Math Is Fun Forum

MathPro101

Member

Registered: 2011-06-08

Posts: 18

Probability Question

Hello all,

I am just passing time playing with a deck of cards and thought of a probability question. It is a very straight forward problem, but for some reason I cannot figure out the solution.

If I have a deck of 52 cards, shuffled, and I turn over one card at a time, what is the probability that I will see 2 of the same number in a row? (i.e., flipping over a 9, followed by another 9).

Let me know what you all think! Thank you.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Probability Question

Hi;

It is a very straight forward problem,

Straight forward yes, easy? No! That problem is very difficult. It could be rephrased as:

Having the multiset:

How many permutations do not have two adjacent numbers the same. As far as I have seen the smaller problem

Has a solution and it is given in Enumerative combinatorics V1 Stanley p 68.

I am not aware of any solution for multisets of multiplicites greater than 2.

I am currently working on that exact problem myself.

Just because there is no analytical solution that I know of does not mean the problem is impossible. There are computational techniques that could provide an answer.

So the first question I ask is just 2 in a row? Is 9,9,9,k,j,2... illegal?

The probability of at least one pair is ≈ .954

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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.
Always satisfy the Prime Directive of getting the right answer above all else.

gAr

Member

Registered: 2011-01-09

Posts: 3,482

Re: Probability Question

Hi all,

So the first question I ask is just 2 in a row? Is 9,9,9,k,j,2... illegal?

Or do we stop flipping the cards when we see two same numbers in a row?

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"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense"  - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

soroban

Member

Registered: 2007-03-09

Posts: 452

Re: Probability Question

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. . . Can that be right?

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MathPro101

Member

Registered: 2011-06-08

Posts: 18

Re: Probability Question

gAr:

The way I looked at the problem, I would stop looking once I hit a pair.

Thank you all for thinking about this one.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Probability Question

Hi;

Is that going to help? Because a Markov chain is usually for with replacement. This is without.

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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.
Always satisfy the Prime Directive of getting the right answer above all else.