# Help me troubleshoot my answer

• Jun 11th 2007, 03:05 AM
CrazyAsian
You are playing a poker game. You have 55 and the board reads 358 giving you 3 of a kind. There are two more cards to come. What's the probability you make a full house or better?

Here's my answer. You can only make a fullhouse or 4 of a kind. So on the first of the two cards to come, you can hit 7 out of 47 cards (3,5,or8). If you miss on the first card, you can hit 10 out of 46 cards. It's 10 because you can not only hit a 3, 5, 8 but also make a fullhouse by getting a pair in the last two cards. So, 1 - (prob(missing both cards)) = 1-(40/47 * (36/46)

But it should also add up to the probability of hitting on the 1st card + probability of hitting on the second card - the probability of hitting on both 1st and second card. So, 7/46 + 10/46 - 6/46. But my two answers don't tie. What am I doing wrong?
• Jun 11th 2007, 05:07 AM
MissTK
I think your answer is definitely correct (I got the same answer doing it a different way than you).

I think it's your checking that's gone wrong! I don't really understand where you're getting your numbers in the second part. The probabilities are

A) Miss first and second (40/47)*(36/46)=1440/2162
B) Miss first, hit second (40/47)*(10/46)=400/2162
C) Hit first (no matter what happens with second) 7/47 = 322/2162

Edit because I think I was wrong:
If you take 10/46+7/47-(7/47*10/46) = 722/2162 you will notice that this is the same as 1-P(A) (what you wanted).
• Jun 11th 2007, 07:01 AM
Soroban
Hello, CrazyAsian!

Quote:

You are playing a poker game.
You have $\{5,5\}$ and the board has $\{3,5,8\}$ giving you 3-of-a-kind.
There are two more cards to come.
What's the probability you make a full house or better?

You are complicating the problem by ordering the draws.

There are: ${47\choose2} = 1081$ possible draws.

Four-of-a-kind
You must get the fourth 5 . . . There is 1 way.
Your fifth card can be any of the remaining 46 cards.
. . There are: . $1 \times 46 \:=$ 46 ways to get 4-of-a-kind.

Full House . . . There are 3 ways.

[1] Draw another 3 and some other card.
There are: ${3\choose1} = 3$ ways to draw another 3
. . and $44$ choices for the fifth card.
There are: $3 \times 44 \:=\:132$ to get "5s over 3s".

[2] Draw another 8 and some other card.
There are: ${3\choose1} = 3$ ways to draw another 8.
. . and $44$ choices for the fifth card.
There are: $3 \times 44\:=\:132$ ways to get "5's over 8's".

[3] Draw a different pair (other than 3s or 8s).
There are: $10$ choices of the value of the pair
. . and ${4\choose 2} = 6$ ways to get the pair.
There are: $10 \times 6\:=\:60$ more ways to get a Full House.

Hence, there are: . $132 + 132 + 60 \:=$ 324 ways to get a Full House.

Therefore, there are $46 + 324\:=$ 370 ways to get a Full House or better.

. . The probability is: . $\frac{370}{1081}$