# Homework Help: Card Probabilities

1. Dec 14, 2012

### jakegoodman

I need some help solving this homework problem.
If you were to randomly select a card one at a time without replacement from a shuffled deck until only face cards remain in the deck, what is the probability that the king of hearts remains in the deck? Assume there are 16 face cards in a 52 card deck.

2. Dec 14, 2012

### Staff: Mentor

IF you have already removed all of the "non" face cards, you now have a deck of sixteen cards. How many of those remaining cards are the king of hearts?

The probability is: [how many king of hearts cards remain] / [cards left in the small deck]

3. Dec 14, 2012

### jakegoodman

I think the problem with your logic is that you don't necessarily have sixteen cards left in the deck.. it could be any number of face cards. Im having trouble trying to figure out how many face cards are left.

4. Dec 14, 2012

### Kolmin

My two cents: Jim McNamara logic is sound.

The problem with your logic is that - if you assume that the tricky side of your problem is not in the way in which it is written, but in some mathematical content - then you have to consider the possibility that you take the 16 face cards from the first sixteen selections... (remote but possible - if you want it's easy to get the probability of that kind of shot).

On the contrary I agree with jim mcnamara: it seems that the tricky side is in the way in which the problem is written down. It gives you an info that it's basically not relevant, while as a solver you should focus only on the fact that in the end you have to compute the probabilities of getting the king of hearts out of the 16 face cards.

5. Dec 14, 2012

### Ray Vickson

There is something wrong with the problem statement. Sometimes all cards would be removed, because the very last card is non-face. So, does the problem mean to ask for a conditional probability that, given only face cards left, the K of hearts is present, or is it asking for the probability that some face cards (but no others) are left, and that one of these is the K of hearts?

6. Dec 14, 2012

### haruspex

I disagree with all other responders so far, radically so with Mcnamara and Kolmin .
It's fairly clear that there is a nonzero chance that you will have no cards left, and that this simply constitutes one of the "HK not in remaining deck" cases.
Try thinking of it working up from the bottom of the deck. See if you can get a recurrence relation on the probability the HK will be in the remaining deck given that the last r cards are other face cards.

7. Dec 14, 2012

### awkward

Suppose we think of the cards as being laid out from left to right, corresponding to top to bottom of the deck. The king of hearts will remain in the deck if and only if all the non-face cards are to its left; the locations of the other face cards are irrelevant. So...

8. Dec 14, 2012

### haruspex

Ha!, there's a much easier way. Think about where the HK needs to be in relation to the non-face cards.

9. Dec 14, 2012

### Biosyn

I'm interested in the solution to this problem too since I'm taking an AP Stats course.

Can it be the probability of not drawing the king of hearts when randomly selecting cards from the deck?
(1 - 51/52)

Because there can be the possibility of having only one card left that is the face card.
I'm not sure whether the question is stating that all of the face cards are left.

Last edited: Dec 14, 2012
10. Dec 15, 2012

### haruspex

No, it's a fair bit higher than that. Try my hint.
No, only that the HK and any number of other face cards are left, but no face cards.

11. Dec 15, 2012

### Biosyn

Wait, sorry I'm confused.

'But no face cards' ?

You stated that only 'the HK and any number of other face cards are left'

12. Dec 15, 2012

### haruspex

Sorry, I meant no non-face cards.

13. Dec 15, 2012

### Biosyn

Oh, okay.

"Try thinking of it working up from the bottom of the deck. See if you can get a recurrence relation on the probability the HK will be in the remaining deck given that the last r cards are other face cards."

I don't know what recurrence relation means.

is it the probability of picking face cards? (16/52) * (15/51) * (15/50) ...

14. Dec 15, 2012

### haruspex

No, I meant my later post: Think about where the HK needs to be in relation to the non-face cards.

15. Dec 15, 2012

### Michael Redei

I'm not sure how one would count all possibilities of finding the HK among the only-face-cards recursively. I'd go about this as follows:

At some point you've removed all 36 non-face cards, plus some number k of face cards (0 ≤ k ≤ 16). You know that card number (36+k) must have been one of the 36 non-face cards, that k of the 15 non-HK face and the remaining 35 non-face cards must be distributed among the first (35-k) places, and that the HK plus (15-k) face cards are distributed among the final (16-k) positions.

Add the resulting expression for 0 ≤ k ≤ 16, divide the sum by 52!, and that should be the probability you're looking for.

16. Dec 15, 2012

### Biosyn

I suppose HK would need to be in the old deck?

I think he meant 12 face cards in a deck.

So, if I removed all of the face cards except for the HK, the probability of HK being in the old deck would be 1/(52-11)? Couldn't I also have removed the HK?

17. Dec 15, 2012

### Staff: Mentor

I agree that the problem statement is confusing. I took it to mean someone created a deck of 16 face cards. Which is wrong.

A more understandable version to me is:
That means you cannot have drawn the KH.

Last edited: Dec 15, 2012
18. Dec 15, 2012

### Ray Vickson

The event we want occurs if the last (k+1) cards are the KH followed by k face cards, for k = 0,.., 15. We can compute the probabilities of these 16 possibilities and sum them. It helps to use a computer algebra system and to keep all results as rational numbers, because doing that produces a very simple final result. The final result is so simple and so obviously related to the number of face and non-face cards that it makes one suspect that a clever and insightful simple solution must exist.

19. Dec 15, 2012

### haruspex

The OP was perfectly clear: keep drawing at random until no non-face cards are left; what is the probability the HK has not been drawn? In contrast, the version quoted above is opaque. I'm not at all sure it's saying the same thing.

20. Dec 15, 2012

### haruspex

Quite so - see post #8.