Question for Math/Gambling/Craps Experts

[quahaug]

I'd rather make it one thousand rather than one dollar but I don't have a Rob Singer bankroll so I have to start small.

[Alan Mendelson]

Alan, I believe you also need to see him or them explain exactly what their interpretation of the OP's question was as presented to the forum. You, I, regnis and maybe others see it as a very clear "what are the odds, now that we know one of the dice is a deuce, that the other die is also a deuce?"

Yes and I said on this forum and on the Wizard's forum that they are not responding to the question, or, they are questioning the question, or, they are applying math to the question that should not be applied to this question.

I think those of us who play craps deal with this type of question every time we play -- and I said this several times. We see one die come to rest on a number and then we say to ourselves "no ____" for whatever number would be a 7-out, or "give us a ____" for the number that would give us a winner. When we say those things we are well aware that it is a one in six chance.

But now we have folks who are asked a similar question but are answering not with a 1/6 answer but with something more complicated, and now I am asking them to explain why it's not 1/6 and why they say it's 1 out of eleven??

I'd be happy if they just prove to me it's not one out of six.

[quahaug]

What does this yield? And the problem only asks about the one die, not both, simply because we are given a value for the first die.

That's the bet. Take it or leave it hotshot. You got a lot of mouth lets see the cash big boy.

[Alan Mendelson]

Hold it. If anyone wants to make a side bet outside of a casino, do it thru a private message. There are still gambling laws and even an office football pool is illegal in California. So please, no gambling or betting here.

Can we stick to the subject? Again, I don't think this bet or any other resolves the question about the answer being 1/11 or 1/6.

[Rob.Singer]

I'll take a bet if it makes any sense. I can't even figure out what he's getting at

[quahaug]

Rob. Have you been told today?

[Alan Mendelson]

I've been reading more of the responses on the Wizard's forum and clearly this response from "ThatDonGuy" illustrates the problem and why they reject the 1/6 response. I reminded ThatDonGuy what the original question was and he wrote this:

Also, if you are serious about this statement:
"The ORIGINAL QUESTION deals with TWO DICE under a cup with at least one of them showing a 2. YOU MUST address the problem using a pair of dice with at least ONE DIE showing a 2."
(which, by the way, is what pretty much every one of us who have figured out that the answer is 1/11 has already done)
then show us the possible rolls of the two dice, each with equal probability, with at least one die showing a 2.
Aren't the possible rolls 1-2, 2-1, 2-2, 2-3, 2-4, 2-5, 2-6, 3-2, 4-2, 5-2, and 6-2?
Am I missing any?

Do you see what is happening here? They are ignoring that one die already shows a 2 and they are referring to the eleven different combinations of dice that show a two. This is why they insist on the 1/11 answer. If they would accept that ONE DIE is already showing a two, then the obvious answer is that the remaining die has SIX faces ONE of which is a two.

[Rob.Singer]

Exactly what I said about them wordsmithing the question to make it more complicated....more theoretical....a problem than it really is. If it were really about keeping both dice in play once the value of one of them was known to be a deuce, the question would have been absolutely clear to that point. But by leaving it loosely stated to these guys but clear to rational thinking people, the mensa-laced math theory door was opened up.

[redietz]

Rob, you should have just kept your mouth shut a little bit and lured them into a real-life wager with one die under each of two cups. Alan could have filmed the action, and we all could have thrown a party with the proceeds.

[Rob.Singer]

The wizard's bunch is indeed a gullible bunch. It doesn't take them much to be lured into making them whine to the admins. to get you suspended. So you're right.

[regnis]

Lets just put one die on 2, roll the other one, and take all the action they want, using their odds, on the second one being a 2.

[regnis]

wonder what they think the odds of neither being a 2 are, even though 1 already is a 2. They are calculating the odds of the second 2 as if there weren't a first 2.

[Alan Mendelson]

Lets just put one die on 2, roll the other one, and take all the action they want, using their odds, on the second one being a 2.

They've already proposed such a bet.

[1in11]

Do you see what is happening here? They are ignoring that one die already shows a 2 and they are referring to the eleven different combinations of dice that show a two. This is why they insist on the 1/11 answer. If they would accept that ONE DIE is already showing a two, then the obvious answer is that the remaining die has SIX faces ONE of which is a two.

I accept that one of the dice is showing a two, but I do not know which die shows a two. This is the distinction here.

If I have the information of which die shows a two, and then you ask what is the chance that they both show a two, then the answer is clearly 1 in 6. But, I don't have that information. I only have the information that at least one of the dice shows a two.

[Rob.Singer]

That's fine 1in11, but it doesn't matter which die shows a deuce as long as we know one of them does. And once we know that info, that deuce-showing die is off the table. Please explain why you believe that's an important issue.

[1in11]

Knowing which die shows a 2 gives us 6 possibilities for the combination of the two dice: 2-1, 2-2, 2-3, 2-4, 2-5, 2-6.

Knowing only that at least 1 die shows a 2 gives us 11 possibilities for the combination of the two dice: 2-1, 2-2, 2-3, 2-4, 2-5, 2-6, 1-2, 3-2, 4-2, 5-2, 6-2.

This is why it is an important issue, as they are two fundamentally different conditions.

[Alan Mendelson]

That's fine 1in11, but it doesn't matter which die shows a deuce as long as we know one of them does. And once we know that info, that deuce-showing die is off the table. Please explain why you believe that's an important issue.

Rob, this is actually the problem with the guys on the Wizard's forum. Another just posted this:

"Alan's group doesn't know the difference between a "certain" die being a 2 and a "random" die being a 2."

Somehow they can't understand that there are only two dice to start with, and it doesn't matter which of the two dice is showing the two. At this point I have to wonder why the Wizard himself hasn't commented -- even to say we don't know how to read and the 1/11 crowd does know how to read.

[Alan Mendelson]

Knowing which die shows a 2 gives us 6 possibilities for the combination of the two dice: 2-1, 2-2, 2-3, 2-4, 2-5, 2-6.

Knowing only that at least 1 die shows a 2 gives us 11 possibilities for the combination of the two dice: 2-1, 2-2, 2-3, 2-4, 2-5, 2-6, 1-2, 3-2, 4-2, 5-2, 6-2.

This is why it is an important issue, as they are two fundamentally different conditions.

I am sorry but you need to look at two dice. When one die is a two -- there are only six possibilities on the other die -- a one, a two, a three, a four, a five or a six. What are you guys smoking?

[redietz]

These ain't Schrodinger's dice.

[1in11]

You are correct that there are only six possibilities for the other die. However, these don't happen with equal probability, given the constraint of the problem.

2-1, 2-3, 2-4, 2-5, 2-6 all happen with 2/11 probability, and 2-2 happens with 1/11 probability.