Solve Card Probability: King of Hearts in Deck

[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.

 

[jim mcnamara]

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]

 

[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. I am having trouble trying to figure out how many face cards are left.

 

[Kolmin]

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. I am having trouble trying to figure out how many face cards are left.

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.

 

[Ray Vickson]

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.

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?

 

[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.

 

[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...

 

[haruspex]

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

 

[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.

 

[haruspex]

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

No, it's a fair bit higher than that. Try my hint.

I'm not sure whether the question is stating that all of the face cards are left.

No, only that the HK and any number of other face cards are left, but no face cards.

 

[Biosyn]

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.

Wait, sorry I'm confused.

'But no face cards' ?

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

 

[haruspex]

Wait, sorry I'm confused.

'But no face cards' ?

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

Sorry, I meant no non-face cards.

 

[Biosyn]

Sorry, I meant no non-face cards.

Oh, okay.

Is this your hint?

"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) ...

 

[haruspex]

Is this your hint?

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

 

[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.

 

[Biosyn]

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

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?

 

[jim mcnamara]

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:

With a deck of shuffled cards, what is the probability of not having drawn the KH,
and having the only cards remaining in the deck be face cards.

That means you cannot have drawn the KH.

 

[Ray Vickson]

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.

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.

 

[haruspex]

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:

With a deck of shuffled cards, what is the probability of not having drawn the KH,
and having the only cards remaining in the deck be face cards.

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.

 

[haruspex]

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.

Quite so - see post #8.

 

[Biosyn]

Quite so - see post #8.

Would you mind taking a look at my previous post?

 

[Ed_Collins]

I'm going to jump in and contribute here. Maybe I know what I'm talking about and maybe I don't.

First, I'm going to assume the original poster meant 12 face cards (four kings, queens, and jacks) and not 16.

As I understand the question, the percentage has to be at LEAST .0833.

Naturally, when nothing but face cards are left in the deck, the king of hearts has an equal chance of being left as any other face card. And 1 chance out of 12 is .0833.

Most of the time, all face cards will be drawn prior to the end of the deck. And when that happens, the entire "sim" or "game" or "trial" can be thrown out, since it doesn't meet the requirements. The requirements state

..."until only face cards remain in the deck."

I wrote a computer program to simulate this. I'm still checking my code for errors, but as of now, after a short, 5 million trial run, 76% of the time I must exit the sim and start over... since I removed all of the face cards PRIOR to running through the deck.

And yet when I DO COME TO A TRIAL when NOTHING but face cards remain in the deck, I can then check the remaining cards. And .1059 percent of the time the King of Hearts is among those few remaining cards.

As I read through the posts, it seems clear to me that not everyone is on the same page, regarding the actual question.

 

[haruspex]

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

Not sure what you mean. I'm saying, suppose we draw from the top always. (It can't matter whether we do that or draw randomly from anywhere within the deck.) Consider the positions of the non face cards and the HK within that deck at the start. For what initial positions will the HK be left when we've drawn all non-face cards? Does it matter where the other face cards are?

I think he meant 12 face cards in a deck.

Whatever. Say there are F face cards, D cards in the full 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?

I'm not following what you're saying there. "Old" deck? Removing them as part of the draw process or throwing them away in advance? But you have the right answer.

 

[Biosyn]

I'm not following what you're saying there. "Old" deck? Removing them as part of the draw process or throwing them away in advance? But you have the right answer.

I meant removing the king of hearts as part of the process of randomly drawing them. So the correct answer is: 1/(52-11) = 1/41 ?

Seems too simple...if someone would please explain indepth why that is.

 

[haruspex]

So the correct answer is: 1/(52-11) = 1/41 ?

Yes, if only 12 face cards. More generally, 1/(D-F+1), where D in the deck, F face cards, one face card of interest.
All that matters is that the HK would be drawn (if we were to continue right through the deck) after all non-face cards have been drawn. Where in the sequence other face cards are drawn is immaterial. So it comes down to the prob that the HK is the last of those D-F+1 cards to be drawn.

