Math Is Fun Forum

Roberto

Guest

Probability Problem

A poker deck, with 52 cards, is well shuffled. Then a card is drawn, without reimmission, until you get:
(a) two aces
(b) five spades
(c) all the 13 hearts

Which is the expected value for the drawn cards for every case?

I really hope that somebody could help me. I need the easiest way to calc (without a pc) these means.

Thanks

Roberto

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Probability Problem

I think that'd be tricky.

For example, the answer to a) is given by

The other two questions are even worse.
This is the kind of thing that a computer loves doing, but is too much work to be calculated by hand.

Is there any reason why you don't want to use a PC?

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Why did the vector cross the road?
It wanted to be normal.

Roberto

Guest

Re: Probability Problem

Thanks for replying.
The reason I cannot use a PC is that this is an exercise of a probability exam. And there I can't use a pc!
How did you get that sum?
Are there other ways to solve the problems?

Thanks Again

Roberto

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Probability Problem

That sum is made up of the probabilities that the 2nd Ace will be drawn on the nth turn.

If you play out the entire deck, there are 52C4 = 270725 combinations of Ace and non-Ace that you can have.

Now, if you want the 2nd Ace in a combination to appear in the, say, 5th spot, then the combination needs to look like:

????A...

Within those 4 question marks there needs to be 1 Ace, and the rest of the deck (...) needs to contain the other two.
There are 4C1 = 4 ways of the first part happening, and 47C2 = 1081 ways for the second part to happen.

So therefore, the probability that the second Ace of a deck will appear 5th is (4*1081)/270725 ≈ 0.016.

In general, the 2nd Ace will appear nth with probability [(n-1)C1 * (52-n)C2]/270725.
Multiply each of these probabilities by n and sum them up to get the expected value.

We're summing from 2 to 50 because it's impossible for the 2nd Ace to appear 1st, 51st or 52nd.

If you don't have a computer, that sum can be evaluated analytically, but it takes a lot of work and I don't think you'd be expected to do that. Particularly since the other two questions will give a similar-looking summation but with a higher-order polynomial involved.

I'd guess there's some simpler method that I'm missing.
I did a) on a computer and it says the expected number of cards is 21.2, which doesn't give me any clues.

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Why did the vector cross the road?
It wanted to be normal.

Roberto

Guest

Re: Probability Problem

Ok, I got your answer, and it is correct!
Anyway, I've found the solution to my problem: it needs to be used the Negative Hypergeometric Distribution! Then everything is very easy

Thanks again
Roberto

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Probability Problem

Hi Roberto;

Oh man, what a treasure trove of info this forum is. Nice sum and explanation by mathsyperson. The method I use is much clumsier.

You are right, Roberto you did find an easier way! Now it is easy to do those problems even without a computer. Thank you both!

b) E(# of cards for 5 spades to show) = 265/14 = 18.93 cards

c) E(# of cards for 13 hearts to show) = 689/14 = 49.21 cards

Last edited by bobbym (2009-07-06 10:30:34)

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
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