## Sunday, September 18, 2011

A reader questioned my approach to calculating the probability of getting quads on the board with the first four cards, in a comment on this post. His objection, in short, is that the cards held by the other players eliminate a bunch of ranks from being possible quads on the board. Since all four of some rank must still be left in the dealer's hand after the pitch, whatever cards are held by other players exclude those ranks from being eligible.

I started to write a reply comment, but it was getting too long. Besides, it's an objection that comes up once in a while when I do a probability calculation, so I thought it would be worth making a separate post about it. It's something that gave me headaches the first several times I pondered poker probability questions. It took me a while to see straight about it, so I'm not surprised that others find themselves confused.

Of course our calculation would change if we knew what other cards were. But if you're going to go down that road, then you're going to drive yourself crazy.

From the point of view of an omniscient being, there is no "probability" about this situation at all. The cards have been shuffled, the order of the deck is set, and we will either get quads on the board on the turn or we won't. It is predestined to either happen or not happen, so to speak of probability makes no sense. In fact, that is true from our non-omniscient perspective, too. It will either happen or it won't, and that determination was made when the deck was shuffled and cut. Asking how likely it is that quads will hit the board is sort of a confusing question, because it has already been settled whether it will happen. We just don't know the answer yet.

Think of it this way: Instead of asking, "What is the probability of getting quads on the board after ten hold'em hands have been dealt out?", instead ask it this way: "Of all the times that in a ten-handed game I know only my own two down cards (and they are not a pocket pair) how often, on average, will I see quads hit the board with the first four cards?" That is actually the question that I answered.

If you start adding assumptions or information about what other cards we know, then you're changing the question, so of course you'll get a different answer, all the way up to the ultimate case where you know the location of every one of the 52 cards, at which point you can ask whether quads will come, and the answer is either a definite yes or no.

Here's another way to think about it. Suppose that I'm alone in my apartment. I shuffle the deck, deal myself two cards, but leave the other 50 in the deck. I look down and see that I've dealt myself a 5 and a king. I take the next 18 cards and move them from the top of the deck to the bottom, without looking at any. Now, if I deal out the next four cards face up, what is the probability that they will be all of the same rank?

Of course, it doesn't really matter if I move 18 cards from the top of the deck to the bottom of the deck. Nor would it matter if, instead of doing that, I dealt those 18 cards two to each of nine imaginary players around my table. I could even burn them in my fireplace, and it wouldn't change the math. I could actually have nine friends (well, if I had nine friends, which is hard for a misanthrope) each pick up their two cards and look at them, not telling me what they are. None of it matters. The answer to the question remains the same, if the question is asked the way I suggested in the fifth paragraph above. In fact, I could deal myself two cards, throw away all but the last four cards in the deck, and ask about the probability that they will constitute quads. Or I could deal myself two cards, then pick four cards at random from the 50 remaining and ask the same question. It's all exactly the same, in terms of the math. It doesn't matter which four cards we pick, or in what order, and it doesn't matter what happens to any of the other 46 cards, as long as we don't know what they are. As soon as we know what they are, then we have changed the question we're asking, and the math must change accordingly.

(I'm feeling very tempted to go off on a tangent about Herr Schrodinger and his poor cat, but I think I'll refrain.)

This issue comes up with every probability calculation in poker, though we usually don't think about it. If you have four to a flush on the turn and ask the probability of making your flush on the river, you don't really stop to think that that is, in one way of looking at it, a nonsensical question--because the card that is destined to hit the board last is not going to change its spots; it's either one of your suit or it isn't. The more logical way of framing the question is, "Of all the times that I am in this situation, what percentage of them will, over the long haul, result in making my flush?" That way you're not assigning a "probability" to an outcome that has already been determined concretely (though we don't know which it is yet).

