Randomly selecting cards until nothing but face cards... probability of KH remaining

You randomly remove a playing card from a regular deck of 52 cards, one at a time (without replacement) from a shuffled deck until only face cards (the kings, queens, or jacks) remain in your deck.

What is the probability that the King of Hearts remains in your depleted deck?

Given that you have 12 cards remaining (kings, queens, and jacks), there's a 100% chance that the king of hearts is one of the 12 cards remaining.

I don't think that's what you're asking :)

Thanks. Maybe it would be clearer if I removed "(the kings, queens, or jacks)" from the question:

You randomly remove a playing card from a regular deck of 52 cards, one at a time (without replacement), from a shuffled deck until only face cards remain in your deck.

What is the probability that the King of Hearts remains in your depleted deck?

Quote: EdCollins

You randomly remove a playing card from a regular deck of 52 cards, one at a time (without replacement) from a shuffled deck until only face cards (the kings, queens, or jacks) remain in your deck.

What is the probability that the King of Hearts remains in your depleted deck?

I'll give this a try!

1.) There is only a 12/52 chance that you would even have anything left at the end of it, because if the deck ends with a non-face card, it's over.

The probability of the deck even ending with a face card is .23076923077, so that's a good starting point.

2.) We're going to go by what the deck ends with, and the total probability should end up being .23076923077 pursuant to #1.

Deck ends with one face card, previous card is A-10

12/52 * 40/51 = .18099547511

The odds of that card being the King of Hearts are 1/12, so: .18099547511 * 1/12 = .015082956259

Deck ends with two face cards, previous card is A-10

12/52 * 11/51 * 40/50 = 0.03981900452

The odds of that card being the King of Hearts are 2/12, so: .03981900452 * 1/6 = 0.00663650075

Deck ends with three face cards, previous card is A-10

12/52 * 11/51 * 10/50 * 40/49 = .00812632745
The odds of that card being the King of Hearts are 3/12, so .00812632745 * 1/4 = .00203158186

Deck ends with four face cards, previous card is A-10

12/52 * 11/51 * 10/50 * 9/49 * 40/48 = .001523686398

The odds of that card being the Kh are 4/12, so .001523686398 * 1/3 = .00050789547

Deck ends with five face cards, previous A-10

12/52 * 11/51 * 10/50 * 9/49 * 8/48 * 40/47 = .00025935088

The odds of that card being the Kh are 5/12, so .00025935088 * 5/12 = .00010806287

Deck ends with six face cards, previous card A-10

12/52 * 11/51 * 10/50 * 9/49 * 8/48 * 7/47 * 40/46 = .000039466438

The odds of that being Kh are 6/12, so .000039466438 * 1/2 = .00001973322

Deck ends with seven face cards, previous card A-10

12/52 * 11/51 * 10/50 * 9/49 * 8/48 * 7/47 * 6/46 * 40/45 = .000005262191

The odds of that being Kh are 7/12, so .000005262191 * 7/12 = .00000306961

Deck ends with eight face cards, previous card A-10

12/52 * 11/51 * 10/50 * 9/49 * 8/48 * 7/47 * 6/46 * 5/45 * 40/44 = .00000059798

The odds of that being Kh are 8/12, so .00000059798 * 2/3 = .00000039865

Deck ends with nine face cards, previous card A-10

12/52 * 11/51 * 10/50 * 9/49 * 8/48 * 7/47 * 6/46 * 5/45 * 4/44 * 40/43 = .00000005563

The odds of that being Kh are 9/12 so .00000005563 * 3/4 = .0000000417223

Deck ends with ten face cards, previous card A-10

12/52 * 11/51 * 10/50 * 9/49 * 8/48 * 7/47 * 6/46 * 5/45 * 4/44 * 3/43 * 40/42 = .000000003973

The odds of that being Kh are 10/12 so .000000003973 * 5/6 = .00000000331

Deck ends with eleven face cards, previous card A-10

12/52 * 11/51 * 10/50 * 9/49 * 8/48 * 7/47 * 6/46 * 5/45 * 4/44 * 3/43 * 2/42 * 40/41 = .00000000019

The odds of that being Kh are 11/12 so .00000000019 * 11/12 = .00000000017

Deck ends with All 12 Faces

12/52 * 11/51 * 10/50 * 9/49 * 8/48 * 7/47 * 6/46 * 5/45 * 4/44 * 3/43 * 2/42 * 1/41 = .000000000004

NEXT POST TO ADD RESULTS

My initial guess is about 2.44%

I end up with a total of .23076923074 for all results ending in a face card, so with 12/52 = .23076923076, errors can be attributed to rounding. Additionally, I lost a full .000000000004 because I was only going to the eleventh decimal place on my adding. Effectively, the twelve face cards = 0.

I get a result of .02439024389, or 1/.02439024389 = 1:41.000000021

Great guess, JB!!!

As always, thanks for responding.

1 out of 41 was the answer in another forum when this question was asked... and I disagreed with it. I disagreed with it because of the way the question was worded. I "read" the question differently, apparently. But I can now see I'm obviously in the minority.

That was the question as I understood it. Note the capital letters:

"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?"

To me, this meant that if you happen to remove all of the face cards first, and there were still cards in the deck, then you didn't have a deck with only face cards, so that didn't count.

To clarify, the stipulation in the problem was to remove cards from a deck until only face cards remain. So if NO face cards remained, well then, you didn't adhere to the stipulation. So you simply try again/toss out that trial. I considered that trial irrelevant and that it shouldn't be considered - it wasn't part of the "equation."

But when you DO remove cards from a deck until "only face cards remain," then at that point you must have between 1 and 12 cards left in your deck. Zero isn't possible.

And, of course, if there was just ONE face card left in the deck, I knew the chances of it being the king of hearts were exactly .0833. (1/12) And naturally, the more face cards that are left in the depleted deck, the greater the chance the King of Hearts is among them. So the answer had to be something greater than .0833.

I wrote a program to simulate this question/problem but of course I wrote it to not include/count any trials where all of the face cards were drawn prior to the end of the deck.

(I hope that's all clear.)

So, again, I can see I'm in the minority - I wasn't including all trials.

Quote: EdCollins

"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?"

To me, this meant that if you happen to remove all of the face cards first, and there were still cards in the deck, then you didn't have a deck with only face cards, so that didn't count.

Say the last card is a non-face card. By the procedure from the question you remove cards until only face cards remain. This includes removing the full deck. Now the original question is not very specific what you do with an empty deck, since you cannot remove further cards from an empty. Your guess was you start the feat again. The other option is you still look for the king of hearts. Since no cards are left, probability of drawing the king of hearts is zero.