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# Colored cards at Morty's

## Question

bonanova    76

It's been far too long, said Alex to no one in particular as
he tossed his hat skillfully onto a coak hook at Morty's
last night.

Ian and Davey knew that was their cue, and they joined
Alex at his favorite table. Not surprisingly, he pulled out a
deck of cards.

I'm feeling generous tonight laddies, he began. I'll play
a game with you that guarantees that you can double your
money. So to make it fair, I'll collect your initial stake
from you, and then you can keep what ever you win.
That way you can assure no worse than to break even.

Jamie had been listening, from the bar, and what he
heard drew him over to join the others. OK tell us,
he said, how does the game work?

I'll shuffle this deck here, and then deal them face up
one at a time. You start with \$1. You can bet any fraction
of your current worth, before I deal the next card, on the
color of the next card. Even odds on each bet, no matter
what.

So how does the guarantee work then? asked Ian.

Drink yer ale, matey! Alex replied, If yer paying attention
at all, you'll know the color of the last card. You can just
wait, and then bet everything on the last card! I'll take
yer stake, you get your winnings, and you break even.

Now, who wants to play?

Jamie and Ian figured that if they were really lucky
they could win every bet and parlay their stake into
fact that if they lost any of the 52 bets, they'd win
nothing, and Alex would get their stake. Seems like a
bunch of coin tosses to me, said Jamie. Not worth my
time just to break even,
said Ian, And the two of them left.

Davey, trance-like for a moment as he stroked his beard,
finally said, I'll play. Deal the cards.

Did Davey have any expectation of winning money?
If so, how much?

## 31 answers to this question

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bonanova    76
The only problem with this if you lose one card you lose ALL your money. Therefore you need to get all 52 cards correct to win ( very slim chance)

Slim chance indeed.

But how big is the payoff?[spoiler=CR has it. ]The OP asks about expectation - the result when averaged over all possible outcomes.

If you bet \$1 and there were 10 equally likely outcomes of

0, 0, 0, 0, 0, 0, 0, 0, 0 and \$20

If you made 10N bets, where N is a large number, you'd expect to get a \$20N return.

The best strategy is to guess a red-black distribution and bet it all on every card.

There are 52C26 red card distributions, and one of them pays off \$252.

The ratio [like the \$20/10 = \$2 above] is \$9.0813....

When Davey gives Alex his stake, he's left with CR's answer.

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This is similar to the St. Petersburg Paradox. If I were actually playing this game, I'm sure I wouldn't go with the bet-it-all strategy, even though that maximizes expectation. I would probably keep the dollar until I hit the end with all the cards the same color, and hope to walk away with a couple bucks.

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At one card per second (assuming Alex bets that quickly), he'll strike it rich once every 817,724,623 years. Inflation will probably kill the value of the winnings by then...

And how many trillions did Alex bring to the bar last night (in neatly stacked ones!)???

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bonanova    76

Davey just gathers 52C26 of his drinking buddies - he's been drinking for a while - and has each of them bet a different one of the 52C26 possible red-card distributions. They agree to work for half their earnings. That is, the winner splits his take with Davey, the others just drink and play.

Davey and one of his buddies get about \$4, Alex gets \$1,

and Morty's closes at the normal time.

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Davey just gathers 52C26 of his drinking buddies - he's been drinking for a while - and has each of them bet a different one of the 52C26 possible red-card distributions. They agree to work for half their earnings. That is, the winner splits his take with Davey, the others just drink and play.

Davey and one of his buddies get about \$4, Alex gets \$1,

and Morty's closes at the normal time.

Well then, I hope Morty's is very, very spacious, or that most of Davey's friends are very, very small!

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Starting with r reds and b blacks, the expected return is

V(r,b) = 2r+b/(r+bCr)

Note that this satisfies

V(r,b) = (r/r+b)*V(r-1,b) + (b/r+b)*V(r,b-1)

Therefore V is a martingale! This means that ANY rational strategy will yield the same expected return. 'Rational' here just means never betting on the rarer colour, and always betting everything when only one colour remains.

Moreover, there exists a risk-free strategy of always betting a proportion (r-b)/(r+b) of current wealth (or likewise (b-r)/(r+b) if there are more blacks remaining), such that one is GUARANTEED a return of \$9.08.

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