# Combinations and Card Drawing

1. Apr 21, 2016

### joshmccraney

1. The problem statement, all variables and given/known data
What is the probability of choosing 1 Jack from a card deck and then one heart (no replacement)?

2. Relevant equations
None come to mind.

3. The attempt at a solution
I was thinking $(4C1)/(52C1)\cdot(13C1)/(51C1)$ but what happens if the jack drawn is a heart? Do I make cases? Then I'm thinking something like $(1C1)/(52C1)\cdot(12C1)/(51C1)$ added to the above probability? Please help.

2. Apr 21, 2016

### haruspex

You cannof simply add the HJ case to what you already had since that includes drawing the HJ.
Yes, you could break it into mutually exclusive cases, but there really is no need. On average, how much is the number of Hearts reduced by the drawing of a J? On average, how much is the number of Spades reduced by the drawing of a J?

3. Apr 21, 2016

### joshmccraney

1/4
1/4

I write 1/4 because the odds of a J being a spade that we draw would be one of four suits, right?

4. Apr 21, 2016

### haruspex

Right. So after drawing a J, how does the probability that the next card is a Heart compare with the probability that it is a Spade (or any other specific suit)?

5. Apr 21, 2016

### joshmccraney

It is the same, regardless of the suit. Sorry, I'm trying to think instead of being spoon-fed the answer but I'm not seeing where you're going with this.

6. Apr 21, 2016

### haruspex

Right. And what must those four probabilities (the suit of the second card given that the first is a J) add up to?

7. Apr 21, 2016

### joshmccraney

The four add to 1. The odds of picking a Jack that is heart, spade, diamond, clover are 1.

8. Apr 21, 2016

### haruspex

So what is the probability that the second card is a H given the first was a J?

9. Apr 21, 2016

### joshmccraney

$1/4 \cdot 1/4$? I'm lost, sorry.

10. Apr 21, 2016

### Ray Vickson

There are two cases: (1) The first drawn Jack is also a Heart; and (2) the first drawn Jack is not a Heart. Is the probability of next drawing a Heart the same in cases (1) and (2)? When you say there is "no replacement", doesn't that mean that the first drawn card is now absent from the deck?

11. Apr 21, 2016

### haruspex

I see Ray is taking you along the path of breaking it into cases, but I shall persevere with showing it is unnecessary. Please do not be confused by the two approaches.
Notation: let the events be JH that first card is a J and 2nd is a H, JS that the 1st is a J and 2nd a S etc. Similarly conditionals, H|J etc.
You agreed that P(H|J)=P(S|J)=P(D|J)=P(C|J), and that they add up to 1. So what is each of them numerically?

12. Apr 22, 2016

### joshmccraney

Nope, they are not the same. So would the probability be $$\frac{1C1}{52C1}\frac{12C1}{51C1} + \frac{3C1}{52C1}\frac{13C1}{51C1}$$

Yep.

P(H|J)=P(S|J)=P(D|J)=P(C|J)=(1C1)/(4C1), since each case is identical!

13. Apr 22, 2016

### haruspex

Yes. What does that simplify to?

14. Apr 22, 2016

### joshmccraney

1/52

15. Apr 22, 2016

### haruspex

So the methods agree, right? Always reassuring.

16. Apr 22, 2016

### joshmccraney

Thanks for the help!

17. Apr 22, 2016

### joshmccraney

And yes, it's reassuring for sure when different approaches yield the same solution.

18. Apr 22, 2016

### haruspex

Fwiw, I've just thought of another interesting way to look at it. Construct a mapping from the actual arrangement of cards in the deck to one in which all the denominations are the same at the various positions (e.g. if the original deck goes Q383A... then so does the mapped deck) but the suits are all rotated around by one position (so if the original deck goes HHSCDS... then the mapped deck goes HSCDS....H.)
The event of drawing a J then a H in the first deck maps to the event of drawing the HJ straight away in the second.
The mapping is clearly 1-1 onto (a bijection).

19. Apr 22, 2016

### joshmccraney

That is interesting, but I'm not seeing the relevance. Could you explain please?

20. Apr 22, 2016

### haruspex

It means that the probability of drawing a J then a H is the same as the probability of drawing the HJ in a single draw, so 1/52.

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