It's also akin to asking, in pre-ultrasound days, whether a baby about to be born is a boy or a girl. It's not 50% boy and 50% girl. It's one or the other. Yet we say, loosely, that the probability of it being a boy is 50%, and the probability of it being a girl is 50%. In reality, it's a very different question than asking, before I flip a fair coin, the probability that it will land tails, because it still could go either way. But the baby is either male or female already. We just don't know which. Our use of the word "probability" for both questions really disguises the fact that the situations are actually different. They sort of look the same superficially, but in one case the outcome has been set already and we just don't know what it is, while in the other the outcome still could be either one.

In spite of those two different realities, the math is identical. If it helps, you can phrase the question analogous to how you might ask about the outcome of the coin flip that has not yet occurred. To set the stage, we specify that after the deck is shuffled and cut, we will deal out ten standard two-card hold'em hands, then discard the top card, put out a flop, discard the next card, and put out fourth street. So the first four cards on the board will be the ones that were at positions 22, 23, 24, and 26 in the cut deck. Now here's our pre-shuffle question: "If I thoroughly shuffle this deck of cards, then do a cut at a random point, what is the probability that the resulting order will have four cards of the same rank at positions 22, 23, 24, and 26?" That's not quite the same question for which I did the math previously, because of the extra condition in the prior example that I knew two ranks that would not work--but it's close. And it gets around the mental stumbling block of asking about the "probability" of an event that has already either occurred or not occurred.

If you want to make the pre-shuffle question truly identical to the one I worked through with the first post, then do this: Remove a 10 and a 4 from a standard deck (representing the two hole cards I had when the situation actually came up). Now ask the question I posed in the previous paragraph. Bingo--the math I worked out gives you the answer.

Whew! That was long-winded. But I hope it clarifies why the answer I gave was correct mathematically, even though it quietly skipped over the problem of speaking of the probability that something will occur, when the real answer is simply either "yes" or "no."

Pete said...

So you agree then its 50-50 either it happens or it doesn't.

Actually my favorite probability question is the one that gets raised after something has happened.

A player flops a Royal and someone asks what is the probability of that happening ... well sir, discounting philosophical questions about what is nature of reality .... I would say that the probability is 100%. After all we just saw it happen.......

Anonymous said...

Part 1 / 3

I'm the commenter that raised the issue on the other post that spawned this reply. First, thank you for taking the time to write such a detailed, thoughtful reply.

I'm actually pretty well versed in probability and understand the concept of unknown cards and how they (generally) don't affect the odds of a certain thing occurring.

I do appreciate your comments re: once the deck is set, this thing either does or does not occur. Given a single deck, the question is binary: it either happens or it doesn't. That was a helpful perspective when thinking about this situation a bit more.

I also understand your point that making other assumptions changes the question (and, thus, the answer). And I think that, as I hopefully demonstrate below, the assumptions I make are valid. In short, the question as you interpret it does not have enough conditions on it to make your answer valid in the context of a game with its constraints. Or, stating it another way, the conditions I place on the situation (e.g. excluding more ranks than you do) cannot be removed. They are true, just as true as you excluding your 10 4.

So, I still think the original calculation is incorrect. It does not take any foreknowledge of the other player's specific hands to recognize that certain ranks are excluded. Also, since you switched to 10-handed from 8-handed for your post, I will as well. Even if you disagree with my reasoning, I think you would agree that this would exclude another rank (2 other players each holding the same pocket pair) from the possible quads on the board ranks in the way I'm calculating things. Also, let's forget any specific examples of cards that others hold. My logic does not require one to know specific holdings.

Also, for the calculations below, rather than a generic objection or another example, if you think these calculations are incorrect, please dissect them directly. I ask this in the spirit of learning.

