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Festus22
09-26-2003, 08:45 AM
I know the numbers aren't exact but it's what I've been using and I'd like someone to confirm.

If a pair flops, there's roughly an 8% chance per opponent that someone will have made trips (assuming you don't have one of course). Considering what the pair is, I may adjust this down for something like a Q-4-4 flop but the 8% per opponent would be the max %. Close enough?

When considering the likelihood of someone matching one card (like when I'm holding K-K and an ace flops), I use 11% per opponent. Again, is this close enough for government work?

Thanks!

irchans
09-26-2003, 09:53 AM
Festus22,
Those approximates look fine to me. The exact numbers are written below

Probability of an opponent holding 1 or more cards matching the flop pair:

91/1081, 178/1081, 261/1081, 340/1081, 415/1081, 486/1081, 553/1081, 616/1081
~= 0.084, 0.165, 0.241, 0.315, 0.384, 0.450, 0.512, 0.570.

Probability of an opponent holding 1 or more cards matching the top flop card:

135/1081, 3874/16215, 1111/3243, 7076/16215, 563/1081, 1934/3243, 10759/16215, 2344/3243
~= 0.125, 0.239, 0.343, 0.436, 0.521, 0.596, 0.664, 0.723.

thylacine
09-26-2003, 10:59 AM
I considered reverse engineering your numbers to see what they were (I see 1081=C(47,2), etc.) but I couldn't be bothered.

What are they?

irchans
09-26-2003, 04:01 PM
<font class="small">Code:</font><hr /><pre>
For the probability of 1 opponent having a card that matches a pair on the folp

probnomatch
= "prob first card does not match" *
"prob second card does not match"
= (52-5-2)/(52-5) * (52-5-2-1)/(52-5-1)
= 990/1081
_
probmatch = 1 - probnomatch = 1 - 990/1081
= 91/1081
_
_
For the probability of any of 2 opponents having a card that matches a pair on the folp
_
probnomatch
= "prob first card does not match" *
"prob second card does not match"*
"prob third card does not match" *
"prob fourth card does not match"
= (52-5-2)/(52-5) * (52-5-2-1)/(52-5-1)
* (52-5-2-2)/(52-5-2) * (52-5-2-3)/(52-5-3)
= 903/1081
_
probmatch = 1 - probnomatch = 1 - 903/1081
= 178/1081
_
</pre><hr />

The rest are similar.

(You can also use combinations as you mentioned. Combinations give shorter formulas. For example, Prob of one opponent having a card matching one of the paired cards is

1 - c[52-5-2, 2]/c[52-5, 2] = 91/1081.

)

thylacine
09-27-2003, 03:31 PM