Sheldon Ross Chapter-2 Question-15b

Question

If it is assumed that all $\binom{52}{5}$ poker hands are equally likely, what is the probability of being dealt two pairs? (This occurs when the cards have denominations a, a, b, b, c, where a, b, and c are all distinct.)

 

my approach is:
selecting a denomination=$\binom{13}{1}$ ways and selecting two suits=$\binom{4}{2}$

so selection of first pair=$\binom{13}{1}*\binom{4}{2}$

Similarly, for the other pair, ways of selecting the second pair=$\binom{12}{1}*\binom{4}{2}$

and the remaining one card can be chosen in $\binom{11}{1}*\binom{4}{1}$ ways

So, probability=$\frac{\binom{13}{1}*\binom{4}{2}*\binom{12}{1}*\binom{4}{2}*\binom{11}{1}*\binom{4}{1}}{\binom{52}{5}}$

I’m getting answer 0.095 but in the book answer is given 0.0475

where am I going wrong?

Comments

I have done like this

$\frac{^{4}\textrm{C}_{1}\times 13\times 12\times ^{3}\textrm{C}_{1}\times 13\times 12\times ^{2}\textrm{C}_{1}\times 13}{52\times 51\times 50\times 49\times 48}$

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@aditi19 say Spade 2, 3 and Diamond 2, 3 are chosen. In your counting you are counting them twice as spade being selected first and spade being selected second are counted seperately.

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but @Arjun sir
it's not necessary that both the pairs will be from the same two suits. it can be spade 2, diamond 2 and hearts 3, clubs 3

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I just told an example. In all cases, whichever you are selecting, there is an alternate selection which will give you the same selection -- you are giving order of selection importance in numerator.

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 you are giving order of selection importance in numerator

got it @Arjun sir, thank u

 this line cleared my doubt

so suppose I selected clubs 2 and diamond 2, then diamond 2 and clubs 2 was also also getting counted but both are same.. right?

so i need to divide the numerator by 2 then?

 

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yes, exactly.

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@aditi19

u selected card first, and then selected suits

right?

But in card playing, we first select a suit and then card from it

isn't it?

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@srestha

But in card playing, we first select a suit and then card from it

u can either ways  also. lets say you have chosen spades out of 4 suits=4C1

now there are 13C1 ways of choosing a denomination of spades=13C1

so choosing a card of spades can be chosen in 4*13 ways

is same as choosing a denomination first and then the suit

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@aditi19

if that is for 1 card, then that is ok.

But here selection is not one.

We selected a,a,b,b,c

Means, in a deck of card four types of cards are Hearts, Spreads, Diamonds , Clubs.

right?

Now take they want to dealt with 2 pairs and one pair is single.

Here u have taken $^{4}\textrm{C}_{2}$ twice. What that really mean?

It means repetition of cards.

Isnot it?

Correct order should be $^{4}\textrm{C}_{1}\times ^{13}\textrm{C}_{2}$ //that is card a,a as they mentioned

and then $^{3}\textrm{C}_{1}\times ^{13}\textrm{C}_{2}$ // Card b,b as here no repetition is allowed

then $^{2}\textrm{C}_{1}\times ^{13}\textrm{C}_{1}$// Card c

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in your solution for a,a you've selected 2 different denominations and one suit -> so you've chosen two cards from different denominations from the same suit -> it is not a, a. suppose I've selected ace and 2 of hearts as per your solution

for b,b, you have chosen one of the remaining 3 suits(different from what has been chosen for a,a) and you choose two different cards from this suit   -> it is not b,b. further you've told that there are 13 ways to do that. so there may be a card of same denomination from my previous choice. Say I choose ace and 3 of clubs

in the last one you choose one of the remaining 2 suits and one card from that suit. suppose I choose ace of diamonds

so my cards are

ace-heart, ace-clubs, ace-diamonds, 2-hearts, 3-clubs

there is no two pairs

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@srestha

for this kind of problems best take images from google like this

and think how can u select this set of cards

P.S. this works :D

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it is not a, a. suppose I've selected ace and 2 of hearts as per your solution
 

No, First we need to choose Heats,then choose two Heart card, that is what I have done .

 it is not b,b. further you've told that there are 13 ways to do that

I havenot got u here. 

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a, a are cards of same denominations from different suits like ace of hearts and ace of diamond

u can't make a pair from the same suit

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ok, I understood, what u have done.

what is meaning of 

probability of being dealt two pairs

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done @aditi19

Ans is in 15. c)

words confusing me :)

http://www.stat.ucla.edu/~rosario/classes/081/100a-2a/100aHW2Soln.pdf

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I agree
I was confused with my own solution
Arjun sir cleared it

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yes.

Actually among 13 denomination, we choose 2 cards at a time.

Now for pairing, among 4 same type cards, we chooses 2 cards for each of that two denominations.

Now select one odd pair and from 4 suit select one suit

right?

Nice question

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yes

Answer 1

I've done in this way.

For selecting 2 pairs, first, we choose 2 types of cards out of possible 13. So, we get 13C2. Now, for each pair, we have to select 2 colours out of 4. So, now, we get 13C2 * 4C2 * 4C2,  as final for selecting the pairs.

But, now for selecting the last card, we will select 1 card from rest 11 numbers, which will be 11C1, but it can be of any 4 colours, so for single card, the number of combinations are 11C1 * 4C1.

Finally, answer :

13C2 * 4C2 * 4C2 * 11C1 * 4C1 / 52C5 = 0.0475