It does not take any knowledge of the specific cards any player holds to know with 100% certainty that the highest number of ranks eligible to make quads on the board is 7 (10 handed). This is derived from the fact that 5 ranks are excluded from consideration due to being in players' hands. The 6th rank is excluded due to the burn card. It is simply impossible to make quads on the board with 11 ranks (12 if you are holding a pocket pair). This is not the same type of calculation as the odds of a 5th flush card showing up on the turn. It is true that the flush might be less likely than we typically calculate (it could in fact be physically impossible to make a flush on a given board if all of the other cards of that suit are distributed among the other players and not in the dealer's hand) but it is possible. In this case, it is not even theoretically possible for 11 ranks to make quads on the board.

Sticking with the original example, we can exclude the 10 and the 4 from possible quads. But we can also exclude 4 other ranks. It's simply not possible to make quads with 11 ranks. So we can't use 11 in the calculation. The fact that we don't know which specific ranks to exclude doesn't mean we can't exclude any.

So using your original calculation, amended with this fact, we would have the following:

50 choose 3 = 19,600. Of those, 7 ranks (not 11) could make trips. 28 flops contain 3 cards of the same rank. So we have (7 * 4) / 19,600 = 0.0014. Still 1/47 for the 4th card = 0.021. Overall odds = 3.04 x 10^-5, or 1 in 32,900.

I could possibly see another angle here, since we don't know which 7 ranks are valid possibilities. But are we in agreement that 11 ranks are not possible? Just as you've excluded the two in your hand from consideration, I can exclude 4 other ranks -- without seeing any other cards.

Anonymous said...

Part 2 / 3

But, let's also pretend that we haven't even looked at our own cards yet. By the logic of your argument, we could use 13 ranks in the calculation of possible quads (since you have no knowledge of any of the cards). But I think you'd agree that we know we can exclude at least one rank from the possibilities (if you hold a pocket pair) or 2 (if you hold the 10 and the 4 as indeed you did). As an aside, this is actually a great case of Shrodinger's cat: the act of observing the situation does change it. If we flip over a pocket pair, one rank is excluded. Otherwise, two ranks are excluded. I presume absent knowledge of whether we hold a pocket pair or not, the odds (on average, as you state in your post as a better way of thinking about this) would be a blended average based on the likelihood of us holding a pocket pair or a non pair. That calculation is for another time.

So the chances in our current situation would be (13 * 4) / 19,600 = 0.0027, with the 4th still 0.021. Overall odds = 5.64 x 10^-5. But as soon as you flip over your hand of 10 4, the odds change to (by your calculation) 4.78 x 10^-5.

Why can we not also exclude the ranks from the other players' hands? I say we must. They change the odds just as you having 10 4 change the odds. Either you having 10 4 does not change the odds (meaning we use all 13 ranks) or it does (in which case we must include whatever knowledge we have about the other players' cards as well). And we know that we can exclude (at least) 4 other ranks.

I would also point out that at least one of your other examples, in an attempt to explain the situation, actually asks a different question. For example, the situation you describe where you deal yourself two cards, then throw away all but the bottom 4 cards in the deck, is actually answering the question of how many times will the bottom 4 cards in a deck be of the same rank after removing 2 cards from the top of the deck. Which is the same as asking how many times the bottom 4 cards on the deck will be of the same rank without taking two cards off the top (since removing the top two cards doesn't touch the bottom 4). So in that example, it doesn't matter what the other players hold -- but it also doesn't matter what you hold. You can't have it both ways. It's a different question.

I will grant that if you know the top two cards, you know that the quads on the bottom of the deck aren't the two cards you hold. But the odds of that situation are actually calculated completely differently: something like 52 choose 4 (270,725) and then of those, 13 can be quads. So 13 / 270,725 = 4.80 x 10^-5. Which is almost exactly what you originally calculated. What that means I'm not sure, other than it seems to indicate you're answering a very similar question.

As to your example about shuffling a standard deck, then identifying the cards at positions 22, 23, 24, and 26, that's an excellent way to think about it. But that is exactly identical to shuffling a deck and checking the bottom 4 cards. I will fully stipulate that selecting, a priori, any 4 positions within a 52 card deck, then shuffling it, then finding if those 4 are the same, are 100% identical. But that's not the same as what was asked (even when you include removing a 10 and a 4).

