[unreality]

Your friend staggers up to you with a grin on his face.

"Want to play a game?" he asks nonchalantly. "Just ten bucks. If you win, you get three hundred dollars back!"

You know him pretty well, so you're reluctant. "What's the game?"

Your friend pulls out a deck and shuffles it absentmindedly, then hands it to you. You inspect it; it's a standard deck, 52 cards, no jokers, all cards look the same and feel the same from the back. You give it a couple good shuffles and hand it back.

"You saw the two black aces in there, right?" your friend asks.

"Of course."

"Well, here's the game. You're going to take the deck, and one at a time, flip over the top card, placing it on the table face-up. Easy enough?"

"Sounds good," you say.

"Well here's the catch. You're going to predict when the first black ace will turn up. Anywhere you want to pick in the deck... 1 being the top card, 52 being the bottom. Remember that it's been shuffled and re-shuffled, so each card has a 1/52 chance. If you accept the bet, you pay in $10, but get $300 if you win."

Do you accept the bet?

[Guest]

But that is irrelevant, since the first black Ace will always turn up on the top of the deck. That is, afterall where you flip the cards from. Thank you for the $300. Unless I'm wrong.

[unreality]

Nope, sorry

[HoustonHokie]

but I think you'd win if you took the bet. As Grayven noted, the chances of any one flip of the cards being a black Ace is 2/52. So let's say that you wanted to get the game over quickly (your friend staggered up to you remember - I think you want to get the money out of him before he pukes or passes out) and you guess position 1 every time. If your chances are 1/26 that a black Ace is in that position, you'd put in $260 on average before you win $300. So, unless I'm missing something, you'd win, on average, $40 for every 26 times you played the game.

[unreality]

The odds are not 2/52

The first black ace is different than any old black ace

edit to clarify: There is a point in the deck where the odds are in fact 51/1326 (which is 2/52), but not all points are the same... your goal is to find the ideal point in the deck, and then determine what the odds are and thus if you would take the bet

Edited November 9, 2008 by unreality

[Mekal]

The odds are not 2/52

The first black ace is different than any old black ace

edit to clarify: There is a point in the deck where the odds are in fact 51/1326 (which is 2/52), but not all points are the same... your goal is to find the ideal point in the deck, and then determine what the odds are and thus if you would take the bet

Okay... i am probably reading into this too much... but here goes nothing...

The first black ace would probbly refer to the ace of spades because in a brand new pack the ace of spades is the first card after the jokers... but i personally would NOT take the bet...

[Guest]

the odds would not be 2/52, because its the FIRST black ace. so the second one would have to be after it,(it couldnt be last) so it would be a 1/51 chance, right? maybe? [im not sure if you'd have to factor in the second card, i dont feel like thinking very hard.]

[unreality]

Mekal: the pack was shuffled

adrenalgirly: yes, the fact that the FIRST black ace is the first one flipped & seen changes it, and marks an ideal point in the pack that you should predict beforehand... and the goal is to discover this point and find whether or not you should accept the bet, and if you DO accept the bet, what your prediction is

[unreality]

any more takers?

Edited November 10, 2008 by unreality

[Guest]

You should choose the first card in the deck. Every card has the same chance of being one of the two black aces, but if that card is the second ace to come up then it won't work. The only card that has no chance of being the second ace is card #1.

As to whether you should take the bet: You have a 2/52 chance of being correct with card #1. So if you win once every 26 times, then you would pay $260 but recieve $300.

So you should take the bet.

I realize you've said that the chance wasn't 2/52, but I think that it is for the first card.

[Prof. Templeton]

Your first black ace can be in one of 51 places. It can't be the 52nd card because you would have come across the other black Ace before you got there. As the first black Ace gets closer to the back of the deck it limits where the second one can be. Let's look at a set-up with only four cards, Aces and Eights let's say. Here are the possibilities...

AA88

A8A8

A88A

8AA8

88AA

8A8A

Both the Aces are interchangable and so are the Eights. We can disregard the Aces in the last position since they won't be first. So let's look at that first Ace. It appears in position No.1 - 3 out of 6 times, it appears in position No.2 - 2 out of 6 times, and position No.3 - 1 out of 6 times. It seems that the farther you go from the front of the deck the less likely the first Black Ace is to show up. I guess I'd put my money one the first card, but i don't have the proper probablity yet.

[unreality]

Good work so far

You are correct!! The answer is pretty counterintuitive, or at least I think so - your best odds are indeed guessing that the first black ace will show up first, and the odds are 2/52 - when I said they weren't before, that was in response to someone that said the probability is for 2/52 for ALL card positions, but that's false. The only card with 2/52 chance is the top card

top card: 51/1326 (or 2/52)

second card: 50/1326

third card: 49/1326

... (and so on)

[HoustonHokie]

Good work so far

You are correct!! The answer is pretty counterintuitive, or at least I think so - your best odds are indeed guessing that the first black ace will show up first, and the odds are 2/52 - when I said they weren't before, that was in response to someone that said the probability is for 2/52 for ALL card positions, but that's false. The only card with 2/52 chance is the top card

top card: 51/1326 (or 2/52)

second card: 50/1326

third card: 49/1326

... (and so on)

All makes sense now - I did all my work assuming that I'd put my money on position 1! Should have gone into more detail...

[unreality]

yeah Well it was a lucky guess in that case; as the top card does in fact have the best chances

[unreality]

What if the problem was ANY ace? I'm not sure how much that would change it

[Prof. Templeton]

What if the problem was ANY ace? I'm not sure how much that would change it

Still the first, just any color so now we're working with four, or any ace anywhere in the deck?

[HoustonHokie]

Still the first, just any color so now we're working with four, or any ace anywhere in the deck?

Odds would be 4/52 placed on the first card of the deck and would go down from there, but generally the same principle as before. Expect to win $300 every 13 times you play, for net winnings of $300 - $130 = $270 / 13 = $20.76 per bet placed. Comparison with original: expect to win $300 every 26 times you play, for net winnings of $300 - $260 = $40 / 13 = $3.08 per bet placed. Much better winnings (increased ~675%) this time!

[unreality]

I've devised a formula that covers general cases, I'm pretty sure it's correct

x = number of black aces

y = total number of cards

c = position within deck (1=top, y=bottom)

when c=1, formula is x/y

other c values:

(x/y)(y-c)(1/[y-1])

it just so happens when c=1 that the second two cancel each other out and it simply becomes x/y

edit: looks better as:

(x/y)(y-c) / (y-1)

Edited November 15, 2008 by unreality