common error in poker logic?

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JonRock
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common error in poker logic?

Postby JonRock » Tue Jan 13, 2009 4:27 am UTC

Many poker books, articles, TV commentators etc. often make the following conclusion:

i.e.
the flop in a texas hold'em game comes something like [ A 7 7 ]

now you can calculate a certain probability that your opponent holds one of the two remaining sevens in his two holecards. (lets put aside for that calculation things like the various likelyhoods that players play certain hole cards and just assume all possible holdings are equally likely)

Okay now the turn comes another 7

The books now conclude that it becomes less likely that your opponent held a 7 and I think that is a wrong conclusion.
If your opponent holds a 7 its of course less likely that there will be the case 7 on the turn. But once the 7 has hit the turn you cant tell wether it was more or less likely to come and thus I think you cant conclude that it was less likely your opponent got dealt the 7 at the start.



I dont know much about maths and how you can really prove or disprove this theory and so Id be happy if some of you guys who really know what they are doing could take a look at the problem and help me out.

thanks in advance,
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Re: common error in poker logic?

Postby skeptical scientist » Tue Jan 13, 2009 4:32 am UTC

Bayes' theorem: P(hole card is 7 given turn card is 7)=P(hole card is 7 and turn card is 7)/P(turn card is 7). This is less than P(hole card is 7) since they are negatively correlated, so you are wrong and the poker books are right.
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Re: common error in poker logic?

Postby qinwamascot » Tue Jan 13, 2009 4:38 am UTC

Your problem seems to be that you assume the 7 appears independently of the opponent's card (like in the monty hall problem). The probability that a 7 would be picked is different if your opponent holds one, so it's not the same. The poker books got it right here.
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Re: common error in poker logic?

Postby GreedyAlgorithm » Tue Jan 13, 2009 6:57 am UTC

The turn comes 7, the river comes 7. Now do you think it's less likely your opponent holds a 7?
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Re: common error in poker logic?

Postby SlyReaper » Tue Jan 13, 2009 4:08 pm UTC

GreedyAlgorithm wrote:The turn comes 7, the river comes 7. Now do you think it's less likely your opponent holds a 7?


Slaps down a pair of sevens on the table. Six of a kind! I win! :mrgreen:
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Re: common error in poker logic?

Postby gnuoym » Tue Jan 13, 2009 7:58 pm UTC

SlyReaper wrote:
GreedyAlgorithm wrote:The turn comes 7, the river comes 7. Now do you think it's less likely your opponent holds a 7?


Slaps down a pair of sevens on the table. Six of a kind! I win! :mrgreen:


No... he has the uh, royal sampler.

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Re: common error in poker logic?

Postby Mighty Jalapeno » Tue Jan 13, 2009 8:11 pm UTC

SlyReaper wrote:
GreedyAlgorithm wrote:The turn comes 7, the river comes 7. Now do you think it's less likely your opponent holds a 7?
Slaps down a pair of sevens on the table. Six of a kind! I win! :mrgreen:

By real Texas Hold Them rules, you would then be strung up from an oak tree by your Two of Clubs.

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Re: common error in poker logic?

Postby skeptical scientist » Tue Jan 13, 2009 10:05 pm UTC

SlyReaper wrote:
GreedyAlgorithm wrote:The turn comes 7, the river comes 7. Now do you think it's less likely your opponent holds a 7?


Slaps down a pair of sevens on the table. Six of a kind! I win! :mrgreen:

Silly Reaper, don't you know that Texas hold'em hands are made from exactly five cards?
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Re: common error in poker logic?

Postby Token » Tue Jan 13, 2009 10:13 pm UTC

And there's no allowance in the rules for more than four of a kind. Since there's already an ace on the table, all your extra 7s are going to do is hurt you.
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Re: common error in poker logic?

