Two dice are tossed independently until each pair has appeared at least once. Find the expected number of tosses needed and the standard deviation of this number of tosses.

My attempt

Let X be the random variable that counts the number of tosses needed until each pair appears at least once. Let $X_1$ be a random variable that counts the number of tosses until the first "new" pair appear, $X_2$ counts the number of adittional tosses, after the first "new" pair until the second "new" pair appears and so on until $X_{36}$. Then,


We can see that each $ X_i$ is a geometric random variable with $p=\frac{1}{36}$ (that count the number of trials).And since the tosses are independent, I thought that I could say that they are independent. But I am not sure about the last affirmation. If they are indeed independent, I could say that $X$ is negatove binomial, right?

Thanks in advance!

  • 1
    $\begingroup$ Several interpretations of the problem seem possible, depending on what a "pair" is. You have taken it to be any ordered pair of outcomes of two distinguishable dice. It would also be interpreted to be any unordered pair or even any outcome in which both dice show the same result. In the first and third interpretations, this is just the Coupon Collector Problem with equal probabilities, which has been extensively discussed at stats.stackexchange.com/questions/48396. In the second interpretation it has unequal probabilities, but the solution methods are similar. $\endgroup$
    – whuber
    Commented Sep 23, 2014 at 23:00
  • $\begingroup$ Searching here on coupon collector problem turns up a number of related posts, one or two of which have the odd interesting bit of additional information. $\endgroup$
    – Glen_b
    Commented Sep 24, 2014 at 1:53

1 Answer 1


$X_i$ is a geometric random variable with $p_i = \left(36 - (i - 1)\right)/36$.

$X_i$ is the number of tosses until the next new pair; any new pair counts. Until $X_i$ you have observed $i - 1$ distinct pairs, and you are waiting for any of the $36 - (i - 1)$ remaining pairs to roll up.

For example, take $i = 1$, then $p_1 = 1$, because in the very first toss any pair is a new pair.

So, no! this is not negative binomial. Yes, the $X_i$ are independent random variables, but the $p_i$ are not the same.

  • $\begingroup$ Indeed, your are totally correct. But how to calculatethe standard deviation (or the variance) using this fact? $\endgroup$
    – Giiovanna
    Commented Sep 23, 2014 at 21:45
  • $\begingroup$ Never mind, they are independent! $\endgroup$
    – Giiovanna
    Commented Sep 23, 2014 at 21:47
  • 1
    $\begingroup$ @Giiovanna you have a summation of independent variables, and the expected value of the sum is sum of expected values, and using the independence, similarly for variance. $\endgroup$ Commented Sep 23, 2014 at 21:47

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