Probability question

[MathBabe]

This came up with Rogue23's four K's in the last league game, and in another post where all four of a rank were in play.

The question is, what's the probability that all four A's, for instance, are in play?

The answer depends on how many players are in the game. Let's assume a full table of 10; that gives us 20 hole cards and 5 on the table by the river. That's 25, almost half the deck.

I think my math here might be a bit dodgy, but this is what I did to calculate it:
I started by figuring out how often the first four cards off the deck are of the same rank. The first card is unimportant; the second card has a 3/51 probability of matching it; the third has 2/50, and the fourth has 1/49. That gives a combined probability of 1/20825 of the first four cards being the same.

Now for higher numbers of cards, I consider how many combinations could be four the same. If I deal five cards, for instance, there are four ways the four cards could be arranged within the five. (This is the 5 choose 4 thing again). That makes it four times more likely, or 1/5206.

For 25 cards out in play, 25 choose 4 is big - 12650. Multiplying that by the 1/20825 we get .60, making it a 60/40 split - it will happen 60% of the time, actually making it a favourite! Note that this is the probability that "four cards will be the same" - we don't care which rank they are.

To consider just one rank, say Aces, we divide by 13 again. So the odds of all four Aces being in play in a full ring game is 4.7%, or roughly 1:20.

Three-handed, by comparison, I calculate 1:62 odds of four cards the same being dealt; and only 1:818 of all four Aces (or whatever) being out.

Sorry, this is a bit hard to follow... certainly lacking in elegance!

And, of course, it's not that useful. For any given rank, you have to consider how likely it is that whoever was dealt the card stayed in with it - a bigger variable than Chance. And because you see an Ace on the flop, you know the odds that all four Aces are in play is more likely - but the odds that someone was dealt AA or a single A are the same as always, regardless of that flop Ace, since their cards were dealt first.

[Talking Poker]

Actually, that's not true. If there is an Ace on the flop, that means there are only three other Aces available to be distributed around the table. That makes it LESS likely that one of your opponents is holding one - purely mathematically speaking. It doesn't matter which event (hole cards or flop) happened first. There can still be a maximum of 4 Aces in play, and there are still only 52 cards in the deck.

Take an extreme example (I find it's always easier to think logically when you consider extreme examples): Say the flop is AAA. That means there are 49 unknown cards and only one of them is the case Ace. So the odds of that being in play at a 10 handed table are 1/49*10=20.4%. Had the flop contained ZERO Aces, for comparison sake, the odds of a single Ace being in play would be: 4/49*10=82%. Huge difference!

I suppose a much better and non-mathematical example would be this: IF the flop is AAA, what are the odds that someone is holding AA? Very obviously, they are 0%. With 0 Aces on the flop (or 1 or 2), those odds are obviously > 0%.

[PunkyChikn]

Side note, just out of curiosity...what's the probability in Hold 'Em of someone making Quad Aces against Quad Kings in the same hand? I'd figure it out for myself but I'm unsure of how to figure out the math.

[MathBabe]

Well, we need all four Aces in play, and all four Kings. So if I'm right that there's a 4.7% chance of either happening, multiplying those gives .22%... 1 time in 500.

It's actually quite a bit less than that, because we can't have the Aces or Kings split between two hands. Let me get back to you on that tomorrow!

[MathBabe]

I think you can get in a lot of trouble thinking of it this way. Yes, if you can see 3 aces on the board, logically nobody has AA. But that's eliminating possibilities, not calculating probabilities.

I disagree - it DOES matter what happens first. If the flop came first and had an Ace, then yes, the odds that anyone at the table is holding an Ace are mathematically less. But the fact that the hole cards are dealt first means that every person at the table has a 15% chance of being dealt at least one Ace, and you have to figure that into your decision-making.

[Talking Poker]

I beg to differ. I believe it is YOU who could get into a lot of trouble thinking this way.

PREFLOP, you are correct. But once the flop comes, you are absolutely, positively, 100% incorrect. Think about the AAA flop example, and I'm confident you won't tell me that with 10 opponents at a table, every 22 hands someone at the table will be holding AA... because you could deal that out 1 million times, and I promise you that not ONCE will someone be holding AA if the flop is AAA (assuming a legal deck, of course).

Poker is all about using all the information available to you when putting your opponent on a range of hands. Sometimes, we discount hands like 83o when our opponent makes a preflop all in reraise from middle position, because MOST people don't play 83o like that. Mathematically, it's just as likely that they are holding 83o as it is that they are holding AKo, but I promise you, they will show you AKo there FAR more often than they will show you 83o.