Anonymous said...

Part 3 / 3

The next step, removing a 10 and a 4, then shuffling, then looking where the flop/turn would be, is helpful. We are certain that the quads can't be a 10's or 4's. But just as you removed your (known) 10 4 from the deck, you must remove 18 other (random) cards, then shuffle. If you don't, it doesn't answer the question. If you remove 18 more cards, then shuffle, then look at positions 2, 3, 4, and 6 (where the flop/turn would now be) you get a different chance of getting quads than if you don't remove the 18 cards. But remove the cards you must, if you wish to answer the original question. Since the cards removed are random, we must run this trial millions of times (Monte Carlo simulation?) to come up with the average chance of this occurring. But I think we can come close mathematically.

I think now it is also clear why I changed the 50 choose 3 to (8 handed) 36 choose 3. Since we're talking 10 handed now, let's update those calculations. 52 cards - 20 dealt - 1 burn = 31 cards eligible for the flop. (I apologize for forgetting about the burn card earlier.)

The possible flops are 31 choose 3 = 4,495. So we have (7 * 4) / 4495 = 0.0062. The 4th card is 1/28 or .036. Overall odds = 0.0062 * 0.036 = 2.2 x 10^-4, or 1 in 4,516. I think this most accurately answers the original question. This is a very interesting result, because it is an order of magnitude more likely than the original result. I think this can be explained by the fact that excluding certain ranks is more than outweighed by selecting the remaining quads from far fewer cards.

Back to the original situation: you hold T4 and the flop/turn are 5555. Your question: "I have seen a board like this on rare occasions before, but this time it occurred to me to wonder how frequently it will occur. Let's find out."

Your restated question (which I agree is a more clear way of stating it) was as follows: "Think of it this way: Instead of asking, "What is the probability of getting quads on the board after ten hold'em hands have been dealt out?", instead ask it this way: "Of all the times that in a ten-handed game I know only my own two down cards (and they are not a pocket pair) how often, on average, will I see quads hit the board with the first four cards?" That is actually the question that I answered."

I'm not sure exactly how to say this. Your revised question, while more clear, is inaccurate. The game could have been heads up, 6 handed, whatever, and you would have come up with the same answer. So I would strongly argue that in fact that question is misleading. The 10 handed part is irrelevant information. So the question you claim to have answered should be stated as "Of all the times that in a Hold'em game I know only my own two down cards (and they are not a pocket pair) how often, on average, will I see quads hit the board with the first four cards?" If you think I'm mistaken on this, then please show your calculation for a 6 handed game. If it's the same (which it must be since the fact that the game was 10 handed never entered into your equation) then 10 handed doesn't matter. I suppose the crux of our disagreement is that the number of players matters. You (implicitly) claim it doesn't. I claim it does. Either way, I do think the answer was incorrect, and I think I have proven as much above.

In short, I think what you answered could be more accurately stated as "Of all the times I only know my own two down cards (and they are not a pocket pair), how often, on average, will 3 cards appear sequentially at position X in the remainder of the deck, followed by a blank (burn card), followed by the 4th of that rank?"

Bottom line: That will occur 1 out of 20,936 times. Placed in the context of a ten-handed game, quads on the flop/turn will occur 1 out of 32,900 times, or possibly 1 out of 4,516 times.

Rakewell said...

Anon:

I think this is the crux of our disagreement: "The next step, removing a 10 and a 4, then shuffling, then looking where the flop/turn would be, is helpful. We are certain that the quads can't be a 10's or 4's. But just as you removed your (known) 10 4 from the deck, you must remove 18 other (random) cards, then shuffle. If you don't, it doesn't answer the question. If you remove 18 more cards, then shuffle, then look at positions 2, 3, 4, and 6 (where the flop/turn would now be) you get a different chance of getting quads than if you don't remove the 18 cards. But remove the cards you must, if you wish to answer the original question."

