Question for Math/Gambling/Craps Experts

[Rob.Singer]

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.

That's what's called an incessant need to take theory to a more complicated, unnecessary level. The OP's question clearly eliminates one of the die from the calculation--random or not. Putting the other die back into the issue would lead to an F from any math professor at any university.

[redietz]

Rob, take the action. If it's in Las Vegas, I'll drop by for back-up.

[synergistic]

I'm shithoused, so I might rescind in the morning, but a quick Excel run tells me that 1/11 is accurate. Mind going through my steps above and making sure they accurately simulate the question you posed?

[synergistic]

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?

I agree and yet my spreadsheet does not. I've now run through literally millions of simulations and it is apparent that your friend not telling you WHICH die is a two somehow has a huge effect on the chances of the other die being a two as well.

I don't get it but I keep going back to something someone said about you having a great chance to win a hand of poker with pocket kings, until you see your opponent's pocket aces. I think there's something distinctly Schrodingery about the situation but my mind hasn't sorted out the why yet.

[1in11]

That's what's called an incessant need to take theory to a more complicated, unnecessary level. The OP's question clearly eliminates one of the die from the calculation--random or not. Putting the other die back into the issue would lead to an F from any math professor at any university.

The original question does not give any clear information on one die - it does not say that this die is a 2. The original question only gives information on the pair of dice - at least one of them is a 2.

[Alan Mendelson]

Synergistic instead of Excel take a pair of dice and figure it out the old fashioned way. Make one die a 2 and then do the math. Then change the dice putting the 2 on the other die and figuring the math again.

1in11 you do the same thing.

[Alan Mendelson]

Synergistic your application of Excel was just wrong.

[Dan Druff]

I got into a rather heated exchange over on the Wizard's forum on this 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?

My answer was 1/6 or one out of 6. The Wizard had a different answer and so did just about everyone else.

What's your answer?

Haven't read the rest of the thread yet, but it's not 1/6.

The key is "at least".

Saying "At least one of the dice is a 2" is different than "The first die is showing a 2".

If it's "the first", the chance of the second one showing 2 is indeed 1/6. That makes them two completely independent events, so the first doesn't effect the second, much like flipping a second coin completely ignores the results of the first coin's flip.

But "at least" changes everything, because it means that the situation MUST be either a 2 on the first, second, or both.

Let's look at the probabilities if you don't have any information:

Probability of first die with 2 = 1/6
Probability of second die with 2 = 1/6
Prob of first die NOT having a 2 = 5/6
Prob of second die NOT having a 2 = 5/6

So:
Prob of NEITHER having a 2 = 5/6 * 5/6 = 25/36
Prob of BOTH having a 2 = 1/6 * 1/6 = 1/36

So "at least one 2" is basically every combination except for neither, meaning -> 1 - 25/36 = 11/36

Since the chance of both having a 2 is 1/36, comparing it to the chance of "at least one 2" of 11/36 is a difference of a factor of 11.

So it's 1/11 that both show a 2.

[Rob.Singer]

The original question does not give any clear information on one die - it does not say that this die is a 2. The original question only gives information on the pair of dice - at least one of them is a 2.

Which in turn eliminates that particular die, regardless of it being the one on the left or the one on the right, from the equation. Immediately and unequivocally. Choosing to keep it in play is not within the parameters of the original question--which only asks about the die that as yet has no known value.

[regnis]

Rob-Alan--don't you guys understand that if the left die is a 2 then the odds are different than if the right die is 2. Come on--this ain't that hard.

Seriously, these guys can't understand simple English. We have a given fact as stated that one die is a 2. So we have only the other die to consider. 1 of 6!!!!

The question was not what are the odds that one will be a 2. We are given the fact that one die is a 2--it doesn't matter if it is the first or second die. If one is a 2, the odds of the other being a 2 are 1 of 6.

Red--"at least" only changes it if we are not given that 1 is a 2. If the question just said "what are the odds that at least 1 would be a 2", without having already told us that 1 is a 2, then we could all agree to your theory.

I rest my case.

[redietz]

The original question said somebody is peeking at one of two dice under a cup, and it's a two. Okay, just do the real world wager. Roll two dice in a cup and have somebody peek. When one of them is a two, then make the wager. We need an honest peeker -- I volunteer.

Bring lots of cash. Rob usually has lots of cash.

[regnis]

The original question said somebody is peeking at one of two dice under a cup, and it's a two. Okay, just do the real world wager. Roll two dice in a cup and have somebody peek. When one of them is a two, then make the wager. We need an honest peeker -- I volunteer.

Bring lots of cash. Rob usually has lots of cash.

Can you give me a marker?

[Rob.Singer]

Haven't read the rest of the thread yet, but it's not 1/6.

The key is "at least".

Saying "At least one of the dice is a 2" is different than "The first die is showing a 2".

If it's "the first", the chance of the second one showing 2 is indeed 1/6. That makes them two completely independent events, so the first doesn't effect the second, much like flipping a second coin completely ignores the results of the first coin's flip.

But "at least" changes everything, because it means that the situation MUST be either a 2 on the first, second, or both.

Let's look at the probabilities if you don't have any information:

Probability of first die with 2 = 1/6
Probability of second die with 2 = 1/6
Prob of first die NOT having a 2 = 5/6
Prob of second die NOT having a 2 = 5/6

So:
Prob of NEITHER having a 2 = 5/6 * 5/6 = 25/36
Prob of BOTH having a 2 = 1/6 * 1/6 = 1/36

So "at least one 2" is basically every combination except for neither, meaning -> 1 - 25/36 = 11/36

Since the chance of both having a 2 is 1/36, comparing it to the chance of "at least one 2" of 11/36 is a difference of a factor of 11.

