Question on calculating probabilities with multiple unknown players

Hello:

I have a technical question on poker probabilities in Texas Hold-em:

- Suppose you can calculate the probability of winning (and losing and tieing) with a specified two pokcet cards, against one unknown opponent, at some stage in a Texas Hold'em game. Call the probability of winning 'pw'.

- Suppose instead of one unknown opponent, you now have 'n' (n > 1).

- Would a reasonable estimate of the probability of your pocket winning against all n opponents be: (pw)^n?

- Statistically, each opponents' hand is not quite independent as we consider multiple wins (against each opponent), but I suspect the difference is so minor in
ever possible case of hand and exposed (or not exposed) flop, turn, river, etc., that you can safely just use this
simplification.

Thoughts from the math-inclined out there? Safe in general, or maybe there are hand up-card combos where this may not be accurate?

[AaronBrown]

No, it's not safe. There are a lot of minor reasons why not, like the one you mention, but the biggest problem is your own hand.

For an extreme example, suppose you hold suited six/seven. If you get a flush or a straight, you're likely to win against every hand. Your probability against N hands is not much worse than against 1. If the board has two or more sixes and sevens in it, you're also likely to win against any other hand likely to be in the pot. But if neither of these things happen, you're very unlikely to win against any hand. Again, your probability of beating N hands isn't much worse than your probability of beating 1.

At the other extreme, holding say King/eight unsuited, you're likely to be in the middle of N hands a lot. Your chance of beating N hands could be much worse than your chance of beating 1 raised to the Nth power.

Thank you Aaron for replying on this math question... I appreciate your taking the time to provide this excellent feedback.

I see what you're saying, but in both cases the probabality 'pw' is such that the Nth power can still leave you drastically in favor, even with high 'N'.

In the straight or flush example, suppose your chance of winning against one hand is .95. With 5 oppponents, (.95)^5 = 0.77... still very high, as expected.

In the other example, suppose your change of beating one hand (like k/8) is .5 (50%... in the 'middle'). with 5 opponents, .5^5 = 0.03. Much lower, as expected.

One way to resolve this objectively is to write a monte carlo simulation with multiple unknown hands, and compare to the one-unknown-hand probability (raised) approximation formula. I may explore this some more, across a broad range of test pockets, like the ones you posed.

I tend to agree that it may be unsafe with some hands, but I'd like an objective mathematical case/analysis or some sim data to really prove the point one way or the other.

Thanks again.

[AaronBrown]

Fortunately, it's already been done, and Andrew Prock (http://www.pokerstove.com/) gives it away for nothing.

Suited 7/6 has 0.4536 chance of beating one random hand. 0.4536^9 = 0.0008. But suited 7/6 has 0.1219 chance of beating 9 random hands.

Unsuited K/8 has 0.5602 against a random hand, better than suited 7/6. 0.5602^9 = 0.0054, 7 times the predicted chance of beating 9 hands as suited 7/6 if things were independent. But only only does it have less chance of actually beating 9 random hands (0.0942) than suited 7/6, it has less than the 0.1 average chance. It other words, unsuited K/8 better than a random hand head to head, but worse than a random hand if there are nine random hands. That's what I mean about it being likely to end up in the middle.

Thanks again Aaron for the input and data! Poker Stove is exactly what the doctor ordered, and the perfect tool to solve this. I had started to look at the PokerSource (open source) project prior to this post as a way to look into this more. I see that PokerStove is in fact using that code (or a non GPL precursor) as base.

Well, the question is answered... the exponential approximation is a bust idea! Now, the reason I thought of it was to avoid a 'monte carlo' simulation to solve these exact cases (with lots of players). Using PokerStove, I was only able to replicate your 9 player example (with no board) using it's own 'monte carlo' simulator, rather than an exhaustive approach. I presume you used the same button/option in your sim with 9 players?

This all makes sense now, since with 9 players and pre-flip, the enumeration of hands (even using PokerStove/PokerSource's improved 'zealot' model) is impossible (sim immediately failed for me with errors. Worked fine with one player. Monte Carlo with 9 players works).

So, my actual original goal was to come up with a way to avoid 'empirical simulation approximation' (or otherwise called Monte Carlo) to calculate actual odds with many players. I think PokerStove has shown that this may be impractical pre-flop and/or with many players, due to huge number of exhaustive permutations required.

Back to the drawing board.

[AaronBrown]

There are two simple models. One is independence, which you started with. The other is thinking of n independent random variables from the same distribution. The odds of the first one being the biggest is 1/n, not (1/2)^(n-1).

For an intermediate case, suppose I have a hand that beats a random hand with probability p. I model my hand as a random draw from a uniform distribution from p-x to p+x for some x, then compare it to n-1 hands drawn from uniform 0,1. The odds of my hand being the highest number is:

[(p+x)^n - (p-x)^n]/2nx

For small x, this approaches p^(n-1). So if there's little uncertainty about your hand, as is the case after all the cards are dealt, the independence approximation works pretty well. For the largest possible x (p or 1-p depending on whether p is more or less than 0.5) you get:

(2p)^(n-1)/n for p<=0.5 and
[1 - (2p-1)^n]/[2*n*(1-p)] for p>=.5

So, for example, if you have 0.4 chance of beating one hand, your chance of beating 8 hands could be as high as 32*0.4^8 instead of 0.4^8.

I don't know if this formula will help you or not.

Thanks again Aaron... this is very enlightening. I'm afraid the application of the statistics you used is a bit beyond me, but I see what you're doing in general and do like it (ack... why didn't I take 3rd year stats!). I'm not sure how I would apply this approach in general, since I can't make any assumptions about the required 'x' range, but I may be misunderstanding your derivation. Plus, it's still only an upper end (or bounding range) on the probability (if I understand correctly), leaving the actual probability uncertain. Hence, possibly still the need for a monte carlo sim (blah).

I suppose the independent probability, and the 'max x' probability from your equation derivation, provides a bounding box for the actual probability for n players.

Thanks again Aaron... your expertise in this area is very helpful.

[AaronBrown]

You're welcome.

I'm just suggesting you could put in some value for x and get a quick estimate. With a bit of fooling around, you could probably come up with a value that works pretty well for most hands.