1

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?

KCRJ
  • 11

1 Answers1

1

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.)

Ethan Bolker
  • 95,224
  • 7
  • 108
  • 199
  • Doesn't probability change with every picked domino? I would think we need to multiply probabilities that a particular piece is not picked during first, second, third try and so forth. – Vasili Mar 26 '19 at 18:39
  • The double "1" was the last domino after the 40 others were picked. The "odds" of it being picked at each draw was 1 in 41, but the probability of it being picked last would be much smaller. – KCRJ Mar 26 '19 at 21:19
  • @Vasya No. See my edit. – Ethan Bolker Mar 26 '19 at 21:29
  • @EthanBolker: I think I've got it. The number of ways to pick remaining $41$ dominoes is $41!$ and if we put one-one aside, it's $40!$. So the probability that one-one will remain is $\frac{40!}{41!}=\frac{1}{41}$ – Vasili Mar 27 '19 at 00:57
  • @Vasya Yes, that is one way to see it. You can see it without the formulas since there's no way any particular one of the $41$ is more likely to be last, so the probability for each one must by $1/41$. (If the answer satisfies you, you can accept it (the check mark) and upvote it.) – Ethan Bolker Mar 27 '19 at 01:23