Old Problem Revisited - Gambling and Probability

**BruceZ** · 08-08-2012, 05:19 PM

Try to do this without performing any calculations (or simulations).

Starting from the top of a full deck, I will turn over the cards one at a time until I turn over the first ace. Then I will turn over one more card. Is it more likely that card will be the ace of spades or the deuce of clubs? What are the probabilities for each?

I answered this years ago, and I recently happened across it again. I was struck at how it affords an extremely simple and elegant solution that is related to the solutions to some other problems that we've had here lately.

TomCowley · 08-08-2012, 05:47 PM

Spoiler:

Should be 1 in 52 for both. Relative to the order of the other cards in the deck, any one card is equally likely to be in one of 52 possible slots (after 0..51 cards), and one is a winner and the rest are losers.

heehaww · 08-08-2012, 08:13 PM

Spoiler:

The first ace could have been the ace of spades, so the chance that it's the card after the first ace is only (3/4)*(1/51). Whereas the deuce of clubs has a probability of 1/51.

But I have my doubts, since I feel like you wouldn't be asking this if that were the answer.

heehaww · 08-08-2012, 09:29 PM

Oh I didn't take into account the location of the first ace. Extreme example of how location affects the probabilities: if the first ace isn't until the 49th card, the final 3 cards must be all be aces.

So I guess the challenge now is to handle that in the simplest way possible.

David Lyons · 08-09-2012, 06:52 AM

The As always has exactly a 0.75 chance of being in the stub.

The 2c always has exactly a 1/51 chance of being the "next card after first Ace".

I'm struggling to figure out an elegant solution but am guessing the As is a big favourite.

David Lyons · 08-09-2012, 07:04 AM

Hold on.

I just imagined a 3 card deck. As 2c Ac.

There's six ways to arrange these three cards. And two arrangements have 2c in middle (after first ace) and two arrangements have As after Ac (with the 2c either first or last).

So it's the same, or have I horribly oversimplified?

Frallan10 · 08-09-2012, 09:18 AM

It seems kinda hard to do without any calculations, but since its possible there has to some sick simple logic solution.

I'd think that given that the 2 of clubs is still in the deck once we've reached the first ace, 3/4 P( 2clubs )= P( Ace of spades ).
So the problem kinda boils down to whats the probability for the 2 of clubs to be thrown out of the deck before the first ace is reached which.

If we give the cards a number between 1 ( bottom of deck ) and 52 ( top of deck), whats the probability the 2 of club have a higher number than all of the four aces?
I think its 1/5, allthough Im not sure which is kinda embarassing since it shouldnt be too hard to figure out.

If we assume its 1/5 I think we'll get P( 2clubs ) = 4/5 * 4/3 P( Ace of spades ) = 16/15 P( Ace of spades ) which means getting the 2 of clubs is slightly more likely.
However I dont think this is the way to go, feels like Ive missed something and made it too easy to myself =)

Frallan10 · 08-09-2012, 10:34 AM

Once you know x in P( 2clubs ) = x * P( Ace of spade )
I think u can determine the probabilitys using
48*P( 2clubs ) + 4*P( Ace of spade )=1

ExaMeter · 08-09-2012, 11:53 AM

we know that after the first ace there are 3 aces remaining in the deck.

what we really try to figure out is whether there will be more or less than 3 deuces left in the deck (on avg) after the first ace is turned. if its more than 3 then the 2c is more likely otherwise the As is more likely.
since were not supposed to calculate anything this cant be right though.

edit: the chance that we first draw an ace or first draw a deuce are the same. however we dont stop drawing when we draw a deuce.
therefore the 2c is slightly more likely since there there must be slightly more deuces remaining in the deck.

cyberfish · 08-09-2012, 02:11 PM

Quote:

Originally Posted by David Lyons

I just imagined a 3 card deck. As 2c Ac.

This leads you to the correct solution (however with calculations):

Spoiler:

As -> 1/3
2c -> 1/3

Add one random card (Kh for instance):

As -> 1/4 (=2/4*1/3+1/4*1/6+1/4*1/6)
2c -> 1/4 (=3/4*1/3)

You can add cards until a full deck -> 1/52

heehaww · 08-09-2012, 02:22 PM

Tom and cyberfish, I don't see how either card can be 1/52.

We're not just talking about a spot in the deck. It has to be after an ace. Focusing on the 2c, we already know that a non-2c has been removed from the deck (since it was an Ace), so if anything it would be 1/51.

