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Give your answers in the form '1 in x' or 'y to 1'.
1)a) In a 3-handed game, given the fact that all 3 people have cards of the same suit, what is the chance of 3 of that suit flopping?
You don't have to account for any straight flush possibilities.
b) Does anything change if it was a 9-handed game?
2) If 3 people have pocket pairs in a 3-handed game (none shared) what is the chance of all 3 people flopping a set? -
Im pretty rubbish at poker, but ok at maths, so:
1a) 7 remaining cards in the suit, 46 unknown cards -- 7*6*5/(46*45*44)= 0.0023
1 in about 430
b) Unless you have any more information about the other hands then nothing changes.
2) 6*4*2/(46*45*44) = 0.0005
about 1 in 1900 -
^^ He pretty much nailed it.
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Doing it like that orders the cards as well - initially there are 6 possible cards to make a set, then once one of them is done there are 4 more for the remaining 2 sets and finally 2 cards to make the last set.
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When I answered, I looked at the possibilty of this happening at the start of the hand. Flush over flush over flush in a 3 handed game. Basically 12/51*11/50*10/49*9/48*8/47*7/46*6/45*5/44 which comes to 19958400/2567428617600. This is approximately 1,312,698 to 1 in a 3 handed game.
For the set over set in a 3 handed game. 1/17 (i wont use exact # for odds of a pocket pair but it is approximately that) * 1/17*1/17*1/8 (once again apprimate for hitting a set) *1/8*1/8 which is 1 in 2,515,456. Once again the 3 handed stuff is easier to do.