 

[Ed_Collins]

Guys, I'm confused. Someone please educate me. I'd appreciate it. Thanks.

We've got a depleted deck of cards. We know there are nothing but face cards left in this depleted deck. Thus, we know this depleted deck contains at LEAST one card, but no more than 12 cards. We want to know what the probability is that the king of hearts is in this depleted deck.

Isn't that the question? It is as I understand it, and maybe that's where I'm going wrong:

"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?"

If you were to tell me that there was just ONE face card left in the deck, I know the chances of that are exactly .0833. (1/12)

And naturally, the more cards that ARE in the depleted deck, the greater the chance the King of Hearts is among them.

What's interesting is that there aren't going to be very many cards at all in this depleted deck. Yes, it COULD contain as many as 12 cards, the chances of that happening are incredibly, incredibly small. (To arrive at that, all 40 of the non face cards would have to have been randomly removed before ANY faces cards at all.)

Even 11 cards in this deck is small. And 10. And 9. As you get closer and closer to 1, the chances grow. Most of the time there is just a card or two, and that's why the percentage should be not too much more than .0833.

Below are the results from my last computer simulation:

Number of total trials: 10,000,000
Number of trials to toss out: 76,92,994 (Depleted Deck never had "nothing but face cards.")

Number of valid trials: 2,307,006
Number of times KH was NOT among face cards remaining from valid trials: 2,062,278
Number of times KH WAS among face cards remaining from valid trials: 244,728
King remaining probability: .10608

 

[Ray Vickson]

Guys, I'm confused. Someone please educate me. I'd appreciate it. Thanks.

We've got a depleted deck of cards. We know there are nothing but face cards left in this depleted deck. Thus, we know this depleted deck contains at LEAST one card, but no more than 12 cards. We want to know what the probability is that the king of hearts is in this depleted deck.

Isn't that the question? It is as I understand it, and maybe that's where I'm going wrong:

"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?"

If you were to tell me that there was just ONE face card left in the deck, I know the chances of that are exactly .0833. (1/12)

And naturally, the more cards that ARE in the depleted deck, the greater the chance the King of Hearts is among them.

What's interesting is that there aren't going to be very many cards at all in this depleted deck. Yes, it COULD contain as many as 12 cards, the chances of that happening are incredibly, incredibly small. (To arrive at that, all 40 of the non face cards would have to have been randomly removed before ANY faces cards at all.)

Even 11 cards in this deck is small. And 10. And 9. As you get closer and closer to 1, the chances grow. Most of the time there is just a card or two, and that's why the percentage should be not too much more than .0833.

Below are the results from my last computer simulation:

Number of total trials: 10,000,000
Number of trials to toss out: 76,92,994 (Depleted Deck never had "nothing but face cards.")

Number of valid trials: 2,307,006
Number of times KH was NOT among face cards remaining from valid trials: 2,062,278
Number of times KH WAS among face cards remaining from valid trials: 244,728
King remaining probability: .10608

In theory the probability that a trial must be tossed out is the probability that a non-face card is last; for 12 face cards and 40 non-face cards, this probability is 40/52. Thus, the probability of a valid trial is 12/52, which is a bit less than 1/4. So, your simulation seems OK for that aspect.

For a 52-card deck with 12 face cards the final probability we want is 1/41, which is about 0.02439. Your simulation gives 244,728/10,000,000 = 0.02447, which is near the theoretical value.

For a 52-card deck with 16 face cards the final probability we want is 1/37. In general, if there are F face cards and N non-face cards, the probability we want is 1/(N+1).

 

[Ed_Collins]

Thank you. Yes, I noticed that .02439 was very close to .02447.

But if the trial is to be tossed out, (40/52 times) in my mind you shouldn't consider it a trial. Only the times when face cards remain should be considered a trial. That's what was asked.

Of course, maybe this is just a matter of semantics or how I interpret the question.

Thanks again.

 

[haruspex]

we know this depleted deck contains at LEAST one card,

No, it may be empty.

but no more than 12 cards. We want to know what the probability is that the king of hearts is in this depleted deck.

Isn't that the question? It is as I understand it, and maybe that's where I'm going wrong:

"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?"

