here about million+ hand VP simulations, using all different strategies.

This dice problem seems easier to simulate, but nobody in the 11/1 crowd has offered one.

Pages and pages of debate on at least 2 different forums, but what does the computer simulation show?

---

Sorry, it is not a computer simulation problem. It's a simple arithmetic problem. You only need to count to six on one die.

Quote:

---

Originally Posted by Alan Mendelson

Quote:

---

Originally Posted by arcimede$

No one is twisting words. You are simply denying the obvious fact that the peeker has to see both die making Regnis statement nonsense and showing you will say anything to protect your ego.

---

I see we have a real problem here with the English language. So I will answer this... slowly.

The peeker only needs to see one die with a 2 to say, truthfully, "at least one die is a two." It is not necessary to see both dice. The peeker only needs to see one die.

If the peeker sees a 2 and a 6, he can truthfully say "at least one die is a two."
If the peeker sees a 2 and never stops to look at the second die, he can truthfully say "at least one die is a two."

The term "at least" means, according to Dictionary.com: "at the lowest estimate or figure" and they give this example: "The repairs will cost at least $100."

All the peeker has to do is see one die.

Now, why are you even raising this? We know from the original question that truthfully one of the dice is showing a 2. Challenging the presence of a 2 serves what purpose?

There are two dice. There is a two on one of those two dice. That die with a two cannot change. Deal with that reality.

I can't wait to see what regnis says.

---

Alan, your lack of anything approaching logic is quite hilarious. You even admit .... "If the peeker sees a 2 and a 6, he can truthfully say "at least one die is a two."" In other words, he cannot ALWAYS make the determination until he sees both die when one of them is not a 2. If he sees the 6 first then he has to look at the other die. The fact there are a few times he could make the determination by seeing only one die is NOT sufficient to ALWAYS make the determination. Hence, just looking at one die will not allow him to ALWAYS make the claim "at least one die is a 2".

This entirely negates Regnis claim.

Alan, you really are making a complete fool out of yourself.

Quote:

---

Originally Posted by Alan Mendelson

You are told TRUTHFULLY that at least one of the two dice in the cup is showing a 2. So again, what are the odds that the second die is also showing a 2? It is still 1/11 for a six sided die? Or is it 1/6?

Respond to the question and its conditions.

The condition is one die is known to be a two. It is not the same as throwing two dice simultaneously and knowing that there are 11 combinations showing at least one two. Repeat: it is not the same as throwing two dice simultaneously and knowing that there are 11 combinations showing at least one two.

This is a reading comprehension issue for those of you who think the answer is 1/11.

---

Once again here is the original problem:

You have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."
What is the probability that both dice are showing a 2?

That is the original question.

NOT

"what are the odds that the second die is also showing a 2?"

Quote:

---

Originally Posted by coach belly

Quote:

---

Originally Posted by Alan Mendelson

It's a conspiracy launched by me with the cooperation of regnis and a few others.

Maybe it's to show just how much logic it takes to understand two physical dice.

---

I'm surprised that nobody has run a computer simulation for this.

I have read here about million+ hand VP simulations, using all different strategies.

This dice problem seems easier to simulate, but nobody in the 11/1 crowd has offered one.

Pages and pages of debate on at least 2 different forums, but what does the computer simulation show?

---

Mickey did a real life simulation and it showed 1 in 11.

Quote:

---

Originally Posted by Alan Mendelson

You guys keep making this ridiculous statement:

Quote:

---

Originally Posted by a2a3dseddie

Since we don't know which dice has landed on the 2, we need to consider all combinations

---

If you have two dice, and you know one die is a 2, how many possible combinations could there be? There are only two dice. One is a 2.

It doesn't matter which of the two dice is showing the 2.

Maybe on a chart showing the combinations of dice it matters, but when you are dealing with two physical dice it doesn't matter which is showing a two.

Take your minds off your charts and graphs and put two dice in front of you. Set one of the dice as a 2. What's the answer to the problem? It's one out of six.

---

Alan, you shoot a lot of craps.

Are you telling me that when the dice are thrown, and at least one of them lands on a 2, you see the hard 4 get paid off 1 time in 6?

