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#1
10-11-2007, 11:34 PM
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Poker Math
I've been getting a lot more into poker theory and from my notes I have the following formula for working out the percent to win when you have 10 outs to the situational nuts. Unfortunately I didn't include the workings behind the fomula and I was hoping someone could help me with this?
10/47 + 37/47 * 10/46 ~ 38.4% 
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#2
10-11-2007, 11:57 PM
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Re: Poker Math
I like using the rule of 2/4. If you have 10 outs after the flop, you have 10*4 ~ 40% chance. After the turn, 10*2~ 20% chance. This isn't exact, but a way to quickly calculate approximate chance of hitting your hand vs. pot odds/implied odds, etc...
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#3
10-12-2007, 12:35 AM
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Re: Poker Math
[ QUOTE ]
Unfortunately I didn't include the workings behind the fomula and I was hoping someone could help me with this? 10/47 + 37/47 * 10/46 ~ 38.4% [/ QUOTE ] Not surprisingly, this exact calculation has come up recently. You can split up the chance of hitting a 10 out draw into the events that you hit on the first card, and that you miss on the first card but hit on the second card. Since these events are exclusive (they can't both happen), the probability that either happens is the sum of their probabilities. The 10/47 is the probability that you hit on the first card, since you have 10 outs in 47 unseen cards. The 37/47 is the probability that you miss, and 10/46 is the conditional probability that you hit on the second card given that first card misses. The probability that you miss on the first card and hit on the second is the product of these. Keep in mind that hitting the obvious draw may not be your only way to win, and may not be enough to win.
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