Bayesian? Probability of 2 Aces

[panamajack]

Auntbeast : It is pretty likely that your whiner will get an unmatched deuce in a hand. Analysis follows. It’s just as likely he’ll get an ace, too, of course, but you probably knew that.

The number of other players has no effect on the cards one person gets. If you have trouble with this, consider that it ought to make no difference if you’re dealt 7 cards straight off the top, or dealt the 1st, 8th, 16th, etc. The cards are random either way.

• In a stud game, you can see the other cards and have some idea of what cards you may get, but that only applies to that particular game. It also doesn’t affect the final outcome, only what you can expect it to be while playing. I’ve ignored it here.

Assuming just one deck (i.e. you can’t get dealt the same card twice), the formula you use is ‘sampling without replacement’.
The ‘choose’ function to get combinations is useful for notation: C(n,m) means 'the number of ways to choose m items out of n total; written out it is

n!

m!(n-m)!

(n! means n-factorial, the product of all numbers from 1 to n).

Suppose we have a deck of N cards. There are D deuces in the deck.
We’re dealt n cards, and we want to know the probability of exactly k
cards that are deuces. We have this formula:

(n-card hands that have k deuces)

---

(all possible n-card hands)

which becomes

(number of ways to choose k cards of D deuces) * (number of ways to choose the other cards (n-k) from the rest of the deck (N-D) )

---

(total number of ways to be dealt n cards from a deck of N)

which is

C(D,k)*C(N-D,n-k)

---

C(N,n)

I wasn’t sure if you were asking about the whole hand or just the first three.

To get exactly one deuce in a 7-card hand :

C(4,1)*C(48,6)

---

C(52,7)

= 0.370 which you can see is high enough it’s almost sure to happen in a night.

For the first three:

C(4,1)*C(48,2)

---

C(52,3)

= 0.204 which is still about 1 in 5 hands.

[Indistinguishable]

@LinusK

I think that’s it exactly. Try this:

Suppose you go to the doctor, and he tells you one out of a thousand people have a particular disease. He gives you a test, and the test is 99.9 percent accurate.

The test comes back positive.

What are the chances you have the disease?

Well, I suppose it depends on what “99.9 percent accurate” means. If it means the test is correct 99.9 percent of the time when giving a positive result, then the answer is, of course, 99.9 percent. But clearly you have something else in mind for what “99.9 percent accurate” means.

[Snarky_Kong]

@Rysto

Is it just me, or have you all read the question wrong? I read it as:

1. Hold’em Player is dealt 2 cards from a standard deck of cards
2. Player looks at one card; sees that it’s an ace(he does not note the suit)
3. Player shuffles his 2 card hand so that he doesn’t know which card is which
4. Player looks at one card; it’s an ace.

Given all that, what is the probability that the player was dealt Pocket Rockets?

It’s 50%, right? Because he has a 50% chance of seeing the card he saw the first time.

You read the question correctly, your reasoning is wrong. Half the time you have seen 2 aces, half the time you’ve seen 1 ace and have a 3/51 chance of having another. Do you follow?

[Dead_Cat]

@Auntbeast

Mind if I hijack a bit?

We have a seven card stud player who bitches every time you give him a deuce. I am math challenged and can’t figure out the chances he would get a deuce, with 7 players, being dealt out three cards to start, there are 21 cards on the table.

It seems to me the odds of someone NOT getting a deuce in 7 cards is smaller than the chance of them getting one.

There seem to be a few different questions here. Firstly, it is more than likely that someone round the table will get a deuce in their first three cards. If no-one gets a deuce, that implies that all four deuces remain in the deck after 21 cards have been dealt. The chances of this happening are: 48/52 * 47/51 * 46/50 * … * 27/31 = 0.101… , so there is about a 10% chance of no-one getting a deuce. In other words, there is a 90% chance that at least one of the seven players will get at least one deuce in their first three cards. If you want me to explain how I got that, ask and I will try :).

Working out the odds of one individual getting a deuce is more complicated, because it depends on whether they are the first to receive a card, or second, or third, etc. If we assume they are first, then their chance of not getting a deuce in the first three cards is 48/52 * 41/45 * 34/38 = 0.752… , so about 75% chance. In other words, they would expect to get at least one deuce in their first three cards 25% of the time, or about one hand in four, if they are left of the dealer every time. These odds increase slightly as they move closer to the dealer button; when they are on the button they should expect a deuce 28% of the time. This will average out to about 26.5%, of course, over the course of a game.

The chance of receiving a deuce in 7 cards is huge, of course: 75% for the player to the left of the dealer, by my calculations, if all players stay through 7 cards (which is of course highly unlikely). I hope I have all these figures correct, and that some of them are of some use to you :).

[Rysto]

@Snarky_Kong

You read the question correctly, your reasoning is wrong. Half the time you have seen 2 aces, half the time you’ve seen 1 ace and have a 3/51 chance of having another. Do you follow?

:smack:

Yes, of course. I can’t believe I didn’t think of that.

