Guess the formula use - Gambling and Probability

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Guess the formula use

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08-30-2023 , 12:25 PM

statmanhal

Pooh-Bah

Join Date: Jan 2009 Posts: 4,958

I was researching a specific Hold'em situation and came up with the following equation after some algebra.

Probability = (8*R^2 - 38*R+ 46.5)/1225

Can you guess what it represents (I'm assuming it is correct)?

Hint: R is hero's top rank in an unpaired hand

Spoiler:

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09-05-2023 , 03:22 PM

statmanhal

Pooh-Bah

Join Date: Jan 2009 Posts: 4,958

In case anyone is interested, here are the results for all Hero high ranks against 1, 2 and 3 opponents. Simulation was used for the 2 and 3 opponent cases.

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09-06-2023 , 10:57 AM

heehaww

Pooh-Bah

Join Date: Aug 2011 Posts: 5,056

Your numerator is slightly off but appears to be fixed by replacing 46.5 with 45. (Plus it wouldn't make sense for there to be half-combinations.) Also, the formula assumes zero information about Villain's hand, unlike what you said in the spoiler.

Two general formulas are:

C(4r-9, 2n) / C(50, 2n)

C(50-2n, 59-4r) / C(50, 59-4r)

where n=#villains and r is from 3 to 14 (since 2 is impossible with an unpaired hand).

With n=1:

Code:

 0.0024489795918367346
 0.017142857142857144
 0.044897959183673466
 0.08571428571428572
 0.13959183673469389
 0.20653061224489797
 0.28653061224489795
 0.3795918367346939
 0.4857142857142857
 0.6048979591836735
 0.7371428571428571
 0.8824489795918368

With n=2:

Code:

 0.0
 0.00015197568389057752
 0.0014329135909683021
 0.005927051671732523
 0.01683022145028224
 0.038449848024316106
 0.07620495006513243
 0.13662613981762917
 0.22735562310030394
 0.357147199305254
 0.5358662613981763
 0.7744897959183673

With n=3:

Code:

 0.0
 4.405092286683406e-7
 2.907360909211048e-5
 0.00031496409849786355
 0.001707413770318488
 0.00635258358662614
 0.01862787668258793
 0.04633408219902207
 0.10214527994361482
 0.2053165058807982
 0.3836491784502885
 0.6757142857142857

Quote

09-06-2023 , 02:22 PM

statmanhal

Pooh-Bah

Join Date: Jan 2009 Posts: 4,958

The relatively small differences may be due to different assumptions/definitions. It might be due to your reference to “half-combinations.”

I assumed a half win against one opponent occurs when both players have the same top rank and no pairs. Then the better hand as I defined it is the one with the higher kicker. Without any kicker data, I assumed hero has the better kicker half the time. Against 2 or 3 players, there may be more than one tie so I used hero’s win count to be 1/(1+qty. ties) which, admittingly, is not totally accurate but is of very small occurrence probability.

I agree that no assumption is made about hero's hand. The statement was not well written

Of course, a more accurate assessment would be to look at hero’s exact hand - high rank and kicker, much too complicated for me.

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09-18-2023 , 05:58 PM

heehaww

Pooh-Bah

Join Date: Aug 2011 Posts: 5,056

Oh ok, my formula is the probability that Hero has the highest rank, which doesn't include the chance of tying and out-kicking. It also doesn't exclude the chance of Villain having a lower pocket pair, which I'm guessing yours does (since those hands beat Hero's). But what I mean by half-combinations is that your denominator is C(50,2) and yet your numerator isn't an integer.

When heads-up, I believe the chance of having the highest rank OR out-kicking Villain is:

[C(4r-9, 2) + 6(r-3)] / C(50,2)

Code:

0.0024489795918367346
0.022040816326530613
0.054693877551020405
0.10040816326530612
0.15918367346938775
0.2310204081632653
0.3159183673469388
0.4138775510204082
0.5248979591836734
0.6489795918367347
0.7861224489795918
0.9363265306122449

If we exclude the chance of Villain having a lower pocket pair, it's actually the same as my original formula because we subtract 6(r-3) pocket pair combos, leaving us with a numerator of C(4r-9, 2).

More villains would make for some busywork and I don't doubt your sims.

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