It was two years ago today...

[regnis]

Mickey if you want the answer to be 1/11 CHANGE THE QUESTION. Ask: given all of the combinations of two dice that include the number 2, how many combinations can be 2-2. That answer is 1/11. My question is the correct question to get the answer 1/11.
You guys are not answering the question that was asked.

This is the question, the original question, that I answered. "your friend looks under the cup and says "at least one of the dice is a two. What is the probability that both dice are 2's?" This is the question I answered. It is a clear and concise question. I totally understand the question. I DID NOT, repeat, DID NOT, repeat, DID NOT, change the question. And I answered it clearly and concisely. When the only information you have is that "at least one die is a two" you cannot eliminate any of the 11 possible combinations from possibility. Both the dice are 2's in only one combination. Therefore, the answer is 1 in 11.

Mickey-mine was a trick question to show the absurdity of some of this. In my example, you can see both dice. The picture didn't transfer over to where I gave the answer but the original picture showed one die as a 2 and one as a 1. My odd sense of humor was not meant to to anything except highlight the way that the phrasing of the question may lead to this mess.

[regnis]

I have a legitimate question. The guy that peeked under the cup and said one of them is a 2. Would you say that for him the odds that the other one is a 2 would be 1 of 6. And if so, then are you telling me that his odds are different than the other guy's odds?
The guy that peeked under the cup knows which one is a 2. He may also know what the other die is which gets us back to my example. But assuming he only saw the one die that showed a 2, what are the odds for him that the other is a 2?

[a2a3dseddie]

I have a legitimate question. The guy that peeked under the cup and said one of them is a 2. Would you say that for him the odds that the other one is a 2 would be 1 of 6. And if so, then are you telling me that his odds are different than the other guy's odds?
The guy that peeked under the cup knows which one is a 2. He may also know what the other die is which gets us back to my example. But assuming he only saw the one die that showed a 2, what are the odds for him that the other is a 2?

That's an excellent question regnis. It reminds me of a scene in the Martin Scorsese movie "Casino".

It involves a pair of cheaters playing blackjack. In the scene, one player... the "big player" is betting $1000s on each hand. His partner "the spotter" is seated at a table directly behind the dealer of the "big player's" table.

The dealer is "weak" and lifts his hole card too high and the spotter can see what it is. The "spotter" betting just the table minimum then uses an electronic signaling device to notify the "big player" whenever the dealer has a poor card in the hole.

Using this information, the "big player" can stand on stiff hands like 12 through 16 against a dealer upcard of 10 when he knows the dealer doesn't have a pat hand hoping he draws a card and busts. This increases their edge significantly and the pair win several hundred thousand before Robert De Niro figures it out and gives them the hammer.

The "big player's" optimal move minus any additional information is to play basic strategy blackjack. The "spotter" increases their odds of winning by obtaining hole card information and getting his partner to deviate from basic strategy where appropriate.

In the dice question, AFTER the roll, the partner peeking under the cup has information that the other guy does not. For the partner, part of the roll has already been resolved. Using mickeycrimms coloured dice example, the partner KNOWS which dice (red or green) is already a 2. For him, if the remaining die is "spinning" as Alan puts it, relative to him the odds are 1/6 for it landing on 2.

For the guy who has not seen any of the dice, no part of the roll has been resolved relative to him and he does not know which dice (red/green) is a 2 and which one is still "spinning". Relative to him, any of the 11 combinations of two dice where at least one die is a 2 could be under that cup.

[Alan Mendelson]

I have a legitimate question. The guy that peeked under the cup and said one of them is a 2. Would you say that for him the odds that the other one is a 2 would be 1 of 6. And if so, then are you telling me that his odds are different than the other guy's odds?
The guy that peeked under the cup knows which one is a 2. He may also know what the other die is which gets us back to my example. But assuming he only saw the one die that showed a 2, what are the odds for him that the other is a 2?

Warning: if you are ever in court and the other side is represented by regnis, you lost.

(You are brilliant, regnis.)

[arcimede$]

I have a legitimate question. The guy that peeked under the cup and said one of them is a 2. Would you say that for him the odds that the other one is a 2 would be 1 of 6. And if so, then are you telling me that his odds are different than the other guy's odds?
The guy that peeked under the cup knows which one is a 2. He may also know what the other die is which gets us back to my example. But assuming he only saw the one die that showed a 2, what are the odds for him that the other is a 2?

Sorry, the peeker cannot ALWAYS answer the question "at least one die is a 2" unless he sees both die. Hence, once he sees both die then he knows the exact answer already. The odds aren't 1:6 either.

[mickeycrimm]

I have a legitimate question. The guy that peeked under the cup and said one of them is a 2. Would you say that for him the odds that the other one is a 2 would be 1 of 6. And if so, then are you telling me that his odds are different than the other guy's odds?
The guy that peeked under the cup knows which one is a 2. He may also know what the other die is which gets us back to my example. But assuming he only saw the one die that showed a 2, what are the odds for him that the other is a 2?

If all he has seen is one die and he can't distinguish one die from the other then "one die is a two" is all the information he has. Consequently it's 1 in 11. If the dice are color coded, such as one is red and one is green, he has more information and can eliminate five combinations from possibility, putting his chances at 1 in 6.

Note: The dice don't necessarily have to be color coded. If one die has some sort of distinguishing mark on it that he is aware of then he has the same information as if they were color coded.

[quahaug]

I have a legitimate question. The guy that peeked under the cup and said one of them is a 2. Would you say that for him the odds that the other one is a 2 would be 1 of 6. And if so, then are you telling me that his odds are different than the other guy's odds?
The guy that peeked under the cup knows which one is a 2. He may also know what the other die is which gets us back to my example. But assuming he only saw the one die that showed a 2, what are the odds for him that the other is a 2?

