# Homework Help: Probability involving a three card hand.

1. Jan 14, 2014

### Monstrous Math

Ha! First post! Hope you guys can help me. It's not really a homework problem, but it's close enough.

1. The problem statement, all variables and given/known data

What is the probability of drawing a king, a red card, and a jack from a fair deck of 52 cards, in that order, without replacement?

3. The attempt at a solution

If you draw a red king vs. a black king it can affect the probability of drawing a red card next. Similarly drawing a red jack vs. a non-jack from the red suit for the second card will affect the probability of drawing a jack for the third card.

So I divided it into four possible cases, and added together their probabilities:

Red King, Red Suit (non-Jack), Jack:

$\frac{2}{52}$x$\frac{23}{51}$x$\frac{4}{50}$

Red King, Red Jack, Jack:

$\frac{2}{52}$x$\frac{2}{51}$x$\frac{3}{50}$

Black King, Red Suit (non-Jack), Jack:

$\frac{2}{52}$x$\frac{24}{51}$x$\frac{4}{50}$

Black King, Red Jack, Jack:

$\frac{2}{52}$x$\frac{2}{51}$x$\frac{3}{50}$

∴ summing the probabilities yields a solution of $\frac{2}{663}$ I believe.

This makes sense to me, but I have a friend who is arguing with me and telling me this is wrong. I'd like some confirmation either way, and if it is wrong some help with the correct approach.

I tried using WolframAlpha to check, but it wouldn't recognize my question whenever I tried to ask. If anyone knows the correct syntax for asking WolframAlpha (or Mathematica or Matlab), I'd really appreciate it.

2. Jan 15, 2014

### Dick

It looks correct to me.

3. Jan 15, 2014

### Monstrous Math

Thanks for the reply man, I appreciate it.

4. Jan 16, 2014

### haruspex

There is a more direct approach. Clearly it doesn't change the answer if you specify the order as K, J, red.
The probability of drawing a K then a J is 16/(52*51).
Since the K and J are equally likely red or black, on average you will still have equal number of red and black cards remaining, so the probability of now drawing a red card is 1/2.

5. Jan 16, 2014

### .Scott

Looks good to me as well.

6. Jan 17, 2014

### Monstrous Math

Thanks for the replies, they managed to convince my friend. I think he was trying to do it a different way with the tools he knew (using 52 choose 3), but couldn't work it out to his satisfaction.

I wish I knew how to represent it in Mathematica, since there is a sense in the back of my mind that it knows commands for it. I liked the more direct approach, haruspex, thanks for showing me there's almost always a better, simpler method. I suspect it'll come in handy again for me.