Quote:

---

Originally Posted by a2a3dseddie

Quote:

---

Originally Posted by Alan Mendelson

You guys keep making this ridiculous statement:

Quote:

---

Originally Posted by a2a3dseddie

Since we don't know which dice has landed on the 2, we need to consider all combinations

---

If you have two dice, and you know one die is a 2, how many possible combinations could there be? There are only two dice. One is a 2.

It doesn't matter which of the two dice is showing the 2.

Maybe on a chart showing the combinations of dice it matters, but when you are dealing with two physical dice it doesn't matter which is showing a two.

Take your minds off your charts and graphs and put two dice in front of you. Set one of the dice as a 2. What's the answer to the problem? It's one out of six.

---

Alan, you shoot a lot of craps.

Are you telling me that when the dice are thrown, and at least one of them lands on a 2, you see the hard 4 get paid off 1 time in 6?

---

He sees a lot of things that are impossible/improbable. Does 18 11's in a row ring a bell? :-D

Quote:

---

Originally Posted by jbjb

Quote:

---

Originally Posted by a2a3dseddie

Quote:

---

Originally Posted by Alan Mendelson

You guys keep making this ridiculous statement:



If you have two dice, and you know one die is a 2, how many possible combinations could there be? There are only two dice. One is a 2.Here we go.

It doesn't matter which of the two dice is showing the 2.

Maybe on a chart showing the combinations of dice it matters, but when you are dealing with two physical dice it doesn't matter which is showing a two.

Take your minds off your charts and graphs and put two dice in front of you. Set one of the dice as a 2. What's the answer to the problem? It's one out of six.

---

Alan, you shoot a lot of craps.

Are you telling me that when the dice are thrown, and at least one of them lands on a 2, you see the hard 4 get paid off 1 time in 6?

---

He sees a lot of things that are impossible/improbable. Does 18 11's in a row ring a bell? :-D

---

Here we go.

Quote:

---

Originally Posted by jbjb

He sees a lot of things that are impossible/improbable. Does 18 11's in a row ring a bell? :-D

---

Quote:

---

Originally Posted by arcimede$

Alan, you really are making a complete fool out of yourself.

---

arcimede$, in Alan's defence(!) the question doesn't mention multiple dice throws. If just 1 throw is considered, the peeker could have just seen 1 die and truthfully say at least 1 of the dice is a 2.

The peeker might have lucked out, had one of the 11 possible combinations of two dice where at least 1 of the 2 dice is a 2 under that cup and saw that one first when he peeked. ;)

Quote:

---

Originally Posted by arcimede$

Quote:

---

Originally Posted by coach belly

Quote:

---

Originally Posted by Alan Mendelson

It's a conspiracy launched by me with the cooperation of regnis and a few others.

Maybe it's to show just how much logic it takes to understand two physical dice.

---

I'm surprised that nobody has run a computer simulation for this.

I have read here about million+ hand VP simulations, using all different strategies.

This dice problem seems easier to simulate, but nobody in the 11/1 crowd has offered one.

Pages and pages of debate on at least 2 different forums, but what does the computer simulation show?

---

Mickey did a real life simulation and it showed 1 in 11.

---

Here's a link to a billion trial simulation

http://wizardofvegas.com/forum/quest...ice-puzzle/41/

Quote:

---

Originally Posted by coach belly

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Originally Posted by Alan Mendelson

It's a conspiracy launched by me with the cooperation of regnis and a few others.

Maybe it's to show just how much logic it takes to understand two physical dice.

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I'm surprised that nobody has run a computer simulation for this.

I have read here about million+ hand VP simulations, using all different strategies.

This dice problem seems easier to simulate, but nobody in the 11/1 crowd has offered one.

Pages and pages of debate on at least 2 different forums, but what does the computer simulation show?

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I have run a simulation and posted the results 2 years ago or whenever this all started. More interestingly, no one in the 1/6 crowd has the capability of running a simulation. [That was mostly sarcasm, but there's some truth to it.] :)

The way the sim worked is this -- the dice keep getting rolled over and over again, only counting rolls where at least one die was a '2'. Since at least one die was a 2, it then recorded if both dice were a 2 or not both 2.

