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Old 10-06-2016, 01:41 PM
Colophon Colophon is offline
Join Date: Sep 2002
Location: Hampshire, England
Posts: 13,372
Originally Posted by octopus View Post
It's 1/36. There is only one event that results in a 6 being on the other die of a pair when one 6 is removed and that is double 6s. Double 6s is 1/36 chance.
But you are ignoring the fact that 25 out of those 36 chances won't have a six at all, so they can be eliminated from the game.

Edit: that page linked by x-ray vision gives the clearest explanation:

Another way of phrasing this is to explicitly use conditional probability. You are asking for P[both dice are 2|at least one die is 2]. Since the first event implies the second, this probability is equal to P[both dice are 2]/P[at least one die is 2]. Since P[both dice are 2]=1/36 and P[at least one die is 2]=11/36, we have P[both dice are 2|at least one die is 2]=(1/36)/(11/36)=1/11.

Last edited by Colophon; 10-06-2016 at 01:45 PM.