It would be better worded as "no non-face cards remain", but it comes to the same.

If you were to tell me that there was just ONE face card left in the deck, I know the chances of that are exactly .0833. (1/12)

And naturally, the more cards that ARE in the depleted deck, the greater the chance the King of Hearts is among them.

What's interesting is that there aren't going to be very many cards at all in this depleted deck. Yes, it COULD contain as many as 12 cards, the chances of that happening are incredibly, incredibly small. (To arrive at that, all 40 of the non face cards would have to have been randomly removed before ANY faces cards at all.)

Sure. And you can write down the probability that r cards remain: from an original deck of D cards of which F are face cards, there are ^FC_r^D-FC₁ ways in which the last r are face cards but the last r+1 are not. That's out of a total of ^DC_r+1 possibilities for the last r+1. So the prob of having the HK remaining in a final stack of r face cards is (r/F) ^FC_r^D-FC₁/ ^DC_r+1. You can get the right answer by summing those, but there's an easier way.

Number of total trials: 10,000,000
Number of trials to toss out: 76,92,994 (Depleted Deck never had "nothing but face cards.")

There are none to toss out. If no face cards remained then the HK did not.

Number of valid trials: 2,307,006
Number of times KH was NOT among face cards remaining from valid trials: 2,062,278
Number of times KH WAS among face cards remaining from valid trials: 244,728
King remaining probability: .10608

Re-including the omitted failures, 0.0245. That's reasonably close to the correct answer 1/41 = 0.0239.

 

[Ed_Collins]

Quote by Ed_Collins
"...we know this depleted deck contains at LEAST one card..."

Reply by haruspex
"No, it may be empty."

Thank you for responding. But this is where I disagree.

The stipulation is to remove cards from a deck until only face cards remain. If NO face cards remain, well then, you didn't adhere to the stipulation, did you. So you try again/toss out trial. It's irrelevant and shouldn't be considered. It's not a part of the "equation."

When you DO remove cards from a deck until "only face cards remain," you will have between 1 and 12 cards left in your deck. Zero isn't possible.

Just for fun, I'm asking the question in another forum, where many of the members are very math-oriented. It will be interesting to see their responses and how they interpret the question. (I'm not pointing them to this forum.)

Thanks again.

 

[Michael Redei]

Quote by Ed_Collins
"...we know this depleted deck contains at LEAST one card..."

Reply by haruspex
"No, it may be empty."

Thank you for responding. But this is where I disagree.

The stipulation is to remove cards from a deck until only face cards remain. If NO face cards remain, well then, you didn't adhere to the stipulation, did you. So you try again/toss out trial. It's irrelevant and shouldn't be considered. It's not a part of the "equation."

When you DO remove cards from a deck until "only face cards remain," you will have between 1 and 12 cards left in your deck. Zero isn't possible.

Just for fun, I'm asking the question in another forum, where many of the members are very math-oriented. It will be interesting to see their responses and how they interpret the question. (I'm not pointing them to this forum.)

Thanks again.

I'd be most interested to see what "very math-oriented" people say. If they agree with you, ask them please what they would expect if you take the set of integers and remove as many as possible until "only numbers that are both even and odd remain". The wording is the same as with the card example, so if these people accept the empty set as a mathematical concept, they shouldn't agree with your saying "zero isn't possible".

In everyday life, we might say "the guests left the party, until only the hosts remained", and we mean that there actually were some hosts (more than one, as well). Mathematically, this can be understood as "until no one remained who wasn't a host", and that could be just one host, or even nobody (the hosts might have beren called away from their own party earlier).

 

[haruspex]

The stipulation is to remove cards from a deck until only face cards remain. If NO face cards remain, well then, you didn't adhere to the stipulation, did you.

As Michael Redei says, it is normal in mathematics to accept the empty set as a valid subset of any set. The property "only face cards are in the deck" is the same as "every card that is in the deck is a face card". If the deck is empty that becomes vacuously true.

 

[Biosyn]

http://en.wikipedia.org/wiki/Vacuous_truth

Having no cards left is possible.