Card Game Puzzle

[drummerlsu]

I've always found probability problems like this interesting. It is even more interesting to watch other people answer them, which is why I'm posting it here. If this is the wrong forum for a topic like this, feel free to move/delete it.


You are playing a poker game in which everyone is dealt two cards. You are sitting next to a friend.

Hand A is dealt. Before you look at your cards, you ask your friend "Do you have the ace of spades?" He answers "Yes." You fold your cards blind.

Hand B is dealt. Again, before you look at your cards, you ask your friend "Do you have an ace?" He answers "Yes." You fold your cards blind.

Assuming your friend was truthful, which of the following statements is true?
You were more likely to hold an ace in hand A than hand B.
You were more likely to hold an ace in hand B than hand A.
You were equally likely to hold an ace in both hands A and B.

EDIT: If you hold two aces, then you still 'have an ace'.

[Problems]

You are more likely to hold an ace in hand A than hand B. This is because you have said "Do you have the ace of spades?" meaning that he could have the ace of spades and another ace. Then you asked "do you have an ace?" which with my understanding would be only "an ace" not two aces. Maybe it's just my understanding of the wording that you gave to the post...

[Brainmaster07]

Wouldn't it be the same? Assuming they are dealt him, you, him, you.

A) He has the Ace of Spades, his other card has a 3/50 chance of being an Ace, and you have a 3/51 and 3/49 chance for your cards to be an Ace.

B) He has an Ace, his other card has a 3/50 chance of being an Ace, and you have a 3/51 and 3/49 chance for your cards to be an Ace.

In both cases he could have more then one ace, but in either case it's the same probability of the unknown card.

[drummerlsu]

Careful Brainmaster, there are several errors in your reasoning.

[eyespewgreekfire]

You have a better chance in B, because his odds of having AA increases his chance of having a specific ace. If yes is the answer to question be, he sas yes to A 50% of the time with AA but only 25% of the time with AK-A2. If he has AK-A2 you have a 6% chance of having at least one ace. You have roughly a 4% chance of having an ace when he hits AA. There is more math needed for a rigourous answer, but I'm feeling lazy.

[Brainmaster07]

Oh, ignore me, I don't know how to play poker :P.

[talisman]

clearly equal.

[drummerlsu]

I guess everyone is done with this. Eye's solution is correct - you are more likely to hold an ace in the second hand because your friend is less likely to hold AA.

In hand A your opponent has specified one of his cards is a specific ace. He can have AA if the other card is an A - 3 of the remaining 51 cards (3/51 chance).

In hand B he has at least one ace. He can have AA 12 different ways (4*3), Ax 192 different ways (4*48) and xA 192 different ways (48*4). So he has AA 12/396 or 1/33 of the time.

You are obviously more likely to have an A if he has Ax instead of AA (about 12% likely if he has Ax and 8% likely if he has AA).

[psychic25]

What?

In hand A, he can have AA if the other card is an A, 3/51.

In hand B, doesn't the same thing hold true? He has one ace- there is still a 3/51 probability that he holds another one.

Correct me if I'm wrong, but I think the probabilities are equal.

[drummerlsu]

The tendancy is to think the probabilities are equal, which is the goal of the problem.


First, another explanation of why they're not:

It is obvious that the chance of your friend having AA in the first hand is 3/53. The reason why that isn't the case in the second hand is because he looked at BOTH cards and not just one. If you examine all his possible 2-card combinations (386) only 12 of them are AA.

This is similar to me saying I thought of a random 2 digit number, you asking if the number conatins a 5 as a digit, and me saying yes. What are the chances that the number is 55? Not 1/10. It would be 1/10 if I said the FIRST digit was 5. Instead, we examine all the possible numbers (05,15,25 etc and 51,52,53 etc). There are 19 because you can't count 55 twice. So the probability of the number being 55 is 1/19.

Similarly, the chances of the friend's hand B being AA is signifigantly less than 3/53 because his possible hands (AA,AK,AQ,...,A2 and AA,KA,QA,...,2A) have overlapping cases of AA and you can't count them twice. If you incorrectly did, the probability would be 24/424 = 3/51. Instead, you remove the 12 overlapping AA cases: (24-12)/(424-12) = 12/396.


Sure all that is really technical, probably more than I intended it to be. But I find it interesting because on the surface it seems that the two situations are exactly the same. For many people (including myself to some extent) I think there is a sort of psychological problem with realizing the probabilities are different, even after the math behind the correct answer is understood.

[Squeek]

Originally Posted by drummerlsu

The reason why that isn't the case in the second hand is because he looked at BOTH cards and not just one.

It's a good thing you explicitly told us this fact that really mattered in the original question.

Seriously.

~Squeek (/Sarcasm)

[psychic25]

But in your number example, you count 15 and 51 as different numbers, because the order of the digits matters.

In your card example, it appears as though you count A (spades) and K (hearts) as separate than K (hearts) and A (spades).

Actually, you seem to be counting Ax differently than xA throughout your whole solution.

Does the order matter in this case? Is a king and ace really different than an ace and king? I would think not.

[drummerlsu]

Originally Posted by MrESqueek

It's a good thing you explicitly told us this fact that really mattered in the original question.

Seriously.

~Squeek (/Sarcasm)

OK.. technically speaking, he could have looked at just one card, checked if it was an ace, then checked the other only if the first wasn't an ace. But I didn't need to state that in the problem. Do you see why?


Psychic:

Yes, in my solution I was obviously counting each two-card combination twice, once for each order. The solution comes out the same either way though.

If we count each possibile hand only once regardless of the order dealt:

6 cases of AA: AsAh, AsAd, AsAc, AhAd, AhAc, AdAc)
192 cases of AX: 48 cases of AsX, 48 for AhX, 48 for AdX, 48 for AcX where X is any non-A card)

So there are 198 total two-card combinations if we ignore order.
Chances of AA: 6/198 (=12/396 from earlier).

[Squeek]

Originally Posted by drummerlsu

OK.. technically speaking, he could have looked at just one card, checked if it was an ace, then checked the other only if the first wasn't an ace. But I didn't need to state that in the problem. Do you see why?

Because you wanted us to get it wrong?

Whether he looked at one card or both cards made a huge difference, and not noting it made us all seem like idiots when we were guessing.

Not like I even care. If it were a real-life situation, I knew the friend of mine would be a dirty liar. Nobody gets two aces in a row.

~Squeek

[psychic25]

Fine. I concede.

Although I really think there must be something I'm missing :P.

[drummerlsu]

Originally Posted by MrESqueek

Whether he looked at one card or both cards made a huge difference

First of all, I intended it to be obvious that he looked at both cards from the problem statement; I apologize that it wasn't.

Still, we have no reason to rule out the possibility that he looked at both cards. So when calculating the probability of him having AA, we can only rule out hands with no A. All other hands remain viable possibilities. Since there wasn't any specific information given about how many cards he looked at to determine he had at least one A, it's still logical to assume each possible hand is of equal probability.

The only possible way it would come into play is if you assume there is a chance that your friend would have looked at only one card and answered "I don't know" when you asked him whether he had an ace. But I didn't forsee anyone considering that as a possible circumstance. I'll be more clear next time.