NoahT
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I should tell you--in the Hearts puzzle I entered and DiMono answered for this month's contest, I wrote the answer incorrectly.

-Noah

Edit: DiMono now has the correct answer posted below.

[Edited by NoahT on 09-19-2004 at 05:02 AM GMT]

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ErikH2000
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Noah, isn't there a reduced answer? I.e. "a 1 in 526 chance". I am not very knowledgable about math and don't know what the "C" is for. What operation does that perform? Actually, I think Adam Peterson (alefbet) explained this to me once, but... gads, combinatorials, right?

-Erik

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NoahT
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I'm not really sure about the reducing...you'd have to ask DiMono, since he was the one who computed the answer.

-Noah

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DiMono
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I didn't give an answer, I gave a formula from which the answer could be obtained, which Noah then copied basically verbatim in to the puzzle. Amusingly enough, the formula was incorrect.



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bibelot
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I still don't think this is right. First, let's bring out the original statement:

In a regular game of Hearts, what are the chances of you drawing a hand that contains the queen of spades and no hearts, and an opponent drawing exactly three hearts?

So I am interpreting this as follows:
My hand: QS, twelve cards that are not hearts
any of my opponents: 3 hearts, ten non-hearts

Notice that I'm not specifying in advance which opponent has the three hearts.

Now, the chance that my hand is what it is is (38C12)/(52C13), 38 for the 38 non-point cards.

The chance that the person on your left has exactly three hearts is (13C3)*(26C10)/(39C13). Multiplying that by three gives the chance that anybody has three hearts, except that we've overcounted the chance that two people have three hearts.

To find that, there are three ways to choose which two people, and then the chance is (13C3)*(26C10)/(39C13)*(10C3)*(16C10)/(26C13). Subtracting this out, we're done, because it's impossible for all three to have exactly three hearts.

So the answer is:
(38C12)/(52C13)*(3(13C3)(26C10)/(39C13)-3(13C3)(26C10)(10C3)(16C10)/((39C13)(26C13)))
which comes out to 689397137/317506779800, or about 0.00217128, or about 1 in 460.557.

It's interesting to notice that if you have no hearts, the chance that one of your opponents has exactly three is about one-half.

[Edited by bibelot on 09-19-2004 at 07:32 AM GMT]

AlefBet
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ErikH2000 wrote:
I am not very knowledgable about math and don't know what the "C" is for. What operation does that perform?

C stands for combinations. The formula for nCr is n!/(r!(n-r)!). Semantically, it means the number of ways you could select r objects out of a set of n objects. For example, if you have a shuffled pile of 52 distinct cards (since we're on the subject) and you drew (or were dealt) five cards, the number of possible hands you could end up with would be:
52C5, or
52!/(5!47!) or
(52*51*...*1)/(5*4*3*2*1 * 47*46*45*...*1) or
52*51*50*49*48/(5*4*3*2*1) which ends up being 2,598,960 possible hands.

The 52*...*47 above can be thought of choosing an item out of 52, followed by another out of 51 (since one is gone), then another out of 50, etc. This gives you the number of hands you could draw where the order you drew is significant. The 5*4*3*2*1 figure is the number of ways you could rearrange your hand, and you divide by that to get a count of selections where order is irrelevant.

[Edited by AlefBet on 09-19-2004 at 07:38 AM GMT]

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DiMono
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bibelot: The problem with your solution is that we don't care what's in the other two people's hands, only that one of them has exactly three hearts. You're right that my solution was incomplete, because I didn't multiply it by 3 to account for the fact it could appear in anyone else's hand, but since we don't care where the other ten hearts go we don't need to figure it out.

Edit: Though you are correct in that I misread the original question to believe the first hand did not contain the Queen of Spades, instead of having it. 38C13 is different than 38C12, and that must also be multiplied by 1/52. I'm going to pick up from the numbers in my other post.

38C12 is 2707475148. Multiplying 1/52 (to get the Queen of Spades in there) by 38C12/52C13 gives us 8.1993256651130523718613883232157e-5 (about 1/12196), and tripling the odds of any other hand having exactly 3 hearts gives us 0.0071769626980418891829912351370835 (about 1/139). This makes the odds 1/1699343.5

[Edited by DiMono on 09-19-2004 at 12:56 PM GMT]

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bibelot
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DiMono: Oh, I hate to argue over this, but I still think I'm right.

For some reason you multiplied (38C12)/(52C13) by 1/52. This doesn't make any sense, because the 1/52 is already taken care of by virtue of the fact that the denominator is (52C13) and not (51C12); moreover, the chance that your hand contains the queen of spades is 1/4, not 1/52 anyway.

When you calculated the probability that a hand contains three hearts, you divided by (52C13), not (39C13) as you should have. Remember, we've already decided the thirteen cards in your hand, which you acknowledged in the numerator (13C3)(26C10) but not in the denominator. Finally, you can't just multiply this by three to find the probability that the three hearts appear in any of the opponent's hands, because those events aren't disjoint. You have to take into account the situation when two people have three hearts, which changes the answer slightly.

DiMono
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Wow. You're right, and I'm really out of practice. Victory to the bibelot.

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