Probability question: names in hats

[robvalue]

Whateverist: Did you agree with my 3 player model, as I explained with all the diagrams whittling the probability tree down? I have confirmed the answer is correct, just wondering if I explained it properly.

Emjay: The messing about with the person's own name seems weird when calculating probabilities, but you can actually just work it out as if that name isn't there. But this drops the total, affecting the bottom of the fraction.

For example, player 1 has 10 numbers to choose from. But we can ignore his own one. So he actually has just 9, which are 2 up to 10

Each of these has an equal probability, so each one is 1/9

It can be shown explicitly that this is correct, even including the number 1. The probability he picks a particular number can be calculated by him getting it right away, or else picking his own number and then picking the desired one second time. Since these are exclusive events, we can add the probabilities. So to get a specific number (say 3 for example)

Prob(gets 3) = prob(gets 3 on first pick) + prob(gets own number) * prob(gets 3 on second pick)

On the second pick, his own name is no longer there, so we're down to 9 possibilities.

=1/10 + 1/10 * 1/9

So 1/10 of the time he gets a 3 right away; a further 1/10 of the time he pulls his own number, and then 1/9 of the time following this he'll get a 3 as the second pick.

=1/10 + 1/90

=9/90 + 1/90

=10/90

=1/9

So after all that, it comes out as exactly the same as if we had just ignored his number completely. So on a probability tree, people's own picks need not be included, but you must account for the number of possibilities being lessened by removing it.

We can see how player 1s pick affects player 2. If player 1 pulls number 2, then player 2 is left with 1, 3, 4, ... , 10. This is 9 numbers, any of which he can pick. So the probability of picking a particular on is 1/9 again. This is a bit disconcerting as it's the same probability as for player 1 picking a number!

If player 1 takes any other number than 2, player 2 then has his own number plus 8 others left. But we have to exclude his own number. So there's really only 8. The probability he pulls any of the remaining numbers is 1/8, and there will be 8 such branches.

Aractus: Did you end up with 5/36 as the probability for 4 players? I calculated that and it agreed with my coding experiment. I wasn't too sure what your final answer was.

I found there was actually only two ways player 4 ended up with his own number: 2 3 1 4 and 3 1 2 4

All other branches either pick a 4 premuturely or else a player gets his own number.

Probabilities 1/18 and 1/12 respectively; giving a total of 5/36