Forget the 10 and 4 for now. Let's focus on the question of whether what I will call Situation A and Situation B are equivalent. Situation A: From a shuffled deck I remove the top 20 cards. Situation B: From a factory-ordered deck, I remove 20 cards at random (maybe using a random number generator to pick spots in the deck), then shuffle the remaining cards.

Would you agree that there is no difference in the probability that we have ended up with four cards of the same rank at positions (say) 2, 3, 4, and 6 between these two situations?

Assuming you do agree with that, I'll proceed.

Now consider Situation C. We shuffle a deck, remove the top 20 cards, then reshuffle the remainder. Now what is the probability that we have quads in positions 2, 3, 4, and 6? It must be the same as in Situations A and B, right?

If you agree with me there, then I don't see how you can maintain that removing 20 (or 18) cards at random from the deck before shuffling them alters the calculation. As long as the cards removed are random, it doesn't matter if we shuffle before removing them or after or both, or even if we shuffle, remove one, reshuffle, remove another, etc. times 20. Random is random, and no matter the procedure, we end up with a pack of 32 cards that are a random mix from the original 52. Given that, the probability that quads are located at some set of four specified positions in the stub of the deck is the same no matter which random way we got the deck down from 52 to 32.

In other words, the answer is the same whether we remove 20 random cards from the deck, then shuffle and see if we have quads in specified positions, or if we shuffle the deck, then remove the top 20 cards, then check to see if we have quads in specified positions. It's the same thing. Dealing out unknown cards to real or imaginary players doesn't change anything about the math--unless you deal them to so many players around a very large poker table that you don't have even four left in the stub.

Anonymous said...

Addendum: Indeed, the fact that 4.78 x 10^-5 is very close to 4.8 x 10^-5 was indicative that the questions being answered were similar.

The way to get to 4.78 x 10^-5 is 50 choose 4 = 230,300, with 11 ranks making quads = 11 / 230,300 = 4.68 x 10^-5.

Anonymous said...

Grump, I agree with your comment completely. I agree that Situations A, B, and C are equivalent. What I don't agree is that those situations are the same as your original situation.

I think I established that the odds of finding quads anywhere within a deck of 52 cards is 13 / (52 choose 4) = 4.80 x 10^-5. Which is not the same as 4.78 x 10^-5. And it's not a rounding error making them different. So this either means that your original calculation was not correct for that situation, or that A, B, and C are not the same as the original. I'd love to see your work for Situations A, B, and C. What do you think the probability is?

Rakewell said...

Anon:

I'm afraid that I've explained my thoughts as thoroughly and carefully and in as many different ways as I know how to do. It hasn't persuaded you. So be it.

But I'll take one last shot at asking you about your take on it. It is certainly true that any rank represented in either of the hole cards of any player is not going to be one for which we can see quads appear on the board. With that in mind, you seem obsessed with accounting mathematically for the cards that are dealt out to players.

But it is equally true that the rank of, for example, the bottom card of the deck (after it is shuffled and cut) is one that can't make for quads on the board. Same for the next to the bottom card. And the third card from the bottom. And the fourth. Etc. Yet you seen to feel no compulsion at all to account for those ranks being unavailable to generate quads on the board. I don't understand why you feel a need to count how many ranks are unavailable due to one or more of their members appearing the top 20 cards in the deck, but you don't seem to feel a comparable need to count how many ranks are unavailable due to one or more of their members appearing among the bottom 20 cards of the deck.

I am about as certain as I get about anything in this world that my initial post's conceptual approach was correct (granting that an arithmetic error might lurk in it somewhere). I maintain that it cannot make any difference whether we deal cards to 1 player, to 5 players, to 10 players, or to 20 players, nor whether any of those players look at them. It will still be true that if I look at my hole cards and see that they are not a pair, I will see quads hit the board by fourth street on average once in 20,936 times. More generally, after I have removed two unpaired cards from the deck, any random selection of four cards from that deck has a 1/20,936 chance of being four of a kind.

