A Question about overall odds and pot odds
Let's say in hold 'em I was holding AKs and I flop the flush draw. I would normally calculate that as having 9 outs, out of 47 cards to come. So my probability of hitting the flush on the turn would be 9/47 since 47 is prime and doesn't reduce well...9/47 is about 0.1915, so I would assume that's 19.15 percent. This makes sense, because if you were to shuffle a full deck of 52 and were to bet someone that the first card to come was a spade, you would have a 25% chance of being correct, so for the bet to be even money, I would need to get 4-to-1 odds. In my hold 'em proposition, I would be getting slightly less than 25% because 4 of my suit are already out. This makes sense to me; however, after reading the Pot Odds Chapter from Mr. Sklansky's THE THEORY OF POKER, I found that the odds of making the flush are 4.22-to-1. He says the way to calculate your odds of making this hand on the turn are, "the 52 in the deck minus the five cards in your hand [3 on the board and two in the pocket]. If you are holding four of a suit, nine of the 47 unseen cards will give you a flush and 38 won't. Thus, the odds against making the flush are 38-to-9, which reduces to 4.22-to-1 (Sklanski 36)." His math makes perfect sense, but 1/4.22 is about 0.2370 which means we would improve 23.70 percent of the time. So who is right? I assume Mr. Sklanski is correct, but I ask so that I can fix my mistake and correctly evaluate positive or negative expectations.
|
07-09-2005, 01:55 PM
Linear Mode
