Tuesday puzzle

Correction

Alex Y

I mis-stated hand 1. It should read:

Hand 1: I was watching a bridge game, and when Betty picked up her hand and looked at it, I secretly asked her if she had an ace, and if so, to show me one at random. She showed me the Ace of Spades.

Sorry about that!

Re: Correction

Larry Barrett

I originally thought hands 1 and 2 were equally likely, and more likely than hand 3 to contain all 4 aces.

As restated, I think think hand 1 more likely than hand 2, and hand 2 more likely than hand 3 to contain all 4 aces.

Re: Correction

David Weaver

This is a hypergeometric distribution question.

Solution for the one (#2) where you know that the last card is the ace of spades, you know nothing else other than that there are 12 cards in the remainder of the draw, and you want 3 of them to be aces.

The probability of that is (3 choose 3)*(48 choose 9)/(51 choose 12)

I'll have to think about the other two a little.

I think what you know in problem 1 and 3 is the same. You know that barbara, or whatever her name is, has at least one ace. You don't know anything about the rest of her cards.

If we work through the numbers, then I guess you'd have to calculate the universe of probabilities and find out the percentage of those probabilities that is all four aces. I wonder how close it will come to the first answer.

Probability that you draw 13 cards and have one ace:

(4 choose 1) (48 choose 12) / (52 choose 13)

That you have two:

(4 choose 2) (48 choose 11) / (52 choose 13)

That you have three:

(4 choose 3) (48 choose 10) / (52 choose 13)

That you have all four

(4 choose 4) (48 choose 9) / (52 choose 13)

And the chance that isn't included in the answer, that you have zero aces

(4 choose 0) (48 choose 13) / (52 choose 13)

Hopefully, all of those together add to 1.

I don't have a calculator that can do it - i'll have to set up a product in excel and we'll see how things go. I kind of feel like I should get the same answer for all of them, even though I'm arriving at the solution in a different way.

Re: Correction

David Weaver

OK, I think hand 1 and 3 are the same.

Based on what I worked out, hand 2 is approximately 3 times more likely to contain the aces than is hands 1 or 3.

The reason being that for hand 1 and 3, you know she has at least one ace. You don't know what else she has, and she can go to any card in the draw to pull the ace out wherever it is.

On hand two, you know the ace is the last card. That is a less likely scenario than just having an ace anywhere in the draw.

Probabilities from a draw with no conditions are basically 30.4 chance of having no aces, 43.9% of having exactly one ace, 21.3 of having exactly two,4.12% chance that you have exactly three, and .264% that you have all four.

Probability you have four if you are known to have at least one =

0.00264 / (.439+.213 +.00264 + .0412) => approximately .0038 or 1/263 chance

Probably that you have all four if it's known that the 13th card in the draw is an ace of spades is 1.0564% approximately.

Re: Correction

Larry Barrett

Since the problem did not ask for the exact probabilities, I looked at a simpler problem, assuming that the answers to this will also apply to the real problems.

Assume the deck consists of 4 cards, the deal is 2 cards.

The cards can be A(ce) of S(pades), AD, KC, 2C.

The 12 possible deals are:

(AS, AD), (AS, KC), (AS, 2C), (AD, AS), (AD, KC), (AD, 2C),

(KC, AS), (KC, AD), (KC, 2C), (2C, AS), (2C, AD), (2C, KC).

To analyze the third hand (I think this is it, don't have it it front of me), there are 10 hands which contain at least one Ace. Of these, 2 contain 2 Aces, so probability of two aces, given at least one is 1/5.

Hand 2, where the last card dealt is the AS, is possible in only 3 cases, and of these, only one case has both aces, so probability is 1/3.

I will have to re-read the problems to get hand 1; I'll probably come up with a different answer.

Hand 1

Larry Barrett

To continue with my simpler game, for Hand 1 (the restated case), of the 12 possible hands, 10 contain at least one ace, so this is the set of hands that apply to this case. Of those 10, 4 contain the AS and another non-ace card, and Betty would have shown Alex the AS in each case. The other 2 possible hands are the AS, AD and AD, AS; in these cases, Betty would select a card at random, so would show Alex the AS in 1/2 + 1/2 the cases. So overall, she would have shown Alex the AS in 5 of the 10 cases, prob = 1/2.

So for my simple game, Hand 1 has prob = 1/2, Hand 2 = 1/3, Hand 3 = 1/5. I hope that this extends to the real cases, where David is doing the complete analysis.

Re: Hand 1

david weaver

I'm rethinking hand 1

I think hand 1, since we know it's the ace of spades, is done as basically three sets of cards - 1 ace of spades (we know we have it), 3 other aces, and 48 other cards.

We have 51 other cards to draw from, and we know we have the ace of spaces, so we want to know the chance given that we have the ace of spades that the other 3 aces are also in our draw.

That gives (3 choose 3) (48 choose 9 ) / 51 choose 12

(48! / (9!39!)) / (51! / (39!12!))

Still no calculator to do it, and no excel this time!

(48!12!)/ (51!9!)

(12x11x10) / ( 51x50x49)

= 1320 / 12950

that gives: 1.02% chance, and that's actually less than hand 2 but greater than hand 3.

