probability question, cards

**khor** · Dec 27th 2009, 01:42 PM

With 2 card decks, I take 1 card from each one. What is the chance to draw a diamond at least once?

How about with a single deck with 21 cards removed (and you don't know which ones), and then 2 cards are drawn. What is the chance at least one of those is a diamond?

**e^(i*pi)** · Dec 27th 2009, 02:33 PM

Originally Posted by khor

With 2 card decks, I take 1 card from each one. What is the chance to draw a diamond at least once?

How about with a single deck with 21 cards removed (and you don't know which ones), and then 2 cards are drawn. What is the chance at least one of those is a diamond?

1. The chance from 1 pack is $\frac{13}{52} = \frac{1}{4}$ . -facepalm- . EDIT: As below do $1-\left(1-\frac{1}{4}\right)^2$

2. Impossible to calculate unless it's known which cards remain/have been taken out

**Plato** · Dec 27th 2009, 02:37 PM

Originally Posted by khor

With 2 card decks, I take 1 card from each one. What is the chance to draw a diamond at least once?

There is a probability of $\frac{1}{4}$ of drawing a diamond from a deck.
So what is the probability of not drawing a diamond?
Square that number.
That will be the probability of not drawing at least one diamond when drawing one card from two different decks.
So what is your answer?

**khor** · Dec 27th 2009, 03:48 PM

43.75%. Okay I see. Thanks.

**awkward** · Dec 27th 2009, 04:08 PM

Originally Posted by khor

[snip]

How about with a single deck with 21 cards removed (and you don't know which ones), and then 2 cards are drawn. What is the chance at least one of those is a diamond?

Since you don't know which cards were removed, the probability of drawing at least one diamond is the same as if you were drawing from a full deck.

$P(\text{drawing exactly one diamond}) = \frac{\binom{13}{1} \binom{39}{1}}{\binom{52}{2}}$

$P(\text{drawing exactly two diamonds}) = \frac{\binom{13}{2}}{\binom{52}{2}}$

Add these to get the probability of drawing at least one diamond.

**Soroban** · Dec 27th 2009, 05:22 PM

Hello, khor!

The second one has an incredibly long and tedious solution.

How about with one deck with 21 cards removed (you don't know which ones),
and then 2 cards are drawn. What is the chance at least one is a diamond?

Since it depends on how many diamonds were removed,
. . we must consider all possible cases.

Suppose no $\diamondsuit$ is removed. . This probability is: . $\frac{{39\choose21}}{{52\choose21}}$
The deck contains: 13 $\diamondsuit$ s and 18 Others.
The probability that no $\diamondsuit$ is drawn is: . $\frac{{18\choose2}}{{31\choose2}}$
The probability that no $\diamondsuit$ is removed and at least one $\diamondsuit$ is drawn is: . $\frac{{39\choose21}}{{52\choose21}}\left(1 - \frac{{18\choose2}}{{31\choose2}} \right)$

Suppose 1 $\diamondsuit$ is removed. . This probability is: . $\frac{{13\choose1}{39\choose20}}{{52\choose21}}$
The deck contains 12 $\diamondsuit$ s and 19 Others.
The probability that no $\diamondsuit$ is drawn is: . $\frac{{19\choose2}}{{31\choose2}}$
The probabiity that 1 $\diamondsuit$ is removed and at least one $\diamondsuit$ is drawn is: . $\frac{{13\choose1}{39\choose20}}{{52\choose21}}\le ft(1 - \frac{{19\choose2}}{{31\choose2}}\right)$

. . . $\vdots$

We must extend this list up to:

Suppose 11 $\diamondsuit$ s are removed. . This probability is: . $\frac{{13\choose11}{39\choose10}}{{52\choose21}}<br />$
The deck contains 2 $\diamondsuit$ s and 29 Others.
The probability that no $\diamondsuit$ is drawn is: . $\frac{{29\choose2}}{{31\choose2}}$
The probability that 11 $\diamondsuit$ s are removed and at least one $\diamondsuit$ is drawn is: . $\frac{{13\choose11}{39\choose10}}{{52\choose21}} \left(1 - \frac{{29\choose2}}{{31\choose2}}\right)$

The final probability is the sum of these twelve probabilities.

You go on ahead . . . I'll wait in the car.
.

**awkward** · Dec 28th 2009, 06:40 AM

Originally Posted by Soroban

Hello, khor!

The second one has an incredibly long and tedious solution.

Since it depends on how many diamonds were removed,
. . we must consider all possible cases.

Suppose no $\diamondsuit$ is removed. . This probability is: . $\frac{{39\choose21}}{{52\choose21}}$
The deck contains: 13 $\diamondsuit$ s and 18 Others.
The probability that no $\diamondsuit$ is drawn is: . $\frac{{18\choose2}}{{31\choose2}}$
The probability that no $\diamondsuit$ is removed and at least one $\diamondsuit$ is drawn is: . $\frac{{39\choose21}}{{52\choose21}}\left(1 - \frac{{18\choose2}}{{31\choose2}} \right)$

Suppose 1 $\diamondsuit$ is removed. . This probability is: . $\frac{{13\choose1}{39\choose20}}{{52\choose21}}$
The deck contains 12 $\diamondsuit$ s and 19 Others.
The probability that no $\diamondsuit$ is drawn is: . $\frac{{19\choose2}}{{31\choose2}}$
The probabiity that 1 $\diamondsuit$ is removed and at least one $\diamondsuit$ is drawn is: . $\frac{{13\choose1}{39\choose20}}{{52\choose21}}\le ft(1 - \frac{{19\choose2}}{{31\choose2}}\right)$

. . . $\vdots$

We must extend this list up to:

Suppose 11 $\diamondsuit$ s are removed. . This probability is: . $\frac{{13\choose11}{39\choose10}}{{52\choose21}}<br />$
The deck contains 2 $\diamondsuit$ s and 29 Others.
The probability that no $\diamondsuit$ is drawn is: . $\frac{{29\choose2}}{{31\choose2}}$
The probability that 11 $\diamondsuit$ s are removed and at least one $\diamondsuit$ is drawn is: . $\frac{{13\choose11}{39\choose10}}{{52\choose21}} \left(1 - \frac{{29\choose2}}{{31\choose2}}\right)$

The final probability is the sum of these twelve probabilities.

You go on ahead . . . I'll wait in the car.
.

After you carry out this computation, I think you will find the result is the same obtained by the simpler method I previously posted.

Look at it this way: You shuffle the deck and deal out the top two cards. Or you shuffle the deck, cut 21 cards from the top to the bottom, and deal out the top two cards. What's the difference? In terms of the distribution of the two cards dealt, none, really.

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