[Indistinguishable]

@Snarky_Kong

You read the question correctly, your reasoning is wrong. Half the time you have seen 2 aces, half the time you’ve seen 1 ace and have a 3/51 chance of having another. Do you follow?

Actually, given that you see an ace both times, there is a more than 50% chance that you’ve seen the same card both times, as remarked upon above.

(I know that’s counterintuitive, but think about it this way: aces are rare, so very rare that, if you see an ace both times, it’s rather more likely that you’ve just seen the same one twice than it is that you actually saw two different ones. It may help to visualize the situation with a million card deck with just two aces in it).
As a result, the probability of having pocket rockets is NOT 1/2 * 1 + 1/2 * 3/51. It’s 1/9, as noted in several previous posts.

[Auntbeast]

@Dead Cat

The chance of receiving a deuce in 7 cards is huge, of course: 75% for the player to the left of the dealer, by my calculations, if all players stay through 7 cards (which is of course highly unlikely). I hope I have all these figures correct, and that some of them are of some use to you :).

thanks for you replies guys. Evidently, dealers suck at math. I know the whiny bastard will be there tonight and I’ll be sure to throw some ambiguously large numbers at him. Funny though, even though I lack the logical math skills to come up with this number, I noticed that every few hands, he gets a deuce. So the numbers you folks came up with do match my experience.

OnTopic: What you are all forgetting is that the chances of you getting pocket aces increases to 100% if it is a misdeal. And of course, it is always the first decent hand you’ve had ALL DAY and you would have WON for certain. If it hadn’t been for that meddling dealer…

[Chronos]

Working out the odds of one individual getting a deuce is more complicated, because it depends on whether they are the first to receive a card, or second, or third, etc.

It shouldn’t. Every player has the same chance at the same hands as every other player, or the game is crooked.

[LinusK]

@Indistinguishable

Well, I suppose it depends on what “99.9 percent accurate” means. If it means the test is correct 99.9 percent of the time when giving a positive result, then the answer is, of course, 99.9 percent. But clearly you have something else in mind for what “99.9 percent accurate” means.

99.9% accurate means the test is accurate 99.9% of the time. In other words, the test will give a false report once out of every thousand times it’s given, or 0.1% of the time.

[LinusK]

@Chronos

It shouldn’t. Every player has the same chance at the same hands as every other player, or the game is crooked.

If the player ahead of you gets a duece, your chance of getting one goes down. If the player ahead of you doesn’t get one, your chance goes up.
Auntbeast: in my experience, the chances that everybody will fold go up to 100%, if I’m dealt aces. And if they don’t, that just means somebody’s going to suck out. :smack:

[Ximenean]

@Dead Cat

Working out the odds of one individual getting a deuce is more complicated, because it depends on whether they are the first to receive a card, or second, or third, etc. If we assume they are first, then their chance of not getting a deuce in the first three cards is 48/52 * 41/45 * 34/38 = 0.752… , so about 75% chance. In other words, they would expect to get at least one deuce in their first three cards 25% of the time, or about one hand in four, if they are left of the dealer every time. These odds increase slightly as they move closer to the dealer button; when they are on the button they should expect a deuce 28% of the time. This will average out to about 26.5%, of course, over the course of a game.

By that logic, the players later in the order would be more likely to receive any particular card. That can’t be right.

I think your calculation should be 48/52 x 47/51 x 46/50 = 78%, regardless of where the player sits. The order the cards are dealt in is irrelevant. Indeed you could deal them three at at time to each player without affecting the probability.

[MilTan]

@LinusK

99.9% accurate means the test is accurate 99.9% of the time. In other words, the test will give a false report once out of every thousand times it’s given, or 0.1% of the time.

Given the question you were asking, and the answer you were clearly hoping to get, your definition is the one that makes sense :). In that case, the chance that you have the disease is 50%:

Using Bayesian analysis, A is the probability of having the disease, B is the probability the test came back positive. We want to find P(A | B). This is equivalent to:

P(B | A) * P(A) / P(B) by Bayes’ Theorem.

And P(B | A) = .999, P(A) = .001

Figuring out P(B) is a little trickier. The probability the test comes back positive is basically .002. Of a thousand people, one of them will have the disease and 999 won’t. The test will come out positive for 2 people, then: the person who has the disease, and one of the people who doesn’t (because the test gives the incorrect results .1% of the time). The exact value for this is .001998.

Thus, P(B | A) is .5

[mlees]

Wait. I am not understanding the question…

What are the chances both cards are aces?

Your knowledge, or lack thereof, of the cards delt wont change whats actually dealt (from a standard deck). What’s the chance of being dealt two aces? 1/13 (for the first one) x 1/17 (for the second)? 0.452%?

The only thing that is changing by peeking is your knowledge of them. But that doesnt actually change what was dealt…

In other words:

You have a .452% chance of being dealt 2 aces from a 52 standard deck. Period.

