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Subject: Probability Question rss

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Marshall P.
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I've done this before, but now I'm blanking on the method. Probably because I don't do statistics everyday, but anyway here is the question.

Let's say you have a 36 card deck.

2 Cards are labeled "A"
2 Cards are labeled "B"

What is the probability of drawing 6 cards and getting at least 1 "A" and 1 "B" card.

Thanks.
 
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DC
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I would do it this way:

(Probability of drawing at least 1 A and 1 B) = 1 - (Probability of drawing no A's) - (Probability of drawing no B's) + (Probability of drawing no A's and no B's)

You have to add back in that last term because there is overlap between the first two.

Prob. of drawing no A's: (36-2 choose 6)/(36 choose 6) = 1344904/1947792 = 0.6905

Prob of drawing no B's is the same: 0.6905

Prob of drawing no A's and no B's: (36-4 choose 6)/(36 choose 6) = 906192/1947792 = 0.4652

Overall probability = 1 - 0.6905 - 0.6905 + 0.4652 = 0.0842

Edit: this assumes that order of drawing does not matter. Does it?
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Marshall P.
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dcclark wrote:

Edit: this assumes that order of drawing does not matter. Does it?


No it doesn't. Thanks.
 
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Ashfield
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If you want to calculate this sort of thing often, the relevant mathematics is here:

http://en.wikipedia.org/wiki/Hypergeometric_distribution
 
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Christopher Dearlove
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sbszine wrote:
If you want to calculate this sort of thing often, the relevant mathematics is here:

http://en.wikipedia.org/wiki/Hypergeometric_distribution


I don't immediately see how to apply the hypergeometric distribution, as presented in that article, to the given problem to give a simple answer (although clearly you can express the answer using the hypergeometric distribution). Enlightenment welcome.

The previous poster who gave a method was using a special case of the inclusion-exclusion principle, which can be used to solve many problems of this sort. Googling turns up various accounts of it (and the material in the standard text by Feller is better still).

(My thanks to David des Jardins for pointing that out with relation to a more complicated such problem at some point in the fairly recent past.)
 
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Marty Firth
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dcclark wrote:
I would do it this way:

(Probability of drawing at least 1 A and 1 B) = 1 - (Probability of drawing no A's) - (Probability of drawing no B's) + (Probability of drawing no A's and no B's)

You have to add back in that last term because there is overlap between the first two.

Prob. of drawing no A's: (36-2 choose 6)/(36 choose 6) = 1344904/1947792 = 0.6905

Prob of drawing no B's is the same: 0.6905

Prob of drawing no A's and no B's: (36-4 choose 6)/(36 choose 6) = 906192/1947792 = 0.4652

Overall probability = 1 - 0.6905 - 0.6905 + 0.4652 = 0.0842

Edit: this assumes that order of drawing does not matter. Does it?


Good Explanation, the only thing I would add (given that I don't know the OP's mathematical literacy) is that the whole choose thing can be expressed a little differently.

Prob. of drawing no A's =
no A on 1st card * no A 2nd card * ... * no A on 6th card =
34/36 * 33/35 * 32/34 * 31/33 * 30/32 * 29/31 * 28/30 = 0.6905

This works because we start with 36 cards in total and 34 of them are non-A and we reduce both the total and the number of non-A's by 1 each time we draw a card out of the deck.


Prob. of drawing no B's will be identical since there are also 2 B's (or 34 non B's).


Prob. of drawing no A's and no B's =
no A/B on 1st card * no A/B 2nd card * ... * no A/B on 6th card =
32/36 * 31/35 * 30/34 * 29/33 * 28/32 * 27/31 * 26/30 = 0.4652

Here we start with 32 cards that are neither A nor B and still 36 in total, reducing each by 1 as before.


I got the same probabilities as above so the final answer will be the same




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Dearlove wrote:
I don't immediately see how to apply the hypergeometric distribution, as presented in that article, to the given problem to give a simple answer (although clearly you can express the answer using the hypergeometric distribution). Enlightenment welcome.

Here is the probability mass function of the hypergeometric distribution:



Here is David using the same function (with m choose k = 1) to solve Marshall's problem:
dcclark wrote:
Prob. of drawing no A's: (36-2 choose 6)/(36 choose 6) = 1344904/1947792 = 0.6905

Prob of drawing no B's is the same: 0.6905

Prob of drawing no A's and no B's: (36-4 choose 6)/(36 choose 6) = 906192/1947792 = 0.4652

The results are, as you say, chained together via inclusion exclusion... but I think Marshall knew that bit already.

Enlightened?
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John Weber
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I have a probability question for anyone with the time to read this thread.

Roger Federer, who many say may be the greatest male tennis player in history, has reached the semifinals (final 4 of 128 players) in 18 consecutive Grand Slam tennis tournaments.

All other things being equal (i.e. a 50% chance of winning each match), what are the odds of someone duplicating that feat?

Not being a math whiz myself, I would say the odds are 1 over 32 to the 18th, whatever that number happens to be.
 
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Christopher Dearlove
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sbszine wrote:
Enlightened?


Not really additionally. If you are going to use the I-E principle, I think I'd regard introducing the hypergeometric distribution as well as not adding anything. (To a more difficult problem, maybe.) I was hoping there was something else that made it drop out, but I didn't know of anything, and didn't see anything on the page you referenced. But there might have been, and if you don;t ask, you won;t find out. Thanks for the answer.
 
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Marty Firth
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John Weber wrote:
I have a probability question for anyone with the time to read this thread.

Roger Federer, who many say may be the greatest male tennis player in history, has reached the semifinals (final 4 of 128 players) in 18 consecutive Grand Slam tennis tournaments.

All other things being equal (i.e. a 50% chance of winning each match), what are the odds of someone duplicating that feat?

Not being a math whiz myself, I would say the odds are 1 over 32 to the 18th, whatever that number happens to be.


Is this the same Federer who lost to a very retired Sampras in an exhibition match a couple of years ago?

Your statistical reasoning seems to be correct! If you assume that each player has a 50% chance of winning a match and that match results are independent then the probability of reaching the semi-final requires winning five consecutive games = (1/2) ^ 5 = 1/32

Again assuming that results in one tournament to the next are independent, to do this in 18 consecutive tournaments = (1/32) ^ 18 = 1/1237940000000000000000000000

I think that for a "top player" however 1/32 is too low a probability of them making the semi-finals, perhaps we could increase that to say 1/4 (I have never done any stats but it doesn't seem unreasonable that guys like McEnroe, Wilander, Edbeg etc would have reached the semi-finals in about 1/4 of the tournaments they played in).

Anyway the probability a "top player" could then reproduce this feat would be (1/4) ^ 18 = 1/68719476736 which is still incredibly small.

All of this rules out hot streaks, different probabilities on different surfaces etc etc.
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