Postby qinwamascot » Wed Jan 14, 2009 5:51 am UTC

Technically, at least in older variations, 5 of a kind is the highest hand, but is impossible except in wild card games. It doesn't really come into play in texas hold 'em, but it might still be in the rules. 6 of a kind is not a poker hand (although it is impressive that you were dealt 50% more 7s than the deck had)
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Re: common error in poker logic?

Postby The Mad Scientist » Thu Jan 15, 2009 9:40 am UTC

The above posters have said it all, but I may be able to help a bit with your incorrect intuition about probability.

Instead of thinking of the flop, turn, and river coming sequentially, think of them as already determined. If we're playing heads up, I could just deal five board cards face down immediately after dealing the hole cards. Does this help you to realize that whether or not the turn or river card is a 7 has an effect on the likelihood of my holding a 7? Don't think of the turn as coming after the flop, think of it as already determined, but unknown. When we get more information about the cards in the deck, we have to revise our probability estimates.

I hope that helps.

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Re: common error in poker logic?

Postby Mathmagic » Mon Jan 19, 2009 11:01 pm UTC

I figure I'd post this in an existing thread since I don't really feel like creating a brand new topic for this question.

I was playing Texas Hold 'Em over the weekend, and one of my friends managed to get the same set of hole cards three times in a row. By "the same set", I mean it was the same two-card combination of face-values. In this case, he had 8-3 three times in a row, and not necessarily (and actually wasn't) the same suit(s) every time. After the third occurrence of this notably horrid hand, he exclaimed disbelief and facetiously accused the dealer of rigging the deck. I then proceeded to explain that it IS possible to get the same hand three times in a row, despite the unlikeliness of such an event happening.

Now, my probability is a bit rusty (haven't really touched the stuff in school since high school, and have only casually browsed topics on Wikipedia and the board since then), but I think I've worked out the probability (assuming a uniform distribution and perfect randomness) of getting the same set of hole cards three times in a row:

Let E be the event of getting the same set of hole cards three times in a row.

P(E) = P(E1)*P(E1)*P(E1) where E1 is the event of getting a particular set of hole cards.

There are 16 possible combinations of 8-3 you can get (four 8s and four 3s) out of a possible 1326 starting hands (52*51/2). This means that P(E1)=16/1326.

=> P(E) = (P(E1))3 = (16/1326)3 = 512/291434247 ~= 1 in 571428

Which is slightly more likely than hitting a Royal Flush (1 in 649700).

What I'm wondering is what kind of distribution you would get in REALITY, as we all know that the standard shuffling method WITHOUT cutting the deck (which is what happened in this case) isn't perfectly random by any stretch of the imagination. I'm convinced it's actually substantially more likely than the maths lead me to believe.



tl;dr: What kind of distribution would you get with the standard shuffling method WITHOUT cutting the deck? As a follow-up, how would this translate to the probability of being dealt the same set of hole cards (ignoring suits) three times in a row?
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Re: common error in poker logic?

Postby phlip » Tue Jan 20, 2009 12:52 am UTC

Mathmagic wrote:Let E be the event of getting the same set of hole cards three times in a row.

P(E) = P(E1)*P(E1)*P(E1) where E1 is the event of getting a particular set of hole cards.

Not quite... that't the calculation for getting a particular set of hole cards, three times in a row... but really, you'd be just as incredulous if you got any set, three times in a row. The probability of that is approximately P(E1)2 (not exact, 'cause it's different in the case that your recurring hole cards are a pair), which is much more reasonable. Still improbable, but not unfeasable... if you play a lot, you'd expect to see it on occasion.

As for assuming non-uniform randomness in the shuffling... that's a bit out of my league.

