09-06-2001, 03:20 PM
In his "Omaha Holdem Poker" book, Bob wrote that with two cards to come, a twenty-way straight draw will make a straight over 70% of the time. Curious about how much over 70% a 20-outer would hit, I tried calculating the probability myself. I approached the problem by first calculating the probability of missing on both the turn and the river.
p(miss on turn) = (45-20)/45
p(miss on river) = (44-20)/44
p(miss turn & river) = (45-20)/45 * (44-20)/44 = .3030
Therefore, p(hitting) = 1 - .3030 = 69.7%
I'm pretty sure I did this right, but would appreciate confirmation. My math skills are a bit rusty.
p(miss on turn) = (45-20)/45
p(miss on river) = (44-20)/44
p(miss turn & river) = (45-20)/45 * (44-20)/44 = .3030
Therefore, p(hitting) = 1 - .3030 = 69.7%
I'm pretty sure I did this right, but would appreciate confirmation. My math skills are a bit rusty.