Guessing the other dice number, given two dice were rolled, and one of them rolled a 3

Question

An argument i had with a friend followed this question: "Given two dice were rolled, and one of them rolled a 3. What would you bet the other dice rolled?"

The phrasing is just to enhance that there might be some number that has an higher probability of showing on the other dice. One of us said that since 7 is the most likely number to be rolled by two dices, then 4 should be the answer. The other said that once the 3 was set on one dice, we should not be looking at anything other than the other cube - so any number would be an equal guess.

Answer 1

This kind of question is very sensitive to how the roller decides what to say. Maybe the dice are different colors and he will tell you the number on the green die. You know nothing about the other die, so all numbers are equally probable. Maybe he will tell you the higher number. Now the only rolls possible are $13,23,33,31,32$ so you should bet on $1$ or $2$. Maybe he will pick a random die and tell you the number on it. Now he could have rolled $33$ and been forced to say $3$ or rolled $3x$ or $x3$ and chosen to say $3$. You have the same chance of $x$ as $3$ for the other die so bet on anything you like. If you don't know how the roller chooses what to say it is not a mathematical question. There is not enough information to answer.

Answer 2

Ross Millikan is right in saying that it depends on things happening behind the scenes. However, the question is worded similarly to the Boy or Girl paradox, so if we interpret it in that fashion:

Assuming you were told "I rolled two fair dice. One of the dice was a 3." and there is no reason to believe there was any particular pressure to report a particular roll, then:

There are 36 possible rolls of two dice, of which 11 have a 3 as one of the rolls:

$$(1, 3)\ (2, 3)\ (3, 3)\ (4, 3)\ (5, 3)\ (6, 3)\\ (3, 1)\ (3, 2)\ (3, 4)\ (3, 5)\ (3, 6)$$

Each of these rolls is equally likely, i.e. has probability $\frac{1}{11}$. Therefore, the probability that the unreported die is a 3 is $\frac{1}{11}$ (since there's only one (3, 3) pair), while the probability for any other roll is $\frac{2}{11}$ (since there are, for example, both (3, 1) and (1, 3)). So you should pick any number other than 3 to have twice as good a chance of being right.

Answer 3

Since your question doesn't mention any selection criteria or bias in the dice used, the answer is clearly that there is no bias in the values of the second die.

When you roll two dice there are 36 possible permutations of answers that can arise. If the first die shows a 3 then all permutations that involve the first die having any other value are eliminated. You've eliminated 30 possible permutations and there are only 6 possible permutations remaining.

This is because probability is only ever about what you don't already know, or what has not already happened. Flip a coin 9 times. What's the probability of the next flip being heads? 50%. What if all 9 flips were heads? Still the same: 50%. Flip 10 coins together, look at the first 9. The 10th still has a 50% chance of being heads.

And yes, the most common sum rolled on two dice is 7, at 6 out of 36 permutations. 6 and 8 are 5/36 each, 5 and 9 are 4/36 each, and so on. But that's only important when you have two unknowns. When only one is unknown then the probability is still even. You can easily write down the permutations and count them to see what's up.

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Ross Millikan's answer adds additional information, but is correct when he says:

You have the same chance of $x$ as 3 for the other die so bet on anything you like.

ConMan's answer however misses something.

If the selection of which die to reveal is random, and the revealed number is 3, then there are in fact 12 different possible results - 6 for each non-revealed die - with 2 results each for the number on the other die. So each possible number has exactly 1/6 probability. His answer that there are only 11 possible results is ignoring the fact that the [3,3] result would have two chances of being the chosen permutation.

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The big issue is that our intuition is generally quite poor at picking the right answer here. We tend to mix up situations where the probabilities change as we progress - card games, lottery numbers, etc - with ones where they don't. We tend to assume that the probability of an event that happened yesterday is the same as it happening tomorrow, and that things that have happened affect things that will happen regardless of a lack of causal links.

Casinos make bank on us making those mistakes.

Answer 4

Ross has provided a nice answer that highlights the subtleties of deducing probabilities from what someone tells you, where you need to take into account how they decided what to tell you.

ConMan has provided an answer that treats the $11$ results containing a $3$ as equiprobable, but does so again in the paradigm of someone reporting the result, and in my view doesn't explain properly how the report is to be interpreted.

I note that your question doesn't actually mention anyone reporting the result; you merely write “and one of them rolled a $3$”.

That allows us to give an entirely objective interpretation to the question, free of the subtleties of the decisions of people reporting results: What is the conditional probability distribution for “the other die”, conditional on the event that at least one die shows a $3$? To make this interpretation well-defined, we need to define what “the other die” means, but it's intuitively clear enough: If there's a non-$3$, it's the non-$3$, and otherwise it's one of the $3$s (it doesn't matter which); alternatively, we could define its value as the sum of the dice minus $3$ (with the advantage that this defines a random variable in all cases, not just when there's a $3$).

Then, if we let $X$ denote the value of “the other die” and $T$ denote the event that at least one die shows a $3$, we have

$$ \mathsf P(X=3\mid T)=\frac{\mathsf P(X=3\cap T)}{\mathsf P(T)}=\frac{\frac1{36}}{\frac{11}{36}}=\frac1{11}\;, $$

and for $x\ne3$ we have

$$ \mathsf P(X=x\mid T)=\frac{\mathsf P(X=x\cap T)}{\mathsf P(T)}=\frac{\frac2{36}}{\frac{11}{36}}=\frac2{11}\;. $$

Thus, under this interpretation of the question, you should bet on any number other than $3$.

In the reporting paradigm, the corresponding scenario would be that you agree beforehand that the person rolling the dice will tell you whether or not there is at least one $3$, so the person has no decisions to make on what to report.