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Help with a probability problem, please?

O

oLLie

Diamond Member
Hello ATOT,

I have two probability problems I'm having trouble with. The teacher lets us work together as long as we state who we worked with, so if you guys could help me out, I'll gladly put ATOT at the top as my help. 🙂

Here are the problems:

1) Compute the probability that a hand of 13 cards contains the ace and king of at least one suit.
This is how I tried the problem: as far as I can see there is only 1 thing that can vary for the experiment to succeed: the suit.
So I did (4 choose 1) //because there are 4 ways to choose 1 suit multiplied by (50 choose 11) //50 cards left (after the ace and king of the selected suit) to choose the remaining eleven divided by the total number of hands of 13 in a 52 card deck. My logic was that once the suit was chosen, there was only 1 way to get the king and ace of that suit (so imagine a 1 choose 1 and another 1 choose 1 in the numerator, if you want).

(4 choose 1)*(50 choose 11)
______________________
(52 choose 13)

this comes out to be around ~0.235 and the answer in the book is 0.2198
could someone tell me the correct approach?

Thanks everybody,
Ollie
 
The way I understand your problem, there are two ways to achieve the final outcome.

1. Both the King and Ace are of the same suit
2. Both are from different suits

Calculate the probability of each, then multiply the two results.
 
Originally posted by: KMurphy
The way I understand your problem, there are two ways to achieve the final outcome.

1. Both the King and Ace are of the same suit
2. Both are from different suits

Calculate the probability of each, then multiply the two results.
Click to expand...

I'm not sure I understand what you mean. It says "the ace and king of at least one suit", so it sounds like the ace and king from 1 suit, or the ace and king from more than 1 suit, it doesn't sound like different suits.
 
Ah ha... finally got it. I decided to calculate the probably of that situation NOT happening. So it's the sum of several situations.
1) No aces or kings
2) Ace xor king of 1 suit
3) Ace xor king of 2 suits
4) Ace xor king of 3 suits
5) Ace xor king of 4 suits

Case 1) 44 c 13
Case 2) 44 c 12 * 2 (for either king or ace) * 4c1 (for which suit)
Case 3) 44 c 11 * 2^2 * 4c2
Case 4) 44 c 10 * 2^3 * 4c3
Case 5) 44 c 9 * 2^4 * 4c4

1 - sum of all that/52c13 = 0.21978
 
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