There's enough logic and arithmetic by now that I'm sure I've butchered it.

Alex Y

That's the same solution I got by a different route, which gives me some confidence that it is right. I'll post my solution when I get a little time later today.

Good for you for the "meta-reasoning" of why #1 and #3 are the same. My initial post, as Larry observed, had #1 and #2 the same, and I missed the observation you made, so didn't realize that I had "corrected" #1 to make it the same as #3 until I worked out the numbers.

Like the method, but not the answer

Alex Y

Larry, I really like the idea of simplifying the problem like this, but I agree with David's answer of 2>1=3. I'm going to have to spend a little time figuring out:

1) why our answer is wrong;

2) why your answer is wrong, or

3) why moving to smaller numbers changes the relationship.

Re: Like the method, but not the answer

Larry Barrett

I don't see anything wrong with David's logic or calculations, but if I apply his approach to my simpler problem for hand 1, I get prob = 1/3:

(Given AS, problem becomes (1 choose 1)(2 choose 0)/(3 choose 1) = 1/3)

My observation was that the prob for Hand 1 was 1/2.

Often times a simpler problem can give some insight into more complex problems, but not always. I also likely made some mistakes in reasoning.

Re: 

David Weaver

Now I have go reason to myself why I didn't get the same answer for #s 1 and 3 after I "re-reasoned".

sometimes with combinatorics, I have trouble figuring out how the conditional probability affects the setup.

Who here is versed enough to set up simulations to see if they converge to our answers?

I guess I'm still more comfortable on 1=3, because the spade at random means that it's really the same condition - that you know the person has at least one ace, and the only difference is you get them to show you one.

Answer using Bayes

Alex Y

I agreed with David's answer, but said I had approached it from a different angle.

When you have a conditional probability P(A|B), which means "the probability of A, knowing that B is true), Bayes gives us a calculation method that MAY be easier. In this case, it is easier for Hands 2 and 3, but maybe harder for Hand 1.

Bayes says P(A|B) = P(A) * (P(B|A)/P(B)).

When I learned that in school, I thought "What's the point, you still have a conditional probability?" Hand 2 gives a good illustration of when this can be useful. In hand 2,

A = Hand contains four aces

B = Last card dealt to Betty is Ace of Spades.

P(A) = C(48,9)/C(52,13), i.e., the total number of ways to choose 9 cards from the 48 non-aces to go along with the four aces, divided by the total number of ways to choose 13 cards from the 52-card deck. Note that P(A) is the same for all three cases, so we can just treat it as an uncalculated constant in determining the relative probabilities of the three cases.

P(B|A) = 1/13. IF she has all four aces (condition A), then she has the Ace of Spades, and there is a 1/13 chance that it is the last of the cards dealt to her.

P(B) = 1/52. This is the probability that the ace of spades was the last card dealt to Betty, with no assumption about whether she has the aces. The last card dealt to her is as likely to be the Ace of Spades as any other card.

P(B|A)/P(B) = (1/13)/(1/52) = 4, so discovering that the last card dealt to her was the Ace of Spades increased the probability that she has all four aces fourfold!.

The conditional probabilities P(B|A) are trivial for hands 1 and 3, equal to 1/4 and 1, respectively.

P(B) is easy for hand 3, once you realize it is 1 minus the probability that she has 13 non-aces. P(B) is harder for Hand 1, making me think it might be easier to solve the original conditional probability without Bayes, if you didn't get David's insight on the equivalence of the information available in Hands 1 and 3.

I'll stop here, since this sounds a lot more like a solution to a problem in a probability course than a puzzle!

Re: Answer using Bayes

Larry Barrett

Applying the Bayesian approach to my simpler problem gives the same results that I derived for Hands 2 and 3 (1/3 and 1/5). I don't know how to deal with the "pick an Ace at random" part of Hand 1.

Re: Answer using Bayes

David Weaver

I should have just remembered to use bayes.

But I haven't done any probability problems for about 10 years - except for the ones on here. They were, by far, my favorite mathy sort of item in school.

Proofs in a class that was combination mathematics and computer science engineering - not so much.

So whoever doesn't like probability problems here, I'll reside on the trivia board so that Seinfeld's statement is correct ("it all evens out")

Re: Answer using Bayes

Alex Y

Larry, using Bayes on hand 1, I get the same answer as you did for hand 3:

P(A) (both aces) = 1/12

P(B|A) = 1/2

P(B)= 1/12 (prob of 2 aces)*1/2 (prob of choosing the spade ace to show

+ 1/6 (prob of having AS and one of the non-aces)

= 5/24

So the probability we are looking for is 1/12 * 1/2 / (5/24) = 1/5

So your simplified example gives the same ordering, 2>1=3

Re: Answer using Bayes

Larry Barrett

Alex, thanks for the analysis. I think P(A) should be 1/6 instead of 1/12, but oddly enough it does not seem to matter in the answer - still comes out to 1/5. I will have to rethink my earlier analysis that came out 1/2. Thanks for the puzzle.

Re: Answer using Bayes (correction)

Alex Y

Larry said:

I think P(A) should be 1/6 instead of 1/12, but oddly enough it does not seem to matter in the answer - still comes out to 1/5

Sorry.