You don't need a simulation. Roll two dice, don't set one to a two. When at least one two shows and the other one isn't, mark column A. When at least one two shows and the other one is, mark column B. Column A will out number column B by 11 to 1 after a long sequence of trials.

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Originally Posted by a2a3dseddie

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Originally Posted by Alan Mendelson

You are told TRUTHFULLY that at least one of the two dice in the cup is showing a 2. So again, what are the odds that the second die is also showing a 2? It is still 1/11 for a six sided die? Or is it 1/6?

Respond to the question and its conditions.

The condition is one die is known to be a two. It is not the same as throwing two dice simultaneously and knowing that there are 11 combinations showing at least one two. Repeat: it is not the same as throwing two dice simultaneously and knowing that there are 11 combinations showing at least one two.

This is a reading comprehension issue for those of you who think the answer is 1/11.

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Once again here is the original problem:

You have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."
What is the probability that both dice are showing a 2?

That is the original question.

NOT

"what are the odds that the second die is also showing a 2?"

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Amazing to check in to see this still rages on. Eddie's post here and the one in which he stated "arci, in Alan's defense" really defines everything this dice issue is about.

There are absolutely two answers to this "problem" and either is correct depending on how the reader wants to understand the question. Redietz actually identified the complexities of how different individuals might understand the problem, albeit a bit awkwardly as usual.

The "math people" like arci as well as several of these uneducated self-proclaimed "AP's who've been imported from the WoV site, along with the mensa "geniuses" that come out of the infamous pool of libtard atheists, agnostics, queers, and weirdo transgenders over there, all see this problem as a one-way street leading to an 11 to 1 conclusion. In arci's defense, he's a hard core technical person, and he is infinitely trained to see this problem--and its final solution--in only one way: to be an 11:1 result and that is all there is to it. The rest of the 11:1 crew? RS__ & jbjb are simply responding the way they best believe they won't be chided or mocked over on WoV. In other words, they really cannot think for themselves. Mickey however is doing nothing more than taking the opposite position of whatever Alan comes up with because he's always been jealous of Alan and his money, his string of girlfriends, his job, and his basic everyday normal life compared to that of the lowlife, non-productive slug existence mickey endures.

On the other, more grounded side of this resides the people who have no interest in reading anything more or less into the problem than it simply states....while 100% ignoring anything that it may imply. These people seek out an everyday-type answer laced with common sense...and nothing more. And while their adversaries' chief goal is very likely to be making others think how deeply intelligent they can be while the rest of the "dummies" can never attain their special level of being able to dissect and analyze such a problem in ways only a class of Einsteins may begin to understand them, the dumbos know how to "keep it simple, stupid".

So who's got it right? Well, both sides have made an indisputable case for their argument (belief). But it is not really that difficult a problem to figure out--one need not have an engineering degree or be a math or psychology professor at Stanford or Boston College (I had to throw that in :)) to come up with a conclusion. It all depends on your reading comprehension.

HOWEVER, if I were a court appointed arbitrator I would rule solidly in favor of the "6" crowd, while giving the "11" crowd honorable mention for their deep insight, unforeseen complex problem solving abilities, ingenuity, and overall argumentative tenacity. Why? Because NOWHERE IN THE ORIGINAL STATED PROBLEM DOES IT MENTION ANYTHING OTHER THAN A ONE TIME TRIAL. If the peeker sees only one die and it is a 2, or if he sees both dice and at least one of them is a 2--and because he doesn't identify which is the case--"6" wins as the simple, common sense conclusion.

Good luck.

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Originally Posted by redietz

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Originally Posted by arcimede$

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Originally Posted by Alan Mendelson

We fucking went through this before. Of course the peeker has to see a 2 in order to make the statement "at least one die is a two."

You made this ridiculous comment the first time responding to regnis who discussed seeing a 2.

Then you tried to twist your words.

Quit it. You pulled this same shit in your battles with Singer over the years and I'm not going to fall for it.

Fuck off.

And you cheated on your taxes.

