Quote:
Originally Posted by RedJoker
Wrong forum, you should have posted this in the probability forum.
If the probability of them having a better hand individually is 28% then to find the probability of either of them having a better hand you find the probability that they both missed and take it away from 1 (or 100%). So:
1 - (0.72)(0.72) = 1 - 0.5184 = 48.16%
The reason you can't add 0.28 to 0.28 is that it double counts the probability that they both hit. So you can also work that out and subtract it:
0.28 + 0.28 - (0.28)(0.28) = .28 + 0.28 - 0.0784 = 48.16%
For three players it's:
1 - (0.72)(0.72)(0.72) = 1 - 0.373248 = 62.68%
This is close but not 100% correct. The probablility of one of them hitting is not independent of whether the other hit. If one of them hit the ace it's less likely that the other one did as well, so in reality they hit a little bit more than 48.16%.