Question

pick four cards randomly, what is the probability that you will get two distinct pairs?

Answer 1

@julay731 First of all welcome to openstudy :

Given that total observations ( total selections ) ---> 4
In our favour ----> 2

probability -- > in favor / total --> 2/4 --> 1/2

May i know that r u doing Permutation ?

Answer 2

I should add that it is out of a deck of 52 cards. This is for a statistics class.

Answer 3

ok ! may be @Callisto help u best of luck :)

Answer 4

thank you

Answer 5

Since I am not sure with my answer..

Answer 6

Also the point that u added : deck of 52 cards will make no diff. in the ans.

Answer 7

okay no prob. thanks

Answer 8

\[([\left(\begin{matrix}4 \\ 1\end{matrix}\right)*\left(\begin{matrix}13 \\ 2\end{matrix}\right)+\left(\begin{matrix}3 \\ 1\end{matrix}\right)*\left(\begin{matrix}13 \\2\end{matrix}\right)]*4!)\div(\left(\begin{matrix}52 \\ 4\end{matrix}\right)*4!)\]
That is select first colour and select two cards and then select the second colour and select the next two cards ==>(nominator) number of ways of selecting 2 pairs from a deck
denominator==>(sample space) number of ways of selecting 4 cards from the deck

Answer 9

thank you that looks right. i need to review this.

Answer 10

it's not what I get

Answer 11

I would say....... probability is my enemy!
0% sure of my answer :S

Assumption: the cards are (A, 2, 3, 4, 5 , 6, 7, 7, 8, 9, 10, J, Q, K) x 4 = 52

First, pick one card out of 52 cards
=> Probability = 52/52 = 1

Next pick another one to make a pair.
=> probability = 3/51 (only 3 cards with the same number or so are left after picking the first one)

After that, randomly pick another card, condition: not same number as the first two
=> probability = (52-4) / 50 = 48/50 = 24/25

Last, pick a pick to form a pair.
=> probability = 3/49

Probability required = 1 x 3/51 x 24/25 x 3/49 = 72/20825

Epic fail.....

Answer 12

I don't get that either

Answer 13

What do you get?

Answer 14

\[\frac{{13 \choose 2}{4\choose 2}^2}{{52\choose 4}}\]

Answer 15

My Lord, do you mind explaining your answer???

Answer 16

I get everything except why did you square it?

Answer 17

there are 13 types of cards in order to form pairs with...choose 2 of them

then from each pair chose 2 cards from the 4 thus \((\,_4C_2)^2\)

Answer 18

this problem is similar to finding the poker prob of getting 2 pairs with 5 cards...


\[\frac{{13\choose 2}{4\choose 2}^2{11\choose 1}{4\choose 1}}{{52\choose 5}}\]

only here we don't need that last card