It's 2 adjacent spots in the deck. There are 51 pairs of adjacent spots.

edit -- oh, there's also a chance the 2c came before the first ace. I guess you're saying that when you multiply 1/51 by P(not before ace), it comes out to 1/52. I'll have to see if that's true.

edit #2 -- That can't be it either. P(not before 1st ace) = 4/5. That times 1/51 doesn't equal 1/52.

cyberfish · 08-09-2012, 03:08 PM

If you draw an ace as the first card it is 1/51 (for 2c being the second card) but if you draw the first ace later on the probability gets lower for 2c.

Imagine this:
Starting from the top of a full deck, I will turn over the cards one at a time until I turn over a red card. Then I will turn over one more card. Is it more likely that card will be a red card or a black card?

heehaww · 08-09-2012, 04:18 PM

Nevermind, the solution came to me while I was running.

Earlier I stopped halfway through my reasoning, all I had to do was finish my train of thought.

There are 51 ways for the 2c to come right after the first ace,
Out of a possible 52P2 permutations of Ax2c.
= 51 / (52*51) = 1/52

(To be more explicit, one can also multiply the numerator and denominator by 4 since there are 4 possible aces it can be. The denominator can also be 4P2 * 52C2.)

Turning our attention to the ace of spades:
(3 * 51) / (3 * 52*51)
= 1/52

Cool

**BruceZ** · 08-09-2012, 05:43 PM

Spoiler:

Ask yourself, what is the probability that the last card in the deck is the ace of spades? Nobody should have any trouble seeing that this is 1/52. The fact that I will deal down to the first ace has nothing to do with that. For any shuffle, I can always deal down to the first ace, and that doesn't change the fact that the last card has a 1/52 chance of being the ace of spades. But the last card in the deck always has the same probability of being the ace of spades as any other remaining card that comes after the first ace. So the card after the first ace has a 1/52 probability of being the ace of spades. It also has a 1/52 probability of being the deuce of clubs for the same reason. QED

This is similar to the marble problem where the chance that the next marble is red is the same as the chance that the last marble will be red. It's also similar to my explanation of the Monty Hall problem which notes that Monty opening a door to show you a goat is irrelevant to the probability that your door has the prize because he always opens a door to show you a goat. These ways of thinking can elucidate an obvious solution to problems which may otherwise be confusing.

Tom's solution is certainly simple, to the point where it almost makes the problem not that interesting. If you ignore the ace of spades, the 51! possible arrangements of the other cards all have the same probability, and for each one, there are 52 places you can stick the ace of spades, and only one of those has it right after the first ace, so the probability is 1/52. Same for the deuce of clubs. It's interesting that when this problem was posted almost NINE years ago, nobody posted that solution, even though there were a couple of mathematicians in the thread who calculated the right answer but still found the result non-obvious or counterintuitive. That thread generated a ton of opinions, with some arguing vehemently that the probabilities should be different.

The answer to cyberfish's problem isn't as obvious by Tom's method. He asked about dealing to the first red card, and asked for the probability that the next card is red. Again, the probability that the last card is red is 1/2, so the probability that the card after the first red card is red is also 1/2. In fact, you can deal until whatever criteria you like is met as long as that criteria is guaranteed to always be met before the last card, and whatever probability you want to know for the next card will be the same as that for the last card. That observation is similar to one made by a poster named irchans who posted at the very end of the thread nine years ago. It was suggested by one of the mathematicians that he publish this argument. I don't know if that ever happened.

**BruceZ** · 08-10-2012, 11:22 AM

Quote:

Originally Posted by BruceZ

The answer to cyberfish's problem isn't as obvious by Tom's method. He asked about dealing to the first red card, and asked for the probability that the next card is red.

Of course you can do that by Tom's method too. Just find the probability that the next card is the ace of diamonds, which is 1/52, and multiply that by 26 since the red cards are mutually exclusive.

If you have absorbed these methods, then you can certainly answer these:

Starting from the top of a full shuffled deck, I will turn over the cards one at a time until I have turned over 3 aces. Then I will turn over one more card. Is it more likely that card will be an ace or a deuce? What are the probabilities for each?

Spoiler:

1/13 for both.

Starting from the top of a full shuffled deck, I will turn over the cards one at a time until I have turned over 12 spades. Then I will turn over one more card. Is it more likely that card will be a spade or a heart? What are the probabilities for each?

Spoiler:

1/4 for both.

Starting from the top of a full shuffled deck, I will turn over the cards one at a time until I have turned over 2 aces. Then I will turn over two more cards. What is the probability that both of those cards will be aces?

Spoiler:

1/221

Starting from the top of a full shuffled deck, I will turn over the cards one at a time until I have turned over 7 spades, 5 clubs, and 2 aces. Then I will turn over 2 more cards. What is the probability those cards will be 2 black aces?

Spoiler:

1/1326