I am incapable of seeing how it could be otherwise. If I'm wrong, it is apparently at a level that is impenetrable by your arguments, for which I'm sorry.

dfan said...

Rakewell is correct and has been much more patient than I would be in his situation.

But since I haven't wasted any time yet, I'll waste just a little right now.

The main points are these:

1. For computing the overall probability that you get quads, it doesn't matter what happens to the other cards, whether they are dealt face up or face down or discarded or ripped up.
2. If you do deal some other cards face up, you will know that the chance that you will get quads this hand is higher or lower than you originally thought. These all cancel out (exactly) in the long run.

Here's a toy example to make it more clear.

We play with a 10-card deck, with 2 of each rank from 1 to 5, and are each dealt 2 cards. What are the odds that I have a pair?

Method 1: My first card doesn't matter, and of the nine remaining cards, one matches it, so my chance is 1 in 9.

Method 2: After I look at my first card, the chance that the second player doesn't have my needed second card is 8/9 * 7/8 = 7/9 (I need to dodge two bullets). So the chance that he has it is 2/9.

If he has it, obviously I can't, so in that case (2/9 of the time), my probability is 0.

If he doesn't have it, I have two fewer cards left to choose from, so in that case (7/9 of the time), my probability is 1 in 7.

So my total chance is 2/9 * 0 + 7/9 * 1/7 = 1/9, just as with Method 1.

Anonymous said...

Love the blog, but have to ask... are all of the comments but Pete's really Rakewell under 3 different pseudonyms?

Anonymous said...

@dfan Thanks for your attempt to clarify with your simplified example. But I think even Rakewell would agree it's not even close to the same situation. In your simplified example, you're going after the odds of a single player making a pair. In the example we've been discussing, we're actually looking for the odds of everyone making quads. And that makes all the difference in the world.

I'll try one last explanation. Earlier, Rakewell stated (and rightfully so) that if we change the knowledge of the cards, we change the question and, thus, the answer. This is somewhat similar to my first explanation, but hopefully more clear this time.

To recap, Player 1 has 10 4, and 9 other hands have been dealt (we can debate if this is relevant or not, I'm simply stating the situation). In the very first calculation, knowing this hand made us eliminate 2 ranks from the possible ranks for quads. So the probably of quads were calculated as (11 * 4) / (50 choose 3) * (1 / 47) = 4.78 x 10^-5. This is so far considered to be the correct answer.

Let's now flip over Player 5's hand. She's holding the mighty 2 4. Now, I think we would all agree that only 10 ranks can make quads. So the calculation is (10 * 4) / (50 choose 3) * (1 / 47) = 4.34 x 10^-5. Agree?

If you agree with that, I don't see how you can't agree that the probability of making quads in a 10 handed game is (7 * 4) / (50 choose 3) * (1 / 47) = 3.04 x 10^-5.

Yes, in the example of knowing player 1 & player 5's cards, I've changed my knowledge and thus the answer. But if I shuffle and then deal out 10 new hands, I'm just as certain that only 7 ranks can make quads. It can't be otherwise. And there's no "balancing" that evens the odds out. I even showed that shrinking the pool seemed to increase the overall odds. In any case, how can we justify using 11 in the calculation but not 7? We used 11 because we knew something more about the situation. I'm saying we must use 7 because we know something about the situation.

Would you argue that the odds of making quads in our scenario are (13 * 4) / (50 choose 3) * (1 / 47) if we don't look at any cards?

Please recognize that all I've done here is point out that if we have knowledge of the cards (or even their composition) we must factor that into our calculation.

Anonymous said...

(This is not the same Anonymous, BTW) This is comical. I've heard similar explanations to those of the "infamous Anonymous" at the poker table.