When you are "decision making," you need to use all of the information available to you. If the board is showing the Ac, for example, you CAN NOT put any of your opponents on the Ac (I promise you they won't have it), even though preflop, there was a 1/52*2*10=38% chance that that exact card could have been dealt to one of the 10 players at the table. That 38% figure will always be true when we are speaking in terms before the flop, but as soon as the Ac hits the board (on the flop, turn, or river), the new and improved chance that one of our opponents is holding it becomes 0%. The same is true for the AA example, and also for what you wrote initially:



This is an incorrect statement.

[Akverno]

Oh, what's the name of this if I remember right, conditional probability or some such nonsense. The classic stats example is you get to choose door number 1, door number 2, door number 3. Only one door has a prize. After you select 1 the host says let's look behind door number 2. Door number 2 doesn't show a prize. Then the host says would you like to change your choice, statistically you have a 16% better chance of getting the prize if you switch your pick to number 3. Damned statistics.

[Talking Poker]

This is SO not the same thing.

[MathBabe]

I almost posted the other day and said exactly what you said, but I realized later he's actually right. It's a question of conditional probability.

"What's the probability of the fourth Ace in play GIVEN there are three Aces on the board".

A = three aces on the board
B = exactly one hole card ace

P(B|A) = P(A and B)/P(A)
=1/1353 / 1/553
=553/1353
=1/2.45
which is about 40%.

My calculations assume 10 players, therefore 20 hole cards, and the three Aces being anywhere on the board - not just on the flop.

I guess we'll have to agree to disagree...

[MathBabe]

You need to go read this thread:

[Akverno]

I must have missed that one, anyway, as I said earlier conditional probability. I don't remember formulas, don't remember why or how, just remember a teacher discussing cards in a class and this was a similiar example to some problem we did.

[MathBabe]

Absolutely incorrect reasoning.

But I'm too tired to figure out a better way to explain right now.

A demain!

[Talking Poker]

I'll be looking forward to your attempts at "math bending" tomorrow. Cause, well, it's just math, sister. And math doesn't lie.

{This isn't to say that I haven't made an error in my computations somewhere along the way - that's quite possible. But the point I'm trying to make is valid: If there are three Aces on the board, the chances of someone holding AA are 0%. If the 4d is on the board, the chances of someone holding the 4d are 0%. Trying to prove otherwise will be... well... interesting, if nothing else.}

[GTDawg]

I think you skipped right to post-flop while she is talking about pre-flop still.

You guys keep going around in circles talking about different things.

[Talking Poker]

How could she be talking about preflop when she is saying things like: "regardless of that flop Ace, since their cards were dealt first."



I'm pretty sure we're talking about the same thing here.

[MathBabe]

I always pick out the piece I am responding to, so don't think I'm arguing about something I'm not. I was not arguing with this comment:

I am arguing with this:

Just because there are three Aces on the board, you can't go back and re-calculate the probabilities of someone getting an Ace for a hole card. All you can do is eliminate a category of event.

Look at it this way. With 10 players, here's how the scenarios break down:
A: 13% probability that nobody has a hole card Ace
B: 37% probability there is one hole card Ace
C: 35% probability there are two hole cards Ace (may or may not be AA, we don't care)
D: 13% probability there are three hole card Aces
E: 2% probability there are four hole card Aces

If you see AAA on the flop, you know you are in the 50% of the time that there are none or one Aces in the hole cards. But that doesn't make the probability that there is a fourth Ace in play 20%, as you calculated - it's still 37%.

In fact, you could argue that since you've narrowed it down to those scenarios, and their relative probability is the same, there's actually a 26% chance you are in scenario A and a 74% chance you are in scenario B!

I think this is worth niggling about, because if you see three cards of a rank and you think there is a 4:1 chance of something having the fourth card, it's really more like 2:1. It may not happen often enough to affect your winnings, but over hundreds of thousands of hands, it you assume that fourth card isn't out there, you're going to be wrong more than you should.

[Quint]

Before the flop you know what 2 of the 52 cards are (your hole cards). After the flop you know what 5 of the 52 cards are. You've gained information, certainly you can, and should, recalculate the probability of the other Ace being dealt.

Using your logic, if the fourth ace comes on the turn is their still a 37% chance someone has an ace?

[Talking Poker]

I believe MB's response to this will be that we would need to rule out Case B (as we did on the flop with Cases C, D, and E), but she will incorrectly state that Case A remains true - that there is still only a "13% probability that nobody has a hole card Ace" when in reality, we know that this number is 100%.

[Talking Poker]

I'm not sure how I can explain this any better... but what you are saying is just wrong. I thought the examples I've written would help you see the light, but so far, no such luck. Maybe I should think about this some more and post again when I can come up with a different way of explaining this.

