Quote:
Originally Posted by junwagh
Lol you're right that 1/3760 is not less than 1/270725. Dunno what I was thinking. But your final answer is still wrong. I dunno what you did in your calculation but if you multiply the probability of both having aces hu by the different number of combos for players to have aces in a nine handed game that should equal the final probability, no? If so there are 9Choose2 combos (36) and 36*(4/52)*(3/51)*(2/50)*(1/49) does not equal 1/270725.
Damn it! Now I have to answer you. You can use your approach and you will get 72 combinations and guess what? 72*3,760=~270,725. My approach is to consider each pair of aces separately.
The chances of the first pair of aces is 52*51/6/2=1/221, but since there are 9 players who can get the first pair of aces, you have to divide that by 9 which gives you 1/24.55 (rounded to 2 decimal places).
Now consider the chances of the second pair of aces. That is 2/50*1/49=1/1,225, but since there are now only 8 players who can get the second pair of aces, you have to divide that by 8 which gives you 153.12 (rounded to two decimal places). So, you simply multiply 1/24.55*1/153.12 and you get the chances of any 2 of the 9 players getting AA in the same hand as ~1/3,760, (I'm sure you've noticed that divide by 9*divide by 8=divide by 72, so to speak).
When considering only 2 players (either 2 specific players at the 9 handed game or 2 players heads up), you simply multiply 4/52*3/51*2/50*1/49 which does give you 1/270,725. There is no need to divide by either 9 or 8 because there are only 2 players who can get the pairs of aces, so there is only one player who can get each pair.
Seriously though, you have to quit doing this. Otherwise, God forbid, there is actually a chance (a really small chance), that we might end up friends. Permit me to make a suggestion. Go find that Mason guy and insult him. He stuck me as kind of arrogant.