# chances of playing

• Sep 27th 2013, 10:16 AM
parex
chances of playing
I was entertaining myself trying to figure out the following: If my son plays soccer in a team of 11 players, and at any point there are 8 kids playing, and 3 in the bench, what are the probabilities that he'll be playing at any point in time, assuming equal playing time for all players.

If I look at it from the point of view of the players left out, I can reason: For the first player to be seated out he has 1/11 chances of being "chosen". If he's not the one out the first time around, he can still be the second player seated with a probability of 1/10. The third player chosen to rest can turn out to be him with a probability of 1/9. The sum of these probabilities (... OR ...) is 0.294.

Now, if I reason with sample spaces there are 11 choose 8 combinations: 165. And the combinations that include him in the 8 playing combo are 10 choose 7: 120. So the probability is 120 / 165 = 0.7272. And the probability of sitting down is, 0.2727.

Why are the results different with either reasoning?
• Sep 27th 2013, 03:17 PM
parex
Re: chances of playing
... Happily, I got to my own post before anyone...

What's going on is that I am not comparing the same scenarios: In the first method, the knowledge that he is not picked up to sit on the bench on the first or second rounds, increases the probability of being picked up in the third round. It is the Monty Hall set up.
• Sep 27th 2013, 04:14 PM
Shakarri
Re: chances of playing
Edit. Oops I didn't notice you had resolved your problem. It was a good question, it made me think.