As for the OP's question:
Wall of text and numbers spoilered for length.
Spoiler:
The situation: Two-player hold-em. The flop is A 7 7, you don't hold a 7 (we know the position of 5 of the 52 cards). We'll call the event that your opponent holds at least one 7 "O", the event that the turn is a 7 "T".
So:
The chance that the turn is not a 7, your opponent doesn't hold either of the two 7s: P(¬T∧¬O) = (45/47) * (44/46)(43/45) = 946/1081
The chance that the turn is not a 7, your opponent holds either (or both) of the two 7s: P(¬T∧O) = (45/47) * (1 - (44/46)(43/45)) = 89/1081
The chance that the turn is a 7, your opponent doesn't hold the last 7: P(T∧¬O) = (2/47) * (45/46)(44/45) = 44/1081
The chance that the turn is a 7, your opponent holds the last 7: P(T∧O) = (2/47) * (1 - (45/46)(44/45)) = 2/1081

Now, we can figure out the specific probabilities:
The chance that your opponent has a 7, regardless: P(O) = P(O∧T) + P(O∧¬T) = 91/1081.
The chance that your opponent has a 7, given that the turn is a 7: P(O|T) = P(O∧T) / P(T) = P(O∧T) / (P(O∧T) + P(¬O∧T)) = 1/23 (= 47/1081).
The chance that your opponent has a 7, given that the turn is not a 7: P(O|¬T) = P(O∧¬T) / P(¬T) = P(O∧¬T) / (P(O∧¬T) + P(¬O∧¬T)) = 89/1035 (≈ 93/1081).
All three of these we could calculate directly, but it's informative to calculate them indirectly via the probabilities above.

This is basically a long-winded way of doing Bayes' therem, which skep already mentioned. The basic idea is that O and T aren't independent. If they were, then our four probabilities we calculated first would've been nice ratios... P(T∧¬O):P(T∧O) would equal P(¬T∧¬O):(¬T∧O), and ditto for all the other pairs. However, as it happens, the chance of a 7 appearing on the turn is lower if your opponent is holding a 7 (because if your opponent holds a 7, there's one less 7 in the deck that could appear on the turn)... so whether or not a 7 appears in the turn gives you some information about your opponent's hand. Think about the extreme case... say that, somehow, a 7 had a 10% chance of appearing on the turn if your opponent held a 7, but a 90% chance of appearing if your opponent didn't hold a 7. Now, clearly, if a 7 does show up, it makes it much more likely that your opponent isn't holding a 7. Similarly, if a 7 doesn't show up, it makes it much more likely that your opponent is holding a 7. Bayes' theorem is a calculation to say exactly how much more likely it is.

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Re: common error in poker logic?

Postby The Mad Scientist » Tue Jan 20, 2009 2:08 am UTC

Mathmagic wrote:What I'm wondering is what kind of distribution you would get in REALITY, as we all know that the standard shuffling method WITHOUT cutting the deck (which is what happened in this case) isn't perfectly random by any stretch of the imagination. I'm convinced it's actually substantially more likely than the maths lead me to believe.


That sounds plausible, because when you shuffle a deck of cards what you're really doing is shuffling a deck of small groups of cards, with occasional swapping between members of the groups. That is to say, most people don't shuffle with sufficient dexterity to interleave the halves of the deck card by card; instead, as they let the halves of the deck fall onto a surface, they are letting three fall from one half, two from the other, three, four, two, etc.

It's safe to assume that the distribution of "packet" sizes is normal, so we will have a mean packet size of perhaps 4, with some outliers of 1 or 7 or so on occasion. If this weren't the case--if, for instance, the "packet" sizes were always 2, then the probability of being dealt the same hole cards on consecutive deals would dramatically increase. In such a case we could view the deck not as a set of 52 cards, but as a set of 17 sets ("packets") of three cards each. Then your odds of being dealt a particular hand would not be 16/1326, but rather the odds of being dealt the same packet that you were the previous hand: 1/26, a little more than three times greater.