I rarely get into strategy or math talk (or any other talk that would make people better), but this one particular guy was the typical table know-it-all type, who berated others' "bad play", etc. After a particular hand played out, which I won against a third party when I had shoved after flopping a pair with a flush draw and straight draw, he jumped in and insisted that I had played terribly.

To which I replied, "If getting all of the money in as a 55% favorite is horrible, then you're correct." He was absolutely certain that calculations of me winning the hand depended on the number of players dealt into the hand. He was absolutely hell bent that he was correct. Even after I showed him my iPhone based poker calculator, he still insisted that it was flawed because "that doesn't take into account the other hands people had!" When I asked him if he knew what those other hands were, he replied "no". So, I simply asked, "so, what cards would you like me to discount in the calculation?" His reply: "It doesn't matter! You have to change the calculation!"

At the end of it all, he said, "you just don't understand what I'm saying". And, I left well enough alone, and replied, "you're right, I must have been wrong."

I will give the poster here credit for being very civil, actually thinking about the posts, and carefully writing them. But, he/she is still incorrect. Please take the time to learn this concept if you are serious about the game, since it's very foundational knowledge.

Anonymous said...

@OtherAnonymous.

Yes, I have been civil. As has everyone in this thread. Which makes it a very enjoyable/interesting discussion.

The more this goes on, the more convinced I am that I am incorrect. I still can't see where, but I continue to noodle over the problem.

But your point about "which cards to eliminate" seems to be key to this. In the case of the table bully, he wasn't able to state which cards to eliminate. He knew nothing whatsoever even about the characteristics. But this example does seem to be different -- I can definitively state that only 7 ranks can make quads. But yet 11 is used in the calculation. Why?

If someone could poke a hole in my argument directly (invalidating the proof in my most recent post) I would actually be most appreciative, since it would show exactly where the hole in my logic is. The other simplified cases, analogies, etc., aren't cutting it for me. My apologies for "not getting it".

I'm also considering taking this over to the TwoPlusTwo forums, which might be a better place to hold a discussion and may attract alternate viewpoints.

Mark T said...

I'll take a stab, then, if I may.

In calculating card probabilities, it is critical that you know which ranks of quads you are discarding. You cannot say "it doesn't matter - I know that at least 4 more are not able to be made." Sure, at least 4 more are unavailable, but *which* 4 (or more!) are prohibited is material, as there are some combinations of players' hands which might exclude (say) quad 3's, while there are other combinations where that would not be excluded, and there is a probability assigned to each combination of possible players' hands.

At the end of the day, in all probability calculations involving distributions of a fixed set of objects like cards, you simply have to enumerate all the combinations, and count up how many of the various combinations are interesting for your purposes (i.e. have quads on the board). There is no shortcut around this, though there are ways to simplify the resulting math by reducing common denominators.

So, straighten the question: how many ways can a deck of 52 cards be arranged such that Grump holds 10d 4c and the flop and turn form quads?

Imagine laying out the cards in a line, and designating which spots in the line "would be Grump's hand", which "become the flop", which one "is the turn", etc.

Calculating the total number of arrangements is easy: 52! (I'm claiming order is important here - it's clearer, and it will cancel out in the end).

Calculating how many are interesting is a bit less easy: I need to figure out what combos I'm trying to count. The ones I'm trying to count are ones where Grump has 10d 4c and the flop/turn is 2c 2d 2h 2s (and then also count the other orderings of the deuces, all 24 of them together), ones where he has 10d 4c and the flop/turn is 3c 3d 3h 3s (and those other orderings), etc. I also need to count ones where Grump has 4c 10d and then the 24 combos of deuces form the flop/turn, since I'm keeping order important here, and those hands are the same. Note that I am *not* able to count ones where the flop/turn is 4c 4d 4h 4s, or 10c 10d 10h 10s, because there are *no such combinations*. Fine so far?

