Re: Probability of a higher pocket pair
If you notice on Brians page there are three columns and to be exact there should be more columns but the result would be negligible so he didn't include them all .
If you add the columns you should get a number close to 42% .
ie , .3269+ 0.08186 + 0.01186 =~ 42% .
Also , remember that my formula is a very good approximation since it assumes that the events are independent . This isn't entirely true but it is damn close to being independent .
In general , you should use this :
If you hold pocket pair X at an n handed table , then the answer is approximately
1- [1225-a)/1225]^n where a is the number of pairs higher than your pocket pair x . . For instance , if you hold pocket 7's , then there are 7*4c2 = 42 combos .
1- [1225-42]/1225]^9 =~26.94%
Now go to Brians page and add the three columns from the row with pocket 7's and you get 0.2380+0.03218+.0023=~ 27.248%
The answers are pretty close which is why I prefer using the "independence assumption" formula to give me the answers to these questions .
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