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Old 10-10-2007, 12:05 PM
kozyanski kozyanski is offline
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Join Date: Oct 2007
Posts: 2
Default Re: Suited Connectors, Implied Odds, and You (Theory/Math)

Chances of hitting gutshot (and only gutshot) on flop:

Let's say you are holding A2. Flops that give you gutshot should contain 3 & 4, 3 & 5 or 4 & 5. If the flop contains 3 & 4 it shouldn't contain 5 and if it contains 3 & 5 it shouldn't contain 4 but if it contains 4 & 5 it could contain also 3. In that way we avoid double counting. And you also don't want to hit a pair (at least not yet - we'll look at it later) so you don't fancy flops with A or 2. All the possible ways of getting 3 & 4 on flop are:

<ul type="square">[*]3 4 x[*]3 x 4[*]4 3 x[*]4 x 3
[*]x 3 4[*]x 4 3[/list]
x in the first four flops shouldn't be 3 or 4 just to avoid double counting.

Flop | cards you don't want* | number of cards that you don't want

3, 4 A, 2, 5; 3, 4 3*4 + 2*3 = 18
3, 5 A, 2, 4; 3, 5 3*4 + 2*3 = 18
4, 5 A, 2; 4, 5 2*4 + 2*3 = 14

Formula:

p(n) = 2*(4/50 * 4/49 * (48 - n)/48) + 2*(4/50 * (49 - n)/49 * 4/48) + 2*((50-n)/50 * 4/49 * 4/48)

where n stands for the number of cards that you don't want.

Therefore the probability for hitting a gutshot (and only gutshot) with A2 is:

pg = 2*p(18) + p(14) =~ 7,92%


Probability of flopping a pair or better (including gutshot) is:

pp = 2*(3/50 + 3/49 + 3/48) =~ 36,74%

Probability p of hitting gutshot or better is:

p = pp + pg

In this case p =~ 45%


* How to determine the cards you don't want?
First you should write down all of the two cards that help you, starting with lowest two. If you hold for example 98 those cards would be

5 &amp; 6, 5 &amp; 7, 6 &amp; 7, 6 &amp; T, 7 &amp; T, 7 &amp; J, T &amp; J, T &amp; Q, J &amp; Q

You obviously don't want 8 or 9 becouse it would give you a pair which is already counted in pp.
Now start with 5 &amp; 6 and look at all of the other "good two cards" containing either 5 or 6, those are 5 &amp; 7, 6 &amp; 7 and 6 &amp; T. 7 and T are cards you don't want.
5 &amp; 7: The list of other good two cards that contain 5 or 7 is: 5 &amp; 6, 6 &amp; 7, 7 &amp; T and 7 &amp; J, bad cards are 6, T and J.
6 &amp; 7: 5 &amp; 6, 5 &amp; 7, 6 &amp; T, 7 &amp; T, 7 &amp; J but the bad cards are only T and J. 5 is not a bad card becouse in the previous two steps we eliminated flops containing 5, 6 and 7 so we should coun't it now.

And so on

It's really not so complicated as it seems.

Hope it helps!
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