I've had a couple of goes at this without much luck so far.
I'd go about it by saying that the probability should be 1 minus the sum of the probabilities of all branches where a previous player hits player 10's name. Is that correct?
As I understand it, all branches of the tree should add up to 1 but only if they are mutually exclusive... if there's no crossover between events. So the problem I'm finding is how to define the branches of a player choosing. In the simple case, player 1 has a 1/10 chance of hitting player 10's name. But I can't figure out how to phrase this additional choice he has if he hits his own name, or if it even makes any difference.
There's a 1/10 chance that he will hit his own name and a 9/10 chance that he will not. If he does not, he does not choose again, but it's implicit in that he did not that the name chosen is not his own. Conversely, if he gets to choose again it's only out of the other 9 names that are not his own, but in either case the net effect after the choice is that player 1's name stays in the pot and another name is removed. So I wonder if it even matters for this particular question - the probability of a player hitting player 10's name - about this second choice which appears to collapse down to the same thing.
So rob, how would you define the first choice... the probability of player 1 hitting player 10's name? Would you say a simple 1/10 or something like (1/10 * 1/9) + (9/10 * 1/9 or 1/10... I'm not sure what that's supposed to be given that I'm trying to add an implicit choice here that isn't really there just to make sure that the two branches do not overlap). It's very confusing but a fun puzzle
I'd go about it by saying that the probability should be 1 minus the sum of the probabilities of all branches where a previous player hits player 10's name. Is that correct?
As I understand it, all branches of the tree should add up to 1 but only if they are mutually exclusive... if there's no crossover between events. So the problem I'm finding is how to define the branches of a player choosing. In the simple case, player 1 has a 1/10 chance of hitting player 10's name. But I can't figure out how to phrase this additional choice he has if he hits his own name, or if it even makes any difference.
There's a 1/10 chance that he will hit his own name and a 9/10 chance that he will not. If he does not, he does not choose again, but it's implicit in that he did not that the name chosen is not his own. Conversely, if he gets to choose again it's only out of the other 9 names that are not his own, but in either case the net effect after the choice is that player 1's name stays in the pot and another name is removed. So I wonder if it even matters for this particular question - the probability of a player hitting player 10's name - about this second choice which appears to collapse down to the same thing.
So rob, how would you define the first choice... the probability of player 1 hitting player 10's name? Would you say a simple 1/10 or something like (1/10 * 1/9) + (9/10 * 1/9 or 1/10... I'm not sure what that's supposed to be given that I'm trying to add an implicit choice here that isn't really there just to make sure that the two branches do not overlap). It's very confusing but a fun puzzle
