Approximating Multiple Opponents

[bigchips]

If I am facing two opponents, and my hand will beat opponent #1's likely hands 50% of the time, and opponent #2's likely hands 70% of the time, what is the equation for approximating how often my hand will beat both opponent #1 and #2?

If I assume two opponents at 50%, I can deduce that I will win 33% of the time. I *guess* I arrive at that approximation by: (.5 + .5)/3 = 33%. So by way of this equation my anwer would be (.5 + .7)/3 = 40%. Is this answer a good approximation?

I realize that the way to definitively answer the question is to have pokerstove face off against two opponents. But if time is of the essence, is the equation above a good way to quickly and accurately approximate my win% vs multiple opponents or is there a better way?

[gaming_mouse]

[ QUOTE ]
If I am facing two opponents, and my hand will beat opponent #1's likely hands 50% of the time, and opponent #2's likely hands 70% of the time, what is the equation for approximating how often my hand will beat both opponent #1 and #2?

[/ QUOTE ]

.5*.7 = 35%

[ QUOTE ]
If I assume two opponents at 50%, I can deduce that I will win 33% of the time. I *guess* I arrive at that approximation by: (.5 + .5)/3 = 33%.

[/ QUOTE ]

Whoa! Try: .5*.5 = 25%.

HTH,
gm

[bigchips]

Yes I agree that multiplying the two is how my stats professor would suggest doing this. Logically if you pit 3 equivalent hands against each other, you should get a 33% split.

Pokerstove produces the following results:
9c8d 33.181%
9d8h 32.769%
9h8s 34.051%

So the common sense approach seems to be holding up in this instance. I'm trying to understand why this is the case and why multiplying .5 x .5 is not the right answer.

[Stork]

But multiplying .5 by .5 is the right answer. Your method does not count the times where you beat one opponents hand but then lose to the other; that is still a loss. Either you lose to both opponents, you lose to opponent A but beat opponent B, you beat opponent A but lose to opponent B, or you beat opponent A and beat opponent B. Each of those situations are equally likely to occur, but you only win the last one.

In your pokerstove example, those hands weren't 50/50 to beat each other because there was an overwhelming chance to tie. Try comparing a pocketpair to two overcard hands.
From TwoDimes:
4c 4d 360083 26.27
Jc Qd 616219 44.95
5s 6h 387436 28.26

The pocket pair, which is about 50/50 to each hand individually, wins about 25% of the time.

[bigchips]

Yes that explanation makes sense with regards to the ties. I hadn't noticed that so that was probably a poor example by me.

<font class="small">Code:</font><hr /><pre>
equity (%) win (%) / tie (%)

Hand 1: 33.1808 % [ 00.01 00.32 ] { 9c8d }
Hand 2: 32.7686 % [ 00.01 00.32 ] { 9d8h }
Hand 3: 34.0506 % [ 00.02 00.32 ] { 9h8s }
</pre><hr />

And as a matter of practicality, I'm thinking more in terms of ranges of hands. Let's say you have 3 players. Players 1 and 2 only play pairs AA down to 99. Player 3 only plays suited connecters AKs to T9s.

Player 1 vs. Player 2 is a 50/50 proposition.

Player 1 or Player 2 vs. Player 3 is as follows:

<font class="small">Code:</font><hr /><pre> equity (%) win (%) / tie (%)

Hand 1: 68.2245 % [ 00.68 00.00 ] { AA-99 }
Hand 2: 31.7755 % [ 00.31 00.00 ] { AKs, KQs, QJs, JTs, T9s }

</pre><hr />

If we multiply the win % to come up with the projected win% for player 1 we get (.5 x .68) = 34% expected win rate in a three way hand.

When I run the actual 3 way matchup in Pokerstove here is what I get:

<font class="small">Code:</font><hr /><pre> equity (%) win (%) / tie (%)

Hand 1: 39.1855 % [ 00.38 00.01 ] { AA-99 }
Hand 2: 39.1855 % [ 00.38 00.01 ] { AA-99 }
Hand 3: 21.6291 % [ 00.21 00.00 ] { AKs, KQs, QJs, JTs, T9s }
</pre><hr />

So for a range (albeit small) of hands, my simple approximation (.68 + .5)/3 = 39% seems to be giving a superior estimate than (.68 x .5) = 34%.

