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TomasB
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Jason England posted a variation of a problem at

http://www.themagiccafe.com/forums/viewt......2&13

where I also gave an attempt at a solution. I'll re-post it here and hope that someone can point out any errors, because I'm not certain of the answer.

Quote:
--------------------------------------------------------------------------------


On 2003-12-05 03:22, JasonEngland wrote:
A related question:

A poker hand consists of 5 cards.

In my poker hand, I have at least one ace. There is a certain probability that I have another ace.

On subsequent deal, I receive the Ace of Spades. Again, there is a certain probability that I have another ace in that hand.

Are these two probabilities the same? Why or why not?

Jason


--------------------------------------------------------------------------------


I'll write (n k) as short for n!/(n-k)!k! which is the number of ways to take k objects out of n objects when the order doesn't matter.

For the case with any Ace we are interested in the hands (X is no Ace)

XXXXX There exist (48 5) such hands
AXXXX (48 4)*4

There's a total of (52 5) possible hands.

Hands with at least one Ace is therefore (52 5)-(48 5) and the number of hands with at least two Aces is (52 5)-(48 5)-(48 4)*4. The ratio between those is the probability we are looking for, which to three decimals is 0.122.

The same analysis with the Ace of Spades (S) gives (Y is any card...can even be an Ace)

SYYYY (51 4)
SXXXX (48 4) (Just the Ace of Spades and no other Aces)

So the number of hands with the Ace of Spades together with at least one other Ace is (51 4)-(48 3). The ratio this time (three decimals) becomes 0.221.

So it turns out that the probability of having another Ace when the Ace of Spades is present is about 81% higher.

Reservations for that I can be somewhat, but not totally, wrong. Smile

/Tomas
leonard
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Tomas,
I agree 100% with your analysis [one typo: (48 3) should read (48 4)].
leonard

The probabilities are indeed different. I believe this is because knowing we have the Ace of Spades provides more information than simply knowing we have "any" ace. It reduces the five-card problem to a four-card problem, while only reducing the deck to 51 cards.

Perhaps two simpler, yet related, problems will help clarify the issue.

Assume P(Boy) = P(Girl) = 0.5.

A man has two children. One of them is a boy. What is the probability the other one is also a boy?

A man has two children. The youngest one is a boy. What is the probability the other one is also a boy?
dlhoyt
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There is a simpler, more intuitive way of thinking about this problem. Completely enumerate all the possible poker hands that contain two or more aces:

1)*SHCDx 4 Aces
--------
2)*SxCDx 3 Aces
3)*SHxDx
4)*SHCxx
5) xHCDx
--------
6)*SHxxx 2 Aces
7)*SxCxx
8)*SxxDx
9) xHCxx
10) xHxDx
11) xxCDx
---------

As you can see, there are 11 poker hands with two or more Aces. Of these, 7 contain the Ace of Spades, so the ratio is 7/11. So hands that contain two or more Aces should be almost half as common as hands that contain the Ace of Spades and at least one other Ace.

Dale Hoyt

Quote:
On 2003-12-12 13:56, leonard wrote:
Perhaps two simpler, yet related, problems will help clarify the issue.

Assume P(Boy) = P(Girl) = 0.5.

A man has two children. One of them is a boy. What is the probability the other one is also a boy?

A man has two children. The youngest one is a boy. What is the probability the other one is also a boy?



Good problem, leonard, but it is not quite the same as the Ace problem because it depends on the order of birth. In the poker problem we aren't concerned with the order in which the cards are received.

For the children:

1) BB
2) BG
3) GB
4) GG

Probability other child is a boy, given that at least one child is a boy = 1/3

Probability other child is a boy, given that the first born is a boy = 1/2
leonard
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Dale,
I agree with your answers of 1/3 and 1/2.

I also agree with completely enumerating all the possible poker hands, which is what Tomas and I did (albeit in a shorthand fashion).

However, I don't think there is only one poker hand with four aces as in your item #1. (I believe there are 48 such hands.) Nor do I believe there are only 11 hands with two or more aces. (I believe there are 108,336 such hands.) I think you may be considering "types" or "classes" of hands, and I am not sure your analysis is correct. Perhaps Tomas can comment as well.

regards,
leonard
dlhoyt
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Quote:
On 2003-12-12 15:06, leonard wrote:
...
However, I don't think there is only one poker hand with four aces as in your item #1. (I believe there are 48 such hands.) Nor do I believe there are only 11 hands with two or more aces. (I believe there are 108,336 such hands.) I think you may be considering "types" or "classes" of hands, and I am not sure your analysis is correct.

leonard you are absolutely correct! I know better and can only plead a senior moment. The frequencies of each of the hands should be taken into consideration and I didn't. Mea culpa!
If you look at the 2 Ace hands there are an equal no. of hands with AS and without AS, so those cancel out. Its the 3 and 4 Ace hands that tip the scale in favor of the AS. This is a little easier to see for those who are not familiar with combinatorial mathematics.

Thanks for correcting my error!

Dale Hoyt
TomasB
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Thanks for the feedback. I also did some simulations of this and they seem to indicate that the answer isn't totally wrong.

The example with the children is a good analogy actually, as age can be seen as a property just as as the suits of the cards. Imagine just four Aces and you randomly take two of them. If one is a black Ace the probability of the other one being black also is 1/3. If I put the property Spade on one (instead of "youngest" on the child) we get that the other one is black 1/2 of the time.

/Tomas
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