Right then,
The odds of being dealt a pocket pair is 17:1
I can't quite decide what the calculation should be for the odds of two players being dealt a pocket pair, but have found a calculation for a 6 handed table (this was 7 handed), so am going to use that.
0 players dealt a pair is simply:
(16/17)^6 = 0.6951 for 6-handed
and
(16/17)^9 = 0.5795 for 9-handed
1 player dealt a pair is:
6*(1/17)*(16/17)^5 = 0.2606 for 6-handed
and
9*(1/17)*(16/17)^8 = 0.3296 for 9-handed
So 2 or more people being dealt a pair will roughly be:
1 - 0.6951 - 0.2606 = 0.0443 for 6-handed (1 in every 22.6 deals)
and
1 - 0.5795 - 0.3296 = 0.0909 for 9-handed (1 in every 11.0 deals)
So, the odds of two players getting dealt a pocket pair is 22:1
The odds of a third player then being dealt one each of those pocket pair cards is 564:1 (2/48 x 2/47)
And finally, the odds of the remaining case pocket pair cards being flopped is 15,180:1 (91,080 combinations of 46 cards being flopped, with 6 combinations of the two case cards being flopped)
Which would make the total odds 22 x 564 x 15,180 : 1 or 188,353,440 : 1
Seems a big number. Have I missed something obvious ? (Should I be ignoring the first odds, as the probabilty of the second and third part are already based on the first bit having happened ?)
The odds of being dealt a pocket pair is 17:1
I can't quite decide what the calculation should be for the odds of two players being dealt a pocket pair, but have found a calculation for a 6 handed table (this was 7 handed), so am going to use that.
0 players dealt a pair is simply:
(16/17)^6 = 0.6951 for 6-handed
and
(16/17)^9 = 0.5795 for 9-handed
1 player dealt a pair is:
6*(1/17)*(16/17)^5 = 0.2606 for 6-handed
and
9*(1/17)*(16/17)^8 = 0.3296 for 9-handed
So 2 or more people being dealt a pair will roughly be:
1 - 0.6951 - 0.2606 = 0.0443 for 6-handed (1 in every 22.6 deals)
and
1 - 0.5795 - 0.3296 = 0.0909 for 9-handed (1 in every 11.0 deals)
So, the odds of two players getting dealt a pocket pair is 22:1
The odds of a third player then being dealt one each of those pocket pair cards is 564:1 (2/48 x 2/47)
And finally, the odds of the remaining case pocket pair cards being flopped is 15,180:1 (91,080 combinations of 46 cards being flopped, with 6 combinations of the two case cards being flopped)
Which would make the total odds 22 x 564 x 15,180 : 1 or 188,353,440 : 1
Seems a big number. Have I missed something obvious ? (Should I be ignoring the first odds, as the probabilty of the second and third part are already based on the first bit having happened ?)
My head is hurting here trying to work it out and i don't think i'm capable of doing the maths. However I think there are lots of permutations that bring the odds down of it happening.
Things like
Fiive players are dealt pocket pairs then that gives x number of 2 card combinations which could then show up on the flop.
so AA,KK,QQ,JJ,1010 would give ak aq aj a10 kq kj k10 qj q10 j10 then flops containing 2 of AKQJ10
I could just be sprouting shite for all i know though