My family recently had a domino game (double 12 set = 91 unique) in which each of 5 players started with 10 in their hand. We needed to start the game with the double "1", but no one had it. Everyone took turns and continued to draw for the double "1" until it turned out to be the last one drawn from the remaining 41. We all joked that the odds of that happening are probably one in a million. What were the true odds?
2026-03-28 07:57:23.1774684643
Picking a specific domino last
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From the start any domino has the same chance as any other to be picked last, so $1$ in $91$. Once you know the double-$1$ isn't among the first $50$ it has a $1$ in $41$ chance to be picked last from the remaining 41. So not one in a million, once in every $41$ such games.
The chance that no one has the double-$1$ after taking the first $50$ is just $41/91$, so that will happen in just under half the games.
Edit in response to a comment:
Each time you pick a domino and don't get the double-$1$ the probability that you get it the next time increases. When it's the only one left the probability is, of course, $1$. But that does not change the fact that at the moment you have $50$ drawn and not that one, the probability that it will be last is just $1/41$. There is nothing to multiply.
You can do an experiment. Pick six of your favorite dominoes - say 12, 11, 21, 22, 31 and 32 Now draw two hands of one each. Each time you don't see the 11 drawn, turn the other four over one at a time. The 11 will be last in about $1/4$ of the games - do this lots of times so the randomness has a chance to work. (I know those $6$ don't make for a good domino game.)