Probability Question (1 Viewer)

[xavier_eales]

Found in Cambridge 3U Textbook. Chapter 10H, Q22(a).

A poker hand of five cards is dealt from a standard pack of 52. Find the probability of obtaining one pair.

Deceptively simple, can't get it.

The answer is P(e) = 352/833, how did they get that?

 

[rand_althor]

The probability of getting a pair is: . Once we take a card, there are only 3 of the same value in the rest of the deck, and our sample space reduces by one. For the next 3 cards, we need 3 cards of different values (e.g. 4,5,6). The probability of getting this is: . Our sample space starts at 50 as we previously took 2 cards to get the pair, and it reduces by one when we take a card. The amount of cards that we can take reduces by 4 - once we take a card, we have one less card, and we can't take any of the remaining three cards with the same value. Now, we need to consider the different orders in which we can take these 5 cards. If we consider the cards in the pair to be P and the single cards to be S, we want to find out how many ways we can order P, P, S, S and S. This is given by . Thus, the probability is .

Does anyone have another method?

 

[kawaiipotato]

The probability of getting a pair is: . Once we take a card, there are only 3 of the same value in the rest of the deck, and our sample space reduces by one. For the next 3 cards, we need 3 cards of different values (e.g. 4,5,6). The probability of getting this is: . Our sample space starts at 50 as we previously took 2 cards to get the pair, and it reduces by one when we take a card. The amount of cards that we can take reduces by 4 - once we take a card, we have one less card, and we can't take any of the remaining three cards with the same value. Now, we need to consider the different orders in which we can take these 5 cards. If we consider the cards in the pair to be P and the single cards to be S, we want to find out how many ways we can order P, P, S, S and S. This is given by . Thus, the probability is .

Does anyone have another method?

How come we need to arrange it? I didn't think it would matter that it was something like P,P,S,S,S or P,S,P,S,S

 

[rand_althor]

How come we need to arrange it? I didn't think it would matter that it was something like P,P,S,S,S or P,S,P,S,S

Not confident about this, but I think it's because each is considered a unique way of getting the required configuration (1 pair, 3 singles).

 

[braintic]

13 × 4c2 × 12c3 × 4^3 / 52c5

 

[rand_althor]

13 × 4c2 × 12c3 × 4^3 / 52c5

So there's 13 ways of choosing 2 cards from a group of 4 cards with the same value. The 12C3 is then choosing 3 cards of different value from the 12 remaining card values that haven't been chosen. But why the 4³?

 

[braintic]

So there's 13 ways of choosing 2 cards from a group of 4 cards with the same value. The 12C3 is then choosing 3 cards of different value from the 12 remaining card values that haven't been chosen. But why the 4³?

I guess it's too late for you, but for the sake of posterity:

12C3 is NOT the number of ways of choosing three different CARDS - it is the number of ways of choosing three different DENOMINATIONS.
The 4^3 is then the number of ways of choosing a particular card from each of these denominations.