So it's 1/11 that both show a 2.

You've fallen into the same trap of wordsmithing the question as the other math guys have done. When it says "at least one of the dice shows a deuce" we immediately do not care which one shows the deuce or what the other die shows--even if that's a deuce also. The only issue is what the odds are that BOTH the dice, ie. the OTHER as yet unvalued die, shows a deuce. Meaning of course that there is now only one die in question.

[1in11]

You've fallen into the same trap of wordsmithing the question as the other math guys have done. When it says "at least one of the dice shows a deuce" we immediately do not care which one shows the deuce or what the other die shows--even if that's a deuce also. The only issue is what the odds are that BOTH the dice, ie. the OTHER as yet unvalued die, shows a deuce. Meaning of course that there is now only one die in question.

The information that at least one die shows a 2 changes the probability of that "OTHER as yet unvalued die"

[Rob.Singer]

The information that at least one die shows a 2 changes the probability of that "OTHER as yet unvalued die"

There is no information in that statement that changes anything I've said. The probability of that other die showing a deuce is and will always be 1 in 6. Please explain your post.

[redietz]

Time for another Singer wager. Not much equipment required -- two dice, one cup, lots of cash. Let's get the ball rolling. I volunteer as peeker. Alan, how about a filmic opportunity here?

Somebody, as they say in Guys and Dolls, is going to wind up with an earful of cider.

[regnis]

Forget the cup. One is a 2 so just set that die aside. Now let's try to determine the odds that the second die is a 2.

Give me 11-1 all day!!!!!

[1in11]

Please explain your post.

I'll explain it again, and in a different way than I did the last time.

Let's start from the beginning, before there's any peeking. There are 36 possibilities for the dice at this point, each with equal (1/36) probability:
1-1 1-2 1-3 1-4 1-5 1-6
2-1 2-2 2-3 2-4 2-5 2-6
3-1 3-2 3-3 3-4 3-5 3-6
4-1 4-2 4-3 4-4 4-5 4-6
5-1 5-2 5-3 5-4 5-5 5-6
6-1 6-2 6-3 6-4 6-5 6-6

We are then truthfully told that at least one of the dice shows a 2. With this information, we eliminate every combination of dice without a 2, and this leaves us with
1-2
2-1 2-2 2-3 2-4 2-5 2-6
3-2
4-2
5-2
6-2

Each of these combinations has the same probability, as nothing changed about the dice. The only thing that changed was the information that is available to us.
There are 11 equally probable combinations of 2 dice that satisfy the information that at least one of them is a 2. Out of these 11 equally probable combinations, only 1 is the pair of 2s that we are looking for.

If instead we decide that we take the possibilities, and we rearrange them so that the 2 is the first die, and the "OTHER as yet unvalued die" is the second die, the chart looks like this:
2-1
2-1 2-2 2-3 2-4 2-5 2-6
2-3
2-4
2-5
2-6
This still shows the 11 outcomes, but now some are doubled up. 2-1, 2-3, 2-4, 2-5, 2-6 all have 2 distinct ways to happen. 2-2 still only has 1 way. Therefore, 2-1 has a 2/11 probability, but 2-2 only has a 1/11 probability.

[regnis]

By this theory there would be two ways to get a hard 4 in craps:

a 2 on die one and then a 2 on die 2; or

a 2 on die 2 and then a 2 on die 1.

Sheeesh--they have paid it wrong for 100's of years.

[Rob.Singer]

I'll explain it again, and in a different way than I did the last time.

Let's start from the beginning, before there's any peeking. There are 36 possibilities for the dice at this point, each with equal (1/36) probability:
1-1 1-2 1-3 1-4 1-5 1-6
2-1 2-2 2-3 2-4 2-5 2-6
3-1 3-2 3-3 3-4 3-5 3-6
4-1 4-2 4-3 4-4 4-5 4-6
5-1 5-2 5-3 5-4 5-5 5-6
6-1 6-2 6-3 6-4 6-5 6-6

We are then truthfully told that at least one of the dice shows a 2. With this information, we eliminate every combination of dice without a 2, and this leaves us with
1-2
2-1 2-2 2-3 2-4 2-5 2-6
3-2
4-2
5-2
6-2

Each of these combinations has the same probability, as nothing changed about the dice. The only thing that changed was the information that is available to us.
There are 11 equally probable combinations of 2 dice that satisfy the information that at least one of them is a 2. Out of these 11 equally probable combinations, only 1 is the pair of 2s that we are looking for.

If instead we decide that we take the possibilities, and we rearrange them so that the 2 is the first die, and the "OTHER as yet unvalued die" is the second die, the chart looks like this:
2-1
2-1 2-2 2-3 2-4 2-5 2-6
2-3
2-4
2-5
2-6
This still shows the 11 outcomes, but now some are doubled up. 2-1, 2-3, 2-4, 2-5, 2-6 all have 2 distinct ways to happen. 2-2 still only has 1 way. Therefore, 2-1 has a 2/11 probability, but 2-2 only has a 1/11 probability.

Of course the main--and only--reason why you are wrong is because you insist on keeping a total of two dice in play, when the OP's opening statement clearly eliminated one of the die.

This is similar to the Monty Hall problem, where wording and parameters are all-important.