odds: pair v two overcards

[primed]

first post! sorry if this question is stupid, and i should be able to work it out myself but cannot remember how: it is often quoted that e.g. ak v 99, pre flop, is "about a 50-50". logically, it seems to me that ak is effectively looking for one of 6 out of the 50 cards he has not seen, with 99 being safe with the remaining 44, save for the combinations leading to a straight or flush for ak. can anyone tell me how this is calculated? thanks

[GaF]

Re: odds: pair v two overcards The odds of hitting 6 cards from 50 with just one card is 12% (6/50) However if you know that your opponent has 99, then there are only 48 unknown cards and it is 6/48=12.5% The odds of NOT hitting one of your 6 cards with one card is 42/48=87.5% (the inverse of hitting, so 100%-12.5%=87.5%). If you want to measure the odds of not hitting with 5 consecutive cards, then it is ALMOST 87.5% ^ 5 (i.e. 87.5%x87.5%x87.5%x87.5%x87.5%)=51.29% Why almost? Well after you miss with the first card, then there are only 47 cards left in the deck... so the actual chance of missing with 5 consecutive cards is (42/48)x(41/47)x(40/46)x(39/45)x(38/44)=49.7%. If the chance of missing with 5 consecutlive cards is 49.7%, then the chance of hitting one (or more) A or K with 5 consecutive cards is 100%-49.7%=50.3%. Of course, this isn't the chance of winning the hand - you can quite possibly hit the A or K and lose, or you could miss the A and K and win!! If you dont have it, I recommend you download pokerstove - free software that will give you the odds in different scenarios of different cards winning. That tells us that in this scenario (assuming the A,K,J and J are all different suits), then the JJ is a 57.3% favourite v 42.7% for the AKo - the AKo is actually a long way behind for a "coin flip"!!!

[msaban]

Re: odds: pair v two overcards Gaf, that was an excellent response :clap even I learnt something there :tongue2 Thanks Mike