Re: Pot Odds Question
- From: "David Nicoson" <bigdavex@xxxxxxxxx>
- Date: 19 Jan 2006 11:28:20 -0800
muzicman789@xxxxxxxxx wrote:
> How exactly did you get the "1.6 :1 chance of improving" ?
I looked it up in a chart. But we can compute it.
outs = 10
unknown cards = 52 - 2 - 3 = 47
non-outs = 47 - 10 = 37
The probability of missing on both the turn and the river =
(probability of missing on the turn) x (probabability of missing on the
river)
(37/47) x (36/46) = 0.616
Convert to odds notation by dividing the chance of losing by the chance
of winning:
(0.616) / (1 - 0.616) = 1.6
Someone else commented on the 2x outs rule for the percentage. That
works because 47 x 2 is about 100. And the chance of hitting on both
the turn and river is small, so adding the probabilities of hitting on
the turn and river isn't so far off. So to estimate:
10 outs x 4 = 40%.
60% / 40% = 1.5:1
That's certainly close enough for most purposes. This method does get
out of whack when the number of outs gets bigger, though, because the
chance of hitting twice becomes significant. So fudge your chances
back down for large numbers of outs.
.
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