Bayes Theorem

[TStoneMBD]

Could someone give me a quick overview of what the Bayes Theorem is without using complex mathematical equations? Thanks.

[spentrent]

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without using complex mathematical equations?

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Okay, so it's not complex but it does have letters and equals signs...

http://www.seas.upenn.edu/~cse140/Lectures/Bayes.html

[jtr]

The lecture that the previous poster linked to looks good.

But if you want a really short description, how about this: Bayes's Theorem is about how to incorporate new information with old information in a probabilistic world.

For example, you pick up a coin and start tossing it. At first, you have no reason to believe that the coin is anything other than fair. Background knowledge, or prior information, suggests we'd expect 50% heads and 50% tails. If the first few tosses of the coin all came up heads, we wouldn't be very concerned. We'd likely see this as just variance, and wouldn't be immediately questioning our original assumption that it was a fair coin.

But what if the first ten throws were all heads? Nervous yet? The first 50? The first 100? Bayes's Theorem gives us a way of dealing with these questions exactly -- basically quantifying how much confidence we have in the original "fair coin" assumption given the balance of new evidence. That's about it really.

[axioma]

very basically, its used when determining the prob. of something, given that we know something else.

[stigmata]

Could somebody give me a specific example of Bayes theorom being usefully applied to Poker, to illustrate the point?

I have heard it discussed in a vague kind of way on these forums, but never seen it applied statistically to a situation.

[ZeeBee]

You should have a poke around in the probability forum.

Bayes theorem can be used to help understand the likelihood of a player holding certain cards based on his/her actions and the underlying probabilities of being dealt those cards.

So for example, imagine your opponent opens the pot with a raise and you have a pocket kings. You are worried he has a pocket aces and want to know what the chances are.

You remember that the odds of getting dealt aces are 220:1 - are these the chances that he has aces?

Well, no - you have more information - you know that he has open-raised. Using Bayes theorem, based on your knowledge (and smart guesswork) of what range of hands he would open-raise with you can adjust the baseline rate of 220:1 to get
a better estimate of the chances of him having Aces.

You don't do this calculation live (certainly I couldn't) - but you can run through examples "offline" to feed your brain to give it a good feel for different situations. You then adjust in live situations as appropriate.

ZB

PS - You would also adjust for the fact that you know you have 2 Kings, making it more unlikely that he has a King in his hand.

PPS Bayes theorem is oftne confused with the impler application of conditional probabilities.

[jtr]

Here's an example.

NL game. It's folded to us. We complete in the SB. (I know, I know.) Prior probability of the BB holding AA: 0.45%. Then the BB pushes all-in. We now need to revise our estimate of the BB holding AA.

Let's say we know that the probability of the BB pushing if he holds AA is 80%. The probability of him pushing if he holds anything other than AA is (let's say) 5%. Now we can calculate the revised probability that he holds AA given that we've just seen him push. Just as an exercise, take a guess now as to what you think the probability is.

Breaking it down by frequencies can make Bayesian logic easier. So consider that the above scenario happens 10,000 times. 45 times the BB will hold AA. Of those 45 times, he will push on 80% of them, which makes 36 pushes and 9 non-pushes with AA.

The remaining 9,955 times the BB will hold something other than AA. Of those hands, he will push 5% of the time, which makes 498 pushes (497.75 to be exact). And the rest of the time, on about 9457 non-AA hands, he won't push.

Putting those numbers together: over the 10,000 cases, we expect to see 534 pushes. But of those, only 36 will be with AA. So our probability that the BB holds AA, given that he just pushed, is now 36/534 = 6.74%.

That's Bayes's theorem, in a nutshell.

[mmcd]

A simple example:

You open raise UTG w/KQ and and absolute rock 3-bets you. His 3-betting range against your UTG raise is exactly AA, KK, QQ. The flop comes K Q x. The appearance of the K and the Q on the flop makes it more likely that your opponent has AA than KK or QQ (this is essentially Bayes Theorem) so you can figure you are best and play hand relatively fast until he plays back at you hard enough to force you to reasses the situation.

As applied to poker, Bayes Theorem basically means you can use the cards on board (or other factors such as potentially dead cards indicated by a couple of cold-calls from semi-solid players) to adjust the probability of your opponent holding a specific hand within the range of hands you put him on originally.

