Quote:
Originally Posted by junwagh
this doesnt seem correct. Say you were dealt more than 48880 hands. According to your method as you play more hands the chances of getting dealt aces becomes less as you surpass the 48880 mark.
I look at the problem as a binomial distribution which i think is really the correct way to do it. The binomial distribution tells you the chances of gettin n success in x trials given the probability of success is p. that seems like the simplest way to do this.
According to my method, the more hands you play, the more likely it is to happen. In fact, when you are dealt hand #48,882, the probability becomes 1.0, since you now have had 48,881 chances of something that has a probability of 1/48,881, and (1/48,841)/48,841=1. However, the probability of being dealt aces twice in a row says nothing about when it will occur, simply that the more hands you deal (The Law Of Large Numbers, as distinguished from The Law Of Large Lousters
), the closer and closer your actual results will approach the mathematical expression 1/48,881. You could just as easily get AA on the first two hands you are dealt as you could go 48,881 hands without even getting AA once.
Consider a simpler problem, flipping a coin. The probability of heads is 1/2, so mathematically, the probability of heads is 1.0 every two flips. But that is only a mathematical expression of what you can expect to happen over time and doesn't demand that the coin alternate between heads and tails. However, the more times you flip the coin, the more your results will get closer and closer to the probability of heads as 1/2. I see your point, but again, if your chances of getting AA twice in a row are 1/48,881, then, at a ten player table, aren't the chances that one of the players will get AA twice in a row ten times as great, or (1/48,881)/10?
Last edited by LargeLouster; 04-23-2009 at 05:25 PM.