By peeking at the first and seeing an ace, you can now discount all of the combos that do not include at least one ace, so you think you now have a 1/17 chance of having two aces. (But that does not change the fact that the original odds of having been dealt two aces in the first place was .452%…)

By peeking again, you have a 50% chance of seeing the same first card, and 50% chance of seeing the second card. So, on the second peek you have a 50% chance of knowing both cards, or still remaining with the knowledge of what is just the one (first) card… but this knowledge does not change the odds of how they were dealt.

[LinusK]

@MilTan

Given the question you were asking, and the answer you were clearly hoping to get, your definition is the one that makes sense :). In that case, the chance that you have the disease is 50%:

Using Bayesian analysis, A is the probability of having the disease, B is the probability the test came back positive. We want to find P(A | B). This is equivalent to:

P(B | A) * P(A) / P(B) by Bayes’ Theorem.

And P(B | A) = .999, P(A) = .001

Figuring out P(B) is a little trickier. The probability the test comes back positive is basically .002. Of a thousand people, one of them will have the disease and 999 won’t. The test will come out positive for 2 people, then: the person who has the disease, and one of the people who doesn’t (because the test gives the incorrect results .1% of the time). The exact value for this is .001998.

Thus, P(B | A) is .5

It’s the only definition, so far as I know. When doctors, scientists, or whoever talk about something being 99.9% accurate, that’s what they mean - that one test in a thousand is wrong. If there’s another definition, please let me know.

What’s interesting is that xx.x% accurate doesn’t mean what people think it means, and it can make a big difference, when diagnosing rare events - like breast cancer in young women, for example, and AIDS.
But yeah, you’re right. If the test is 99.9% accurate, and it’s a (relatively) rare disease, (1 in 1000 is not all that rare, when it comes to disease), a positive test does not necessarily mean you have anything.

[Indistinguishable]

It’s a matter of conditional probability, mlees. P(getting two aces) is a constant, independent of your knowledge, in some sense, but P(getting two aces | X) varies depending on what that knowledge X is.

E.g.: You’re dealt one card, face down. What’s the probability that it’s the queen of spades? 1/52. Someone looks and tells you it’s a queen. Now what’s the probability that it’s the queen of spades? By convention, we want the probability conditioned on the current knowledge that it’s a queen, so that the answer is now 1/4.

In general, it’s possible for particular new knowledge to change the (appropriately conditioned) probabilities radically, which seems to be what you’re ruling out.

[On edit, it seems you’ve already maybe kinda grasped this, but maybe kinda haven’t.]

[Indistinguishable]

At any rate, the 1/9 answer is indisputably correct. Run some simulations and see. (Or just refer to chrisk’s post, which is basically a comprehensive simulation (don’t forget to note his arithmetic correction))

[mlees]

@Indistinguishable

It’s a matter of conditional probability, mlees. P(getting two aces) is a constant, independent of your knowledge, in some sense, but P(getting two aces | X) varies depending on what that knowledge X is.

E.g.: You’re dealt one card, face down. What’s the probability that it’s the queen of spades? 1/52. Someone looks and tells you it’s a queen. Now what’s the probability that it’s the queen of spades? By convention, we want the probability conditioned on the current knowledge that it’s a queen, so that the answer is now 1/4.

In general, it’s possible for particular new knowledge to change the (appropriately conditioned) probabilities radically, which seems to be what you’re ruling out.

[On edit, it seems you’ve already maybe kinda grasped this, but maybe kinda haven’t.]

Yeah, I was fluttering around the edges of understanding. Youve helped out, though. Thanks!

Partial knowledge of the card is used to help recalculate the odds (by eliminating now impossible combos).

[MilTan]

@LinusK

It’s the only definition, so far as I know. When doctors, scientists, or whoever talk about something being 99.9% accurate, that’s what they mean - that one test in a thousand is wrong. If there’s another definition, please let me know.

I was under the impression that there are usually different false-positive and false-negative rates for most tests. Certainly there are the separate concepts of sensitivity and specificity.

In this case, the specificity and the sensitivity are both 99.9% (so the probabilities of both false positives and false negatives are .001).

[Pasta]

I don’t think the correct answer to the OP as intended has been stated yet. The situation is that you know you hold Ax and that

x=A is the case 3/17ths of the time
x=not-A is the case 14/17ths of the time.

Now you shuffle the cards and see an ace (suits unseen throughout.) The question is:

Given that you see an ace after the shuffle, what is the probability that x=A? As the OP noted, this is a Bayesian problem. Let S=“I see an A after the shuffle” (the conditional part of the question.) Then,

P(x=A|S) = P(S|x=A)P(x=A) / [ P(S|x=A)P(x=A) + P(S|x=not-A)P(x=not-A) ]
P(x=A|S) = (1)(3/17) / [ (1)(3/17) + (1/2)(14/17) ]
P(x=A|S) = (3/17) / (10/17)
P(x=A|S) = 3/10

[MilTan]

@Pasta

x=A is the case 3/17ths of the time
x=not-A is the case 14/17ths of the time.

Given these numbers, your procedure is fine. But where did you get 17 from? Are there only 18 cards in the deck for some reason?