[bonanova]

It's been forever, it seems, since we downed a cold

one at Morty's, but a quick visit last night did not

disappoint, in more ways than one. As usual, Alex

was holding court, over in the corner table.

Gather round me, boys, he began, and try to learn

something. And if it costs ya a quid or two, you'll

remember it all the more.

With that, he took the four Aces from a deck of cards,

shuffled, and dealt them face down on the table. Now

one of you lads grab a couple bottle caps from the

floor and place them, one each, on two of these

here cards. Jamie obliged.

Now here's the deal, Alex continued, there's a certain

probability that the caps are sitting on cards of the

same color [red or black]. I'll bet even money with

anyone who can say what that probability is.

Jamie was first to respond. It's quite simple, ya see?

There are three cases. They're sitting on black cards,

red cards, or mixed cards. The probability of the same

color is clearly 2/3.

You've been at the tap too long, laughed Ian, you

forgot that mixed cards can be red-black or black-red.

There are actually four cases, and the same-color

probability is 1/2. They either match, or they don't.

Davey sat for a moment, stroked his bushy red beard,

and then walked over to Alex and whispered his answer

so none of the others could hear.

Given that only one of the boys had it right, who went

home that night richer than he came?

[Guest]

Neither. The odds of being the same are 1/3.

The first cap picks the suit. Once picked, there is only a 1 in 3 chance of matching that color.

[Guest]

Alex took home more, three guys played against him and he got money from all three but only had to pay out once.

[Guest]

Two possible answers:

If it is a trick question: Neither went home richer as they spent way too much drinking

If it is not a trick question: Ian is right and he went home richer

The case for 1/3 is the probability when you have already put one cap and now want to see what is the probability for same colour cards. 1/2 is the case when you "draw two cards at random" out of the 4 aces

[Molly Mae]

I agree with twhedge. The same principle can be applied to rolling 7's with two six-sided dice. The odds are 1/6 because the first die doesn't matter. So, too, does the first bottle cap not matter.

[Guest]

Alex will come home with more money.

The probability is 1/3.

They will have 1/6 a chance of getting 2 black and 1/6 a chance getting 2 red. After they place the first cap, they will have 1/3 a chance of getting the correct color that matches. The first cap does not matter, since the color does not matter.

Edited September 22, 2010 by superhawk

[Guest]

Obviously Davey goes home with more money because he's the only one capable of stating the correct possibility of 1/3.

[Guest]

1 in 3 because one cap is on a card of a certain colour and since there are 2 of the other colour and one of the same left, the odds of the other cap being on the same colour are 1 in 3.

[Guest]

ALEX is the richer.

[Guest]

THE BARTENDER OF COURSE

[Guest]

there is 50% probability that the 1st card is red and 33% probability that the 2nd one is also red...(thinking)...I think the probability of getting 2 reds/blacks is about 5/6

I'm really not sure that's right, but if it is I can explain it.

Edited September 22, 2010 by kristmark1

[Guest]

The question isn't who got the right answer... it's "...who went home that night richer than he came?"

The answer is that TWO PEOPLE went home richer than they came.

1) Alex paid out to one person, but received money from two people. So part of the answer is Alex.

2) For the other part of the answer...

Picking two cards at random will result in the following possible combinations:

Key: [H]eart, [D]iamond, pade, [C]lub

HD <--pays off

HS

HC

DS

DC

SC <--pays off

only 2/6 (1 in 3) combinations pay off.

So Davey must have whispered the correct answer into Alex's ear (1 in 3 chance). Therefore the correct answer is Davey and Alex went home richer than he came.

[Guest]

Answer is 1/3 so Davey must have gotten it right. I thought of it more in suits (S, C, H, D). There are 12 possible combinations of suits (SC, SH, SD, CS, CH, CD, HD, HS, HC, DH, DS, DC) of which only 4 are of the same color (SC, CS, HD, DH), so the odds are 4/12 or 1/3.

After I posted, I read enrightmcc's response and I have to agree that the answer to the question who went home richer would be Davey & Alex. Well played.

Edited September 22, 2010 by Louie

[Guest]

THE BARTENDER OF COURSE

My thoughts exactly--or Morty(or whoever owns the bar).

I thought Ian would've guessed correctly with 50/50, like a coin toss. There are 4 cards, but only 2 have caps on them and there are only 2 colors--red or black.

[Guest]

My thoughts exactly--or Morty (or whoever owns the bar). Or maybe Davey picked everyone's pocket while they were engrossed in the game and he went home richer.

I thought Ian would've guessed correctly with 50/50, like a coin toss. There are 4 cards, but only 2 have caps on them and there are only 2 colors--red or black.

]I thought Ian would've guessed correctly with 50/50, like a coin toss, but not agreed w/ his rationale. If Alex asked the chance of having 2 reds (or 2 blacks, or 1 black and 1 red) I would've said 1:3 either Red Red, Red Black (or Black Red), or Black Black, but he only asked the probability of them being on the same color. Either they are on the same color or they're not. If he had asked about order that would also change the answer because there would be four possible combinations (Red Red, Red Black, Black Black, Black Red).

I don't understand the whole "bet even" thing, so I don't know what Alex's winnings would be. Besides, no one actually placed a bet, right?