Poker Question - What are the odds?
- Thread starter hapaboy
- Start date
Because I am using all permutations of three pairs (Part B) divided all permutations of possible three two cards hands (Part A). You have to consider the total number of ways for both of your sets. If I am using 52*51*50*49*48*47 for Part A I have to include all possible ways the pairs can come down for Part B.
You reduced Part A by 6*5*4*3*2*1 which only gives you only the number of 6 card combinations and says nothing about the possible number of unique PAIRS. It is like figuring the number of combinations for a giant 6 card hand. If you want to use combinations instead for both parts, you need to change Part A to reflect this, looking like
Part B: 6*13*6*12*6*11
Part A: (52*51/2)(50*49/2)(48*47/2)
This way is more confusing I think which is why I did it the first way in my response to you. The easiest way to think of it is imagine you only dealt 1 hand, what are the chances of getting a pair?
You can do it both ways, possible permutations of getting any pair?
13*6*2 = 156
Possible permutations of dealing two cards?
52*51 = 2652, so 156/2652 = 1/17
Or using combinations,
13*6 = 78
and
52*51/2 combinations of two card hands because each will be counted twice.
= 1/ 17
I'm doing the same thing but with 3 pairs instead.
Agree with the first part, I think the second is (46*45*44)/6*4*2 though like I mentioned above and I'll quote below
In my original post I think I made a calculation wrong and (46*45*44)/6*4*2 should equal 1897.5 instead of 3795. But it looks like I did it right in my final answer which should still be 1 in 9.4 million.