[TStoneMBD]

thank you very much.

[Display Name]

Your post makes complete sense to me; it has to be right. However, when I try to apply Bayes Formula to the facts I a completely different result. I note that this does not factor in at all the 5% of the times that a player would push all in with something other than AA.

Here is my calculation:

P (A | B) = P (B | A) P(A) / P(B)

A = AA
B = All in
P(A) = .45%
P(B) = 99.55%
P(B | A) = 80%

P(A | B) = 80% x .45% / 99.55%
P(A | B) = .36%


P (A | B) = P (B | A) P(A) / P(B)

A = other than AA
B = All in
P(A) = 99.55%
P(B) = .45%
P(B | A) = 5%

P(A | B) = 5% x 99.55% / .45%
P(A | B) = 1106%

No wonder I suck at poker. Help!

[jtr]

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Here is my calculation:

P (A | B) = P (B | A) P(A) / P(B)

A = AA
B = All in
P(A) = .45%
P(B) = 99.55%
P(B | A) = 80%



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OK, so here's a small problem. It's fine to label the state of holding AA as event A. And then the act of moving all-in can be event B, OK. Then the prior probability of holding AA, P(A), is indeed 0.0045 or 0.45%. But the complement of that is the prior probability of the bad guy holding anything other than AA, usually expressed as P(~A) or "probability of not-A". That should be P(~A) = 99.55%. Instead you have this as a value for P(B), the prior probability of an all-in move.

That prior, P(B) is something we need to calculate here as an intermediate step (see below).

So then the conditional you've expressed is OK, probability of pushing if he holds aces -- P (B|A) -- is indeed 80%. But this needs to be accompanied by the other conditional, the one you know you've left out. That's P(B|~A), the probability of him pushing if he has a hand other than aces.

P(B|~A) should be 5%.

To repeat your formula: P (A | B) = P (B | A) P(A) / P(B).

This is fine. But first we need to know what P(B) is -- the probability of seeing a push. Well, we're going to see two kinds of pushes. One when he has aces, and one when he doesn't. The probability of seeing a push with aces is the probability of aces themselves, multiplied by the conditional probability of the guy pushing when he has them. That covers the first kind of push, and the notation is P(A) * P(B|A).

We also care about the non-AA pushes. Similarly to the above, that would be P(~A) * P(B|~A) -- the probability of a non-AA hand times the probability of him pushing when he has such a hand.

So the working for p(B) = ( P(A) * P(B|A) ) + ( P(~A) * P(B|~A) ) is (0.0045 * 0.8) + (0.9955 * 0.05) = 0.053375 or about 5%.

That gives us what we need to plug stuff correctly into your original formula.

We want to know the probability of aces given the push, which is P(A|B). This is indeed P(B|A) * P(A) / P(B), or 0.8 * 0.0045 / 0.053375 = 0.06744 or 6.74%.

So you weren't far off, you just need to be clear about what A, B, ~A, and ~B are before applying the formula you've got.

[SenecaJim]

try this

http://arnoldkling.com/apstats/bayes.html

[Display Name]

Thank you so much!

My calculation lead me to conclude that there was an 1100% chance that villian had something other than AA. That felt just a tad high to me /images/graemlins/grin.gif

[Cyrus]

You are invited to calculate the odds of only 1 coin coming up heads in the next roll when a few coins will get flipped. You ask the flipper how many coins he will be flipping and he informs you "3 coins".

You calculate those odds to be precisely 37.5% but right before you get a chance to announce your calculations, the flipper states that 2 of the coins were found to be biased so he will be flipping only 1 coin.

You immediately change your prediction about 1 coin coming up heads to 50%.

The change in the probability is Bayes at work.

[barongreenback]

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My calculation lead me to conclude that there was an 1100% chance that villian had something other than AA.

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That's online poker for you. /images/graemlins/smile.gif

[BottlesOf]

What does the vertical line mean, and where is it on my keyboard?

[Huskiez]

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What does the vertical line mean, and where is it on my keyboard?

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It means "given" or "such that", and it's SHIFT + \ (which is right above the enter key).

P(A|B) means probability event A occurs given event B occurred.

[BottlesOf]

That's what I thought it meant. Thanks.