The analysis is, of course, complicated by the fact that the size of a packet will not be constant. Thus, my left hand drops a packet of 3 cards, my right a packet of 2, my left a packet of 1, my right a packet of 3, my left a packet of 2, my right a packet of 1. So, assuming that we view the deck as an ordered 17-tuple of ordered 3-tuples plus one odd card (guessing that the mean of the packet size distribution is 3), the following transformation occurs:

((Ace of Clubs, 2 of Diamonds, King of Spades), (8 of Hearts, 6 of Hearts, Jack of Clubs), (Queen of Diamonds, 9 of Diamonds, 4 of Spades), (7 of Spades, 2 of Spades, King of Hearts))

becomes

((Ace of Clubs, 2 of Diamonds, King of Spades), (8 of Hearts, 6 of Hearts, Queen of Diamonds), (7 of Spades, 2 of Spades, King of Hearts), (9 of Diamonds, 4 of Spades, Jack of Clubs))

My intuition says that whether or not shuffling by hand is truly random depends on whether or not the distribution of packet sizes is both normal and goes from 1 to 52. Since this can't really be the case (no one shuffles by dropping packets of 20 cards as often as they do packets of 3 cards), my guess would be that shuffling by hand does not truly randomize the order of the cards in the deck. This problem seems like it would be amenable to some algebraic techniques, but I am not an algebraist, so rather than try to give an awkward proof for my conjecture, I will leave it to those more suited to the task.

Edit: Edited for coherence. :lol:
Last edited by The Mad Scientist on Tue Jan 20, 2009 5:57 am UTC, edited 1 time in total.

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Re: common error in poker logic?

Postby Mathmagic » Tue Jan 20, 2009 4:43 am UTC

phlip wrote:Not quite... that't the calculation for getting a particular set of hole cards, three times in a row... but really, you'd be just as incredulous if you got any set, three times in a row. The probability of that is approximately P(E1)2 (not exact, 'cause it's different in the case that your recurring hole cards are a pair), which is much more reasonable.
Could you explain how my calculation wasn't correct and how you got the result you think is correct?
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That makes sense to me. I guess since you're just moving packets of cards around (either up or down in the deck, and sometimes back and forth), there's a greater chance of getting the cards you had the previous hand than getting cards that were in the bottom half of the deck the previous round. I'd like to know by what magnitude this increases the odds.
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Re: common error in poker logic?

Postby phlip » Tue Jan 20, 2009 5:16 am UTC

Mathmagic wrote:Could you explain how my calculation wasn't correct and how you got the result you think is correct?

As I said, it's not so much that your calculations were wrong. It's more that it's the right answer but for the wrong question... your calculation was "What's the chance of getting an 8 and a 3 for three consecutive hands", but I'm claiming the right question is "What's the chance of getting the same values for three consecutive hands (regardless of what those values actually are)".

Getting the right question is often the hardest part, in complex problems in probability or statistics...

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Re: common error in poker logic?

Postby The Mad Scientist » Tue Jan 20, 2009 5:53 am UTC

What he said. You always get two hole cards the first time, so your conditional probability should only involve the next two hands.

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Re: common error in poker logic?

Postby Diadem » Tue Jan 20, 2009 4:22 pm UTC

There are 52 cards. 4 of them are 8s. So that is a 4/52 probability of getting dealt an 8 as your first hole card. And a 4/51 probability of being dealt a 3 as the second hole card. But you could also get an 3 first and an 8 second. so total probability is 4/52 * 4/51 * 2 = 32 / 52 / 51 = 8 / 13 / 51 = 8 / 663 ~= 8 / 640 = 1/80

This needs to happen twice in a row (not three, as has been pointed out before). So odds of this happening are roughly 1 / 6400

[edit]This is assuming two different holecards. If they are not different the odds are smaller. Namely 4 / 52 * 3 / 51 (no more * 2), that squared. 4 / 52 * 3 / 51 = 1 / 13 / 17 = 1 / 221. Total chance 1 / (221^2) = 1 / (22^2 * 100 + 1 + 2 * 220 * 1) = 1 / (48400 + 440 + 1) = 1 / 48841

So the odds of getting a pair three times in a row is a lot smaller.
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