Sorry, the peeker cannot ALWAYS answer the question "at least one die is a 2" unless he sees both die. Hence, once he sees both die then he knows the exact answer already. The odds aren't 1:6 either.

How right you are, if the peeker sees two twos The odds are 100% there is another two. If he sees any other number the odds are 0% there is a another two, but that has nothing to do with the original question Regnis' picture notwithstandig. And I don't think I want him representing me against the IRS.

[a2a3dseddie]

Arc, I've done the work. When I have a 2 showing and throw a die, one out of six times it's another 2.

That's the work because that's the question.

No Alan, that's not the question.

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?

Using a craps analogy, let's say you throw the dice down the felt and one of them lands on a 2. The other bounces off the table.

Does the crew let you throw just 1 die in order to complete the roll, add it to the 2 and resolve all the bets on the table?

No. They call it a "no roll" and make you roll 2 dice again. You know why that is right?

While it's still 1 in 6 that you could end up "sevening out" it also becomes 1 in 6 that you could end up with a 3, hard 4, 5, 6 or 8 instead. The chances of you winning a bet on any of those numbers goes up dramatically if you are allowed to just roll 1 die!

By keeping both dice in play for every roll, any of the 36 combinations could come. The chances of a seven out remain at 1 in 6 but the odds for the other numbers 3, hard 4, 5, 6, 8 revert back.

With regards to the original question, both dice are in play for you, but NOT for the peeker. The peeker knows which dice is a 2. You do not.

The question:

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


would elicit different correct responses from the peeker and from the person who has not seen any of the dice.

[Alan Mendelson]

Sorry, the peeker cannot ALWAYS answer the question "at least one die is a 2" unless he sees both die. Hence, once he sees both die then he knows the exact answer already. The odds aren't 1:6 either.

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?

[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?

[Alan Mendelson]

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

If neither die is a 2 then the peeker cannot TRUTHFULLY say "at least one die is a 2."

[RS__]

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

If neither die is a 2 then the peeker cannot TRUTHFULLY say "at least one die is a 2."

Well yeah, duh. That's not quite what I was asking though.....

[Alan Mendelson]

If you are looking at a graph of dice, or if you are examing two dice, YES there are 11 different combinations of the two dice that include a 2, and only ONE of those 11 combinations will show 22. 1/11

But the question asks us to consider what happens when two physical dice are put into a cup and rolled and it is determined that at least one of those two dice is showing a 2 when it comes to rest. That particular question asks us to consider the second die in the problem: one out of six faces will also have a 2. 1/6

As Ive been saying over and over again... the questions are not the same. And the 1/11 people are answering the question about the two physical dice that have been rolled with the graph or thinking that would be used when looking at a graph of dice results or examing two dice.

Is that a little bit easier to understand now?

[arcimede$]

Sorry, the peeker cannot ALWAYS answer the question "at least one die is a 2" unless he sees both die. Hence, once he sees both die then he knows the exact answer already. The odds aren't 1:6 either.

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?

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.

[jbjb]

Using the red/green dice, when "at least" one die is a two, there are ELEVEN possible faces for the other one because the NON-PEEKER does NOT know which two is showing.

Either 1g, 2g, 3g, 4g, 5g, 6g, 1r, 3r, 4r, 5r, 6r OR 1r, 2r, 3r, 4r, 5r, 6r, 1g, 3g, 4g, 5g, 6g.

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

[regnis]

I have a legitimate question. The guy that peeked under the cup and said one of them is a 2. Would you say that for him the odds that the other one is a 2 would be 1 of 6. And if so, then are you telling me that his odds are different than the other guy's odds?
The guy that peeked under the cup knows which one is a 2. He may also know what the other die is which gets us back to my example. But assuming he only saw the one die that showed a 2, what are the odds for him that the other is a 2?

(You are brilliant, regnis.)

My cross to bear Alan. yet the beat goes on here.

[regnis]

Using the red/green dice, when "at least" one die is a two, there are ELEVEN possible faces for the other one because the NON-PEEKER does NOT know which two is showing.

Either 1g, 2g, 3g, 4g, 5g, 6g, 1r, 3r, 4r, 5r, 6r OR 1r, 2r, 3r, 4r, 5r, 6r, 1g, 3g, 4g, 5g, 6g.

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

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.

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.

[Alan Mendelson]

Sorry, the peeker cannot ALWAYS answer the question "at least one die is a 2" unless he sees both die. Hence, once he sees both die then he knows the exact answer already. The odds aren't 1:6 either.

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?

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.

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.

[Count Room]

I have a legitimate question. The guy that peeked under the cup and said one of them is a 2. Would you say that for him the odds that the other one is a 2 would be 1 of 6. And if so, then are you telling me that his odds are different than the other guy's odds?
The guy that peeked under the cup knows which one is a 2. He may also know what the other die is which gets us back to my example. But assuming he only saw the one die that showed a 2, what are the odds for him that the other is a 2?

Warning: if you are ever in court and the other side is represented by regnis, you lost.

(You are brilliant, regnis.)

I'm sorry, but I just have to say I am marshalling a significant portion of my personal restraints not to have an undiplomatic response here.

[Alan Mendelson]

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

Very good, jbjb. If your "question" allows all 12 cards to be in one pile, then the answer is 1/11 because you have picked and identified the ace of spades and all of the other 11 cards are still "in play."

But if there are two piles -- spades in one pile and hearts in a second pile -- and the Ace of spades is the only card you can choose from the Spades pile, then there are 6 hearts that remain "in play." Of those 6 hearts one is the ace of hearts.

(pause)

My goodness, I just reread your question and you said there are two piles. So this means you got the answer wrong to your own question. If the ace of spades has already been chosen from your spades pile and only the hearts pile remains the answer can only be 1/6.