Ok, how about this............:

1. Let's simplify this: Let's just talk about the Ac. I'm confident that with 20 cards being dealt to 10 players preflop, you will agree that the odds of the Ac being in play is 20/52=38%. Right? Now let's substitute that information into this quote of yours, fixing it to match our new scenario (I'll bold what I'm changing, and I'm cutting out the middle part that is irrelevant):

Old scenario:
New scenario:
CLEARLY, our new scenario is incorrect. I mean, the Ac is ON THE BOARD, so we know it's not in someone's hand, but using your logic, you would have me believe that there is a 38% chance that someone is also holding it. And that's obviously wrong. Agreed?

[Talking Poker]

Actually, let's focus on this. If I am reading this right, you are telling me that when three of a rank flop at a 10 handed table, 74% of the time, someone will have flopped quads and 26% of the time, no one will be holding the case card. Is that correct?

Because if it is, I'm pretty confident we can disprove that by running a simulation, and that should put an end to this argument right now.

I too think this is worth niggling about, btw - well, not for me so much, as I'm confident that I'm right - but for you and anyone else who is buying your logic, because I too agree that you are going to cost yourself a lot of money in the long run if you don't understand that the current probabilities of your opponents hole cards change as you gain more information.

Actually, here's yet another example. All in preflop on the WPT: AA vs KK. Normally, they would say the KK is about a 4:1 dog, but when someone at the table says "I folded a King," everyone understands that the KK guy now only has one out instead of two, making him a much bigger underdog. This "new information" has changed the probability that he will make the best hand with 5 cards left to come. I realize this isn't the same scenario as what we are talking about, but the logic is the same: We need to use all of the information available to us (all known cards) when determining probabilities (not probabilities for the last hand or the next hand or at a different street of this hand, but the probabilities right now).

[Talking Poker]

While I'm at it, I'm going to test this, as well:

To test this, I'm going to run 100,000 hands (10 players each) with the AAA flop and simply see how many times someone has flopped quads (meaning they were dealt the case Ace). According to MB, the expected answer here should be either 37% or 74%. I'm not sure - please clarify, MB.

According to me, it should be 20%, as per this:

Whooops. Looking at that again now, I think I made a (big) mistake. This would be if each player was only holding one hold card. Because each player is holding 2 hole cards, I think we need to multiple that all by 2: 1/49*20=40.8%. But that can't be right, because to do that with 0 Aces on the flop, we'd have a greater than 100% chance of someone holding an Ace.

Ok, I've definitely confused myself here. But dammit, I still want to run some simulations!!!

[bunny]

You're both correct in your own way. I initially disagreed with MB's first post. TP's post made more sense to me. But down the line, MB's post made sense too. You both confused me so much so, I had no choice but to look it up. Here's what I found. Hope this helps.

TalkingPoker = Bayesian.

MathBabe = Frequentist.

"There is a history of antagonism between Bayesians(TP) and frequentists(MB), with the latter often rejecting the Bayesian interpretation as ill-grounded. The groups have also disagreed about which of the two senses reflects what is commonly meant by the term 'probable'."

Bayesian statisticians believe that Bayesian inference uses aspects of the scientific method, which involves collecting evidence that is meant to be consistent or inconsistent with a given hypothesis. As evidence accumulates, the degree of belief in a hypothesis changes. WHAT TP WAS TRYING TO EXPLAIN With enough evidence, it will often become very high or very low. Bayesian statisticians also believe that Bayesian inference is a suitable logical basis to discriminate between conflicting hypotheses. Hypotheses with a very high degree of belief should be accepted as true; those with a very low degree of belief should be rejected as false.

Bayesian inference uses a numerical estimate of the degree of belief in a hypothesis before evidence has been observed and calculates a numerical estimate of the degree of belief in the hypothesis after evidence has been observed.

Bayesianism is more popular among decision theorists. Frequentists can't assign probabilities to things outside the scope of their definition. In particular, frequentists attribute probabilities only to events while Bayesians apply probabilities to arbitrary statements.


Frequentists talk about probabilities only when dealing with well-defined random experiments. The set of all possible outcomes of a random experiment is called the sample space of the experiment. An event is defined as a particular subset of the sample space that you want to consider. For any event only two things can happen; it occurs or it occurs not. The relative frequency of occurrence of an event, in a number of repetitions of the experiment, is a measure of the probability of that event.

(thankyou wiki )


Now you have no choice but to agree to disagree. So, shake hands, make up and let's move on.

With love...bunny



(now... back to clearing that "oh, so ever exhausting bonus" at UB)

[MathBabe]

How are you planning to force the flops? If you deal the flop first, I don't think that's a reasonable simulation. The best way to simulate would be to deal some huge number of hands, and only choose the ones where AAA came on the flop.

I think in the end we've arrived at the same answer. I said 40% when I used the conditional probability formula. Your mistake, which I have to admit I didn't catch, when fixed comes up with 40% as well.

That is correct. With more then one card, though, you can't use the easy 4/49*20 calculation - that's where you have to get into the combinations.