You'll notice that although it's true that other players' hands have cards in them in this enumeration scheme, and those cards do indeed "prevent certain quads on board", it doesn't matter -- we are looking through *all possible combinations of all the cards*. The fact that when a combo we're looking at has 7777 on the board it means that no player has a seven in their hand is immaterial; this is merely a combination we are enumerating in our probability calculation. Another way of looking at it: while it's true that in any given combination the cards in the players' hands prevent quads of those ranks from appearing, the *range of players' hands* spans the entire gamut. Without knowing what other players hold, it's equally probable that quad 7's will appear as quad 9's. But quad 4's will *never* appear in an "interesting" combo.

I think this point here is the main crux where you're getting stuck: you have to enhance the fact that some ranks are unavailable with the definitions of *which ranks those are*, and then assign probabilities to those ranks, since you don't know the other players' hands. If you're looking at it as "only 7 are available", then realize that which 6 are unavailable varies depending on what players hold - but the six unavailable ranks *always* include ten and four because we know what Grump holds. So find all combinations of six ranks which include ten and four, versus all combinations of six ranks that don't, then versus other rank combinations that have seven be unavailable, eight, etc., then assign them probabilities. This is incredibly difficult math, but by definition will end up yielding the same number as we get by just enumerating the possibilities.

- TO BE CONTINUED -

Mark T said...

- CONTINUED -

So, back to our calculation. How many combos are interesting? Well, there were 11 ranks where quads were on board (all except 10 and 4, as noted above), each of which had 24 combos and each had 2 ways to form Grump's hand, and all of which left 46 unassigned cards, so there are a total of 11 * 24 * 2 * 46! interesting combinations. (You may have noticed how the 11 came into the picture now, yes?).

This means the probability is 11 * 24 * 2 * 46! / 52! = 11 / (52 * 51 * 50 * 49 * 47) =~ 3.602 * 10^-8, or more than 1000 times less likely than Grump's original calculation.

So why is my number different from Grump's number? Because I'm including the odds of him having exactly 10d 4c in this scenario, and he didn't in his math. So he wasn't dividing by 52!, he was dividing by the number of arrangements of cards which already had his 10d 4c in his hand. That's significantly fewer: 2 * 50!, in fact (again, the 2 is for the two orders of his own cards). Using that as the denominator in our formula yields (11 * 24 * 2 * 46!) / (2 * 50!) = (11 * 24) / (50 * 49 * 48 * 47) = (11 * 4) / (50 * 49 * 8 * 47) = 44 / (19600 * 47) = 4.78 * 10^-5.

I'm not sure that this math is what Grump intended to ask, though: I think he may have meant to ask "What are the odds of the flop and turn forming quads on the board?" In that case, we shouldn't be looking at Grump's cards at all (note that in his original post he writes this off by saying "The answer would be slightly different if I had a pocket pair, but let's not worry about that." Well, let's worry about it, since it's simple. If Grump hadn't looked at his cards yet, the probability that the flop and turn form quads is (13 * 24 * 48!) / (52!) = (13 * 24) / (52 * 51 * 50 * 49) = 4.802 * 10^-5. Yes, this is higher than before - do you see why? Because it's possible Grump has a pocket pair, thus making 12 ranks available.

In fact, let's run the math assuming Grump has looked down and seen he has a pair of 5's (5d 5c). In that case, the interesting combos are 12
ranks of quads, but otherwise our original math still holds, since Grump still has two cards and they can be in either order. Hence the
probability is (12 * 4) / (50 * 49 * 8 * 47) = 5.21 * 10^-5. Notice that the chance goes up significantly if Grump has a pair; this only makes sense, as there is a twelfth rank available for quadding, as it were.

Did this help at all?

Anonymous said...

@Mark T

Thanks, that does help. I think between your explanation and a simulation program I wrote, I now see the validity of the original logic.

Thanks to PokerGrump, Mark T, and Other Anonymous for helping explain this.

Rakewell said...

I returned to this subject in a PokerNews article:

http://pokergrump.blogspot.com/2016/08/pokernews-126.html