It was two years ago today...

[Alan Mendelson]

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.

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

I love this. regnis is a genius attorney.

[mickeycrimm]

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?

[Alan Mendelson]

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.

[mickeycrimm]

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

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.

[Alan Mendelson]

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

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.

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 ??

[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. :-)

[Alan Mendelson]

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. :-)

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

[jbjb]

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.

[Alan Mendelson]

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

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

[jbjb]

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!

[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.

[mickeycrimm]

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.

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?"

[mickeycrimm]

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.

[a2a3dseddie]

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.

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.

[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.

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.

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.

[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?

No comments??

[a2a3dseddie]

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.

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.

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.

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.

[a2a3dseddie]

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.

[Alan Mendelson]

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.

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

[jbjb]

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.

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

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.