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No one is twisting words. You are simply denying the obvious fact that the peeker has to see both die making Regnis statement nonsense and showing you will say anything to protect your ego.

Now you are bringing up my taxes which you obviously don't understand very well either (is anyone surprised?)? So, I guess we can add a complete lack of integrity to your growing list of faults.

So, once again .... How can the peeker answer the question "truthfully" if he only sees one die? If you contnue to avoid the question that is just as good as admitting he can't answer truthfully and therefore looking at one die is NOT an option.

Now, quit your incessant whining and just go away. Isn't that your usual mode of operation?

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Why does the peeker need to see both dice? That is not explicitly stated.

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If the above line was stated "awkwardly," I apologize for no mention of transgenders and libtards.

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Originally Posted by Rob.Singer

HOWEVER, if I were a court appointed arbitrator I would rule solidly in favor of the "6" crowd, while giving the "11" crowd honorable mention for their deep insight, unforeseen complex problem solving abilities, ingenuity, and overall argumentative tenacity. Why? Because NOWHERE IN THE ORIGINAL STATED PROBLEM DOES IT MENTION ANYTHING OTHER THAN A ONE TIME TRIAL. If the peeker sees only one die and it is a 2, or if he sees both dice and at least one of them is a 2--and because he doesn't identify which is the case--"6" wins as the simple, common sense conclusion.

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Originally Posted by Rob.Singer

from the looks of things, the wizard is only paying 9 units for something that has 11-1 odds.

If we're at this point, I'll explain why. When two dice are peeked at and we're informed that one of them is a 2, if that die is removed from the scene then we're left with a single die that no longer has a numerical relationship to the die showing a 2. This is the way I interpret the problem with the wording from the OP, and the odds that 2nd die will be a 2 is 6-1.

However, if you take the WoV interpretation, simply telling you that one of the dice shows a 2 does not remove it from its numerical relationship with the die with the unknown number showing. Whereas rolling 2-2 is a 36-1 possibility, knowing half the outcome, ie. one of the dice is showing a 2, reduces the odds to 11-1 since it is no longer a single die event. You now have to consider all combinations of the two dice where either one of them shows a 2.

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:confused:

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Originally Posted by a2a3dseddie

arcimede$, in Alan's defence(!) the question doesn't mention multiple dice throws. If just 1 throw is considered, the peeker could have just seen 1 die and truthfully say at least 1 of the dice is a 2.

The peeker might have lucked out, had one of the 11 possible combinations of two dice where at least 1 of the 2 dice is a 2 under that cup and saw that one first when he peeked. ;)

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You have to understand this gets back to when Alan stated:

"Yes, a "peeker" can see either one die or both dice and say truthfully and correctly "at least one die is a 2." Do you know how to speak the English language Arc?"

He did this in response to me saying he couldn't "ALWAYS" make a determination unless he sees both die. That was my point and for some stupid reason Alan thought his response refuted what I stated. Obviously, Alan doesn't understand what ALWAYS means or he is just belligerent. Since then I've just been baiting him to see what other nonsense he would spew..

You're right Arc, I don't always understand what you say. So let me say it my way:

The peeker only has to see ONE die to truthfully say "at least one of the dice is a 2" if the first die he sees shows a 2.

The peeker does not have to see the second die to say "at least one of the dice is a 2" if the first die he sees shows a 2.

Is this clear now to everyone?

I'm still waiting for jbjb to clarify the conditions in his question regarding the pile of spades and the pile of hearts.
redietz perhaps you should explain one more time the concept of sequential observation.
Rob Singer you gave the 1/6 answer correctly except that you couldn't resist throwing in your usual insults.

Yes everyone in the 1/11 gang -- one out of 11 is what you have when you throw two dice: there are 11 combinations of those two dice that could show a 2 and 1 of those combinations could show 2-2. That's 1/11.

But after two dice are thrown and a peeker sees that at least one die is showing a 2 the odds that the other die is also showing a 2 is one out of the six sides on that single die. That's 1/6.

You must understand the question. You must use the math (or in this case the ability to count to 6) to answer the question that is asked.

You cannot read a question and then apply math that doesn't answer the question being asked. For the thousandth time.

Read.