I'm not sure why this is the case. Can anyone help reconcile this for me? Ties don't seem to be a significant factor in this example. I would run a larger sample of hands 3-way against one another but my computer is so slow it takes forever.

[Scorpion Man]

OK guys -- this is my first ever post on this site and have not had stats in many years but...i think the reasoning here so far is too simplistic (not in an insulting way...just literally too simple).

the key is the INTERACTION of the three hands together. The problem is you have a guy playing 2 high cards that are VERY frequently being dominated by pairs that match some or BOTH of hte high cards he is playing. So often, he loses. The third player (by that i mean player number TWO) is pulling two cards from the deck that often would have paired number 3 in a heads up scenario.


I would be interested to see what happened is you made the pairs that people play different so that they dont overlap so much. the numbers will change.

[gaming_mouse]

bigchips,

I answered the question given your assumptions. IF you have a 50% to beat hand 1 and a 50% chance to beat hand 2, then (assuming approximate independence, which will not always hold) you have a 25% chance of beating both.

It seems that was not your real question, and that you had a particular hand in mind. Reading your follow up posts, I am still not really clear what your after. Try to clarify.

gm

[bigchips]

Gaming mouse: Honestly I'm just trying to understand why the numbers don't match up. I'm going to reference some numbers from the following two sites:
http://www.jazbo.com/poker/huholdem.html (heads up)
http://1-169pockethands.hitstrade.us/ (hand vs 10 players)

AA heads up: 0.8520371
AA on 10 player table 31%

If your formula holds then 0.8520371^9 = .256 should be the amount that AA wins 10 handed. However if the #s on the site above are accurate, AA actually wins 31%.

0.8520371^9 is only an estimate because it doesn't take into account the intererelationship of the various players' hands, as was pointed out by Scorpion Man, and the example posted by Stork, where a 50% heads up hand ended up winning 45% 3 way due to how the cards matched up.

22 heads up: 0.5033402 ^9 = .00207
22 vs 10 handed table: 11.86%

No I think clearly that multiplying the heads up win percentages simply doesn't accurately approximate how a particular hand will perform against multiple opponents.
My goal is to come up with a simple, quick approximation for multiple players when I know what my win % is heads up vs a single opponent's likely starting cards.

[gaming_mouse]

[ QUOTE ]
No I think clearly that multiplying the heads up win percentages simply doesn't accurately approximate how a particular hand will perform against multiple opponents.


[/ QUOTE ]

That is correct, and I didn't mean to imply that it did. I thought you were asking a theoretical question. It would have been clearer if you had stated the way you planned to use the info.

[ QUOTE ]
My goal is to come up with a simple, quick approximation for multiple players when I know what my win % is heads up vs a single opponent's likely starting cards.

[/ QUOTE ]

As the examples you and others have posted demonstrate, you cannot do this in general. It depends too heavily on how the hands match up. Some hands will be almost independent (in which case multiplying will give you a good approximation) while others will be highly dependent (in which case it won't).

Thinking about this stuff is a good exercise, but I would also say that it's not very practical. Trying to estimate the chance you are beaten with an exact number just isn't realistic. It also encourages you to ignore more important but less easily quantified information, like your opponent's betting patterns and your reads on them.

gm

[bigchips]

[ QUOTE ]
As the examples you and others have posted demonstrate, you cannot do this in general. It depends too heavily on how the hands match up. Some hands will be almost independent (in which case multiplying will give you a good approximation) while others will be highly dependent (in which case it won't).

Thinking about this stuff is a good exercise, but I would also say that it's not very practical. Trying to estimate the chance you are beaten with an exact number just isn't realistic. It also encourages you to ignore more important but less easily quantified information, like your opponent's betting patterns and your reads on them.

gm

[/ QUOTE ]

Thanks for the info. Not the answer I was hoping for, but useful in focusing my attention to the other matters you mention. Thanks for taking the time to read and reply to my questions.

[pzhon]

A simpler counterexample would be to consider AK vs. two low pairs. AK wins under 50% against either, but almost 40% against both.

[Carl_William]

Dear Gamey Mouse

Your are correct -- at least I got same answer as you did of 35%. I used a brute force counting scheme (numerical enumeration). Three players: A, B, &amp; C. A beats B 70% of the time; &amp; A beats C 30% of the time. 100 permutations covers all possible cases.

Evidently you solved it the easy way. I did it using brute force on an excel spreadsheet -- what a waste if time my way was....