RE: Probability question: names in hats
March 14, 2016 at 11:22 am
(This post was last modified: March 14, 2016 at 11:23 am by Whateverist.)
(March 14, 2016 at 11:17 am)robvalue Wrote: That was the very first answer I came up with when I tried this Sadly, it's wrong.
You're on the right track, but the probability for the second person depends on whether or not the first guy took his name already. If he took it, his probability is 8/9 again. If it didn't get taken, it's 7/8. Similarly, each pick will depend on whether his own one has been taken.
I can't get my head around how you account for all the combinations as they can all have different probabilities.
Are you sure? I think it only makes sense to calculate the probability for the second person as 7/8 since there are only 8 names remaining after the first person removed one of the names other than his own. The second person is only drawing from a hat with nine names in it and he too is unable to keep his own. So his probability of choosing a name after the first person has drawn and kept a name other than person ten's should be 7/8 I think.