Poker probability math is incorrect

[6471849653]

I see it in books; they say e.g. that an ace on the flop will decrease the chance that the opponent has an ace. That's not true as it decreases only the chance that the opponent has that specific ace, the reason being that that ace on the flop was dealt AFTER players' hole cards.

That mistake comes from probability math that doesn't necessarily take the order of cards or whatever dealt into calculations and so calculates only a certain kind of an average like the rounded number of 0.5 that could be anything between 0.45 and 0.5499.

I think the cards dealt are similarly more or less non-random when it comes to online poker. One doesn't figure the averages to effect that much, and it's more likely it's less random offline. But having experienced it and knowing the probability math being somewhat incorrect, it's possible, and in parts it can effect more as if the next card coming is calculated from 4 cards rather than three card, it's a huge difference. The mistake in probability math and what one experiences online fits well together.

I am aware of studies about the probabilities of specific cards or numbers coming on specific streets, and it's even enough though some sites just barely. But considering the same mistake being on every street and number, it doesn't prove the deal is random.

[Tom1975]

I like tacos.

[binions]

Question 1:

If you have one ace in your hand (Ax), and the flop comes with one Ace (Axx) heads up, how many possible hands could your foe have with at least 1 Ace in it?

Question 2

If you have KK, and the flop comes with 1 ace (Axx) heads up, how many possible hands could your foe have that have at least 1 Ace in it?

I think you will find that statitistically, it is 50% more likely your foe has an Ace in scenario 2 than scenario 1.

Here's something to get you started. When we see 5 cards (like after the flop), there are 47 unseen cards. This means that the universe of our foe's holdings, on the flop, are 47*46/2 or 1081 possible hands.

Further, there are 16 ways to make AK and 6 ways to make AA, unless there is an A on the flop. Then there is 3 ways to make AA, and 12 ways to make AK (unless there is a K on the flop - then there are 9 ways to make AK). Of course, if there is an A on the flop and an A in our hands, then there is only 1 possible AA hand and 8 possible AK hands (unless a K is on the flop, then 6 possible AK hands).

[bigpooch]

Something mentioned by Barry Greenstein and others is also
enlightening: if you are playing hold'em in a ring game and
several players fold to you on the button, the deck is a
bit skewed towards aces, so not having an ace is a bit worse
than if you were playing three-handed. The effect can be
estimated and is dependent on the ranges of folding hands of
the players that have already folded.

Every combination is equally likely for the player under the
gun, but once players have acted by folding, combinations of
certain hands have different "weight", with those containing
an ace being slightly more frequent than "normal".

[filsteal]

OP,

Go learn about conditional probability.

[AaronBrown]

In one sense you are correct. The other player either has an Ace or he doesn't. What happens on the flop cannot affect that.

What does change is your information about what's in the other player's hand. If four Aces show up on the board, you know he has no Aces.

The reasoning is a bit trickier if one Ace shows up on the flop. Suppose before the flop, you thought he could only have AA or KK, and they were equally likely. Assuming you have neither an A or a K, if he has AA, the probability of getting an A on the flop is 0.12. If he has KK, the probability of getting an A on the flop is 0.23. That means an A on the flop is evidence for KK (of course, it must be weighed with all other evidence).

Suppose 16 hands are dealt this way, 8 in which he has AA and 8 in which he has KK. On average, there will be an A on the flop in 3 of these hands, 1 with AA and 2 with KK. So when you see the A, you figure the chance of AA is now 1/3 instead of 1/2.

The fact that certain cards are more likely on certain streets does not mean the deal is nonrandom. Players are more likely to play high cards than low ones. So low cards are slightly more likely to show up on the board of contested hands. If the deck is shuffled so the board will be AAAAK, it's less likely that players will have the hands to call preflop.

[DiceyPlay]

This is somewhat analogous to the let's make a deal game show:

Suppose you have 3 doors. Behind two of the doors there is a donkey and behind the third door there is a Corvette. You choose one door hoping to get the Vette. The game show host then reveals a Donkey behind one of the other doors. The game show host then offers you the option of keeping your original choice or switching to the other unopened door. What do you do?

You switch - the probabilities have shifted because you now have new information. Your initial guess has a 1/3 change of being the Vette. That doesn't change. But the probability of the Vette being behind the other door is now 2/3.

The same phenomena is at work in the OP. The probability of any one card being an ace before you see any of the cards is 1/13. But once you see an ace (and two other cards) on the flop, the probability that any of the other unseen cards is an ace is 3/47 (assuming you hold no ace) and 3/47 < 1/13.

So what's right in both cases? For the game show, most people want to assign a 50-50 chance after one of the donkeys is revealed. To me that's clearly wrong and what was initially stated above is correct. I don't think it's nearly as clear for what OP is stating.

What do the stat and probability guys think?

[AaronBrown]

This problem is clearer than the Let's Make a Deal problem. Let's Make a Deal depends on the rule the host uses to determine which door to open (or whether to open a door). In this case, we know the flop is dealt randomly.

Suppose we change this to Let's Make a Deal. Two cards are dealt face down, and we win if at least one of them in an Ace. We win 14.93% of the time.

Before we look at the cards, three cards are dealt face up from the rest of the deck. One of them is an Ace. We now revise our probability estimate, and think we have only 11.99% chance of winning.

If the dealer offers to pick up all five cards, shuffle, and deal two cards again, we say "yes." But if he offers to deal two more cards from the top of the remaining deck, we don't care, it doesn't change the odds.

If instead of dealing the three face up cards at random the dealer goes through the deck and picks out one Ace and two non-Aces, seeing those cards makes no difference to us at all. We still think we have a 14.93% chance of winning if we keep our cards (also if the deck is restored and reshuffled). But now we think we have only 11.99% chance of winning if the dealer shuffles the remaining 47 cards and deals two from the top.

[WCGRider]

DiceyPlay, i am by no means an expert but Im fairly certain i disagree with your logic. There is no point in switching, as it has not changed the otehr to door to 2/3. Rather, that percentage is distributed over both remaining doors, changing them both to 1/2. I cant possibly believe that the other door is now more likely to have the vette, just doesnt work like that.

[gaming_mouse]

[ QUOTE ]
DiceyPlay, i am by no means an expert but Im fairly certain i disagree with your logic. There is no point in switching, as it has not changed the otehr to door to 2/3. Rather, that percentage is distributed over both remaining doors, changing them both to 1/2. I cant possibly believe that the other door is now more likely to have the vette, just doesnt work like that.

[/ QUOTE ]

yes, it does:

http://en.wikipedia.org/wiki/Monty_Hall_problem