2
$\begingroup$

This is surely a simple problem, but there are so many probability questions with dice, urns, etc. that I have not been able to find an answer to this specific problem. Say you have a pair of $n$-sided dice, each individually fair, but the pair is correlated $\rho$. What is the probability, given $\rho$, of throwing the dice one time and rolling the same number on both dice?

The equivalent urn problem is: What is the probability of drawing the same color marble from two different urns, each with $n$ marbles of (the same) $n$ colors, with color somehow correlated $\rho$ between urns?

Thanks!

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ To see why this question has no definite answer, consider two fair but correlated d4. In one case, the pairs of results (1,2), (2,1), (3,4), and (4,3) have equal chances. The correlation coefficient is $\rho = 3/5$ and there is zero chance the two results are equal. In another case the outcomes (1,1), (2,2), (3,3), (4,4) each have a chance of $1/5$ and the outcomes (1,4), (2,3), (3,2), and (4,1) each have a $1/20$ chance. Again $\rho = 3/5$ but now there is an 80% chance the two dice show the same value. $\endgroup$
    – whuber ♦
    Aug 31, 2022 at 16:17

1 Answer 1

Reset to default
1
$\begingroup$

Without additional constraints, this question cannot be answered, because there are many different configurations that lead to the same correlation coefficient.

You can think of this problem in terms of a table of probability distribution across all possible pairs of values, e.g. (with many values blank out of laziness)

a=1 a=2 a=3 a=4
b=1 p(1,1) p(2,1)
b=2 p(1,2) p(2,2)
b=3 p(3,3)
b=4 p(4,4)

One way to obtain a strong correlation is to have high probabilities on the diagonal - where both die have the same value. However, another would be to have dice b always having a value one greater than dice a (except when dice a is already at its maximum).

What you can do, however, is specify different probability matrices, and see what correlations they would correspond to.

Update

It's interesting to think about the kinds of constraint you could impose here that would lead to unique solutions. One approach would be to constrain the model so that dice B will either have the same value as A, with probability $m$, or be randomly sampled with equal probabilities from the other $n - 1$ possible values (or from all $n$ possible values).

Improve this answer
$\endgroup$
3
  • $\begingroup$ Let me see if I understand what you're saying. Suppose I come at this backward: I take 2 dice with an unknown probability matrix. I roll the dice a million times and compute the correlation estimate on the data, which is essentially the population correlation. I also compute the proportion of cases where both gave the same number, which is essentially the population probability. You're saying that I could repeat this for k different pairs of dice, come up with the same correlation for every pair, but compute (up to) k different probabilities? $\endgroup$
    – virtuolie
    Aug 31, 2022 at 13:59
  • $\begingroup$ Yes, exactly. Each possible correlation coefficient maps into many possible 2D matrices of proportions, but each possible matrix maps into a single correlation. $\endgroup$
    – Eoin
    Aug 31, 2022 at 18:18
  • $\begingroup$ Thanks! Questions with a false premise are the hardest to look up answers for :) It sounds like the answer to my question is that $p$ is distributed with parameters $n$ and $\rho$, so technically the expected value of that distribution (averaging over all probability matrices, if you will) could be an answer. Depends on whether we're talking probability or statistics. But that answer wouldn't have helped me, because the particular "dice" I'm modeling appear to have an exact probability of $e^z/n$, for Fisher's z. $\endgroup$
    – virtuolie
    Aug 31, 2022 at 22:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.