Quote:
Originally Posted by octopus
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
xray vision gives the clearest explanation:
Quote:
Another way of phrasing this is to explicitly use conditional probability. You are asking for P[both dice are 2at 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 2at least one die is 2]=(1/36)/(11/36)=1/11.
