Probability in a Deck of Cards

[TranscendArcu]

Homework Statement

In a deck of 52 cards, one black card is removed. There are then 13 cards dealt. Show that the probability that they are all red is 2/3.

The Attempt at a Solution

I don't understand why I can't calculate the probability as

That is, the the number of red cards when total cards is 51 divided by this total, times the number of red cards, given that one was drawn previously, divided by the new total of 50, etc. etc.

But this value isn't even close to 2/3! Apparently it's more like 13/(1 740 081)

How can I increase my denominator by several hundred thousand to get 2/3?

 

[tiny-tim]

Hi TranscendArcu!

In a deck of 52 cards, one black card is removed. There are then 13 cards dealt. Show that the probability that they are all red is 2/3.

No, that's ludicrous!

The question must be wrong. ​

 

[Ray Vickson]

Homework Statement

In a deck of 52 cards, one black card is removed. There are then 13 cards dealt. Show that the probability that they are all red is 2/3.

The Attempt at a Solution

I don't understand why I can't calculate the probability as

That is, the the number of red cards when total cards is 51 divided by this total, times the number of red cards, given that one was drawn previously, divided by the new total of 50, etc. etc.

But this value isn't even close to 2/3! Apparently it's more like 13/(1 740 081)

How can I increase my denominator by several hundred thousand to get 2/3?

I get the answer = 38/1740081 (as compared with your 13/1740081), but I don't get anything like 2/3. Whoever told you the answer is 2/3 is either wrong or is describing a different problem.

This is a simple problem in the hypergeometric distribution. We have a deck of 51 cards, 26 red and 25 black. We want to know the probability of getting 13 red in a sample of size 13; I just used the hypergeometric formula.

RGV

 

[TranscendArcu]

When I first did this problem, I calculated C(26,13)/C(51,13). This gives the answer of 38/1740081. I ultimately multiplied out all of the probabilities at each draw because, to me, it seemed more easily visualizable. But I don't know why the two methods should return different results unless there is something fundamentally wrong with the method of multiplying out at each draw.

 

[Ray Vickson]

When I first did this problem, I calculated C(26,13)/C(51,13). This gives the answer of 38/1740081. I ultimately multiplied out all of the probabilities at each draw because, to me, it seemed more easily visualizable. But I don't know why the two methods should return different results unless there is something fundamentally wrong with the method of multiplying out at each draw.

Your product above has 14 factors, not 13.

RGV