Quote:
Originally Posted by greatwhite
I'm not quite sure what the odds are that one of 4 players has 98 when I have one of the 8's
Nobody else knows either, because nobody can know the probability of 98** hands in the range of each of the four players.
However, we can calculate the probability any one player was dealt a hand containing a nine and an eight, and if that probability is small enough, we can multiply by the number of players to come reasonably close to the probability any one of the four was dealt a hand containing a nine and an eight.
Since Hero has one of the eights, there are 3 missing eights, 4 missing nines, and 38 missing other cards (a total of 45 missing cards after the flop).
Using inclusion/exclusion, the probability one opponent was dealt 98** is
1 - C(45,4)/C(48,4) + 1 - C(44,4)/C48,4) - 1+ C(41,4)/C(48,4) =
1 - 148,995/194,580 + 1 - 135,751/19,4580 - 1 + 101,270/194,580 =
= 0.057067 or about 5.7%.
Alas, that's rather large to get a close approximation by multiplying by the number of opponents. That is, 22.8% is too large. (Maybe 20%, something like that, is a fair ball park approximation).
Anyhow, before there's any action on the second betting round, it's unlikely (probably in the neighborhood of four to one against) one of these four opponents has the nut flopped straight. In other words, roughly one time in five at least one of these four opponents will have been dealt a hand with a nine and an eight. And in a very loose game (which this one seems to be) roughly one time in five someone will actually be holding 98**.
Then on he second betting round when SB bets this flop, Hero raises, BTN cold calls a double bet, and SB calls the raise, if we give SB and BTN credit for having a decent connection with this flop, the possibility one of them flopped the nut straight becomes more plausible. In other words, maybe this is the one time in five someone actually holds 98**.
Buzz