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# Quick Draw Aces

## Question

At this weeks friendly game I thought I'd try to get back a little of my money that was still in Coop's chip stack. Here was my proposed set-up: "I'll give you one red and one black ace and I'll take the other two aces. We'll both simultaneously show one ace of our chosing. If both shown aces are black, I'll give you \$10; if the two shown aces are different, you pay me \$5; and if both shown aces are red, I'll give you \$1. Rob, our most philosophical member (read that least mathematical) asked simply, Why?" Coop now had to chime in, "You see there are four possible combinations, black-black, black-red, red-black, and red-red. So 25% of the time I should win \$10, 50% of the time I lose \$5, and 25% of the time I should win \$1. Out of four games then, I should win \$1 on average." I liked the way Coop was already speaking in the first person; it was time to reel him in. "Okay Coop, I'll make that \$2 I'll pay out for every red-red combination." Coop couldn't resist at this point, "You're on but I get to decide when we quit. Huck can keep the tally." Vern just chuckled in amusement. I gave Coop the two pointed aces and the game was on. At least three times I asked Coop in a worried tone if he had had enough but not until we must have played 300 hands or more did he ask Huck what the tally was. When Huck told him he was down \$72 Coop exclaimed, "You're drunk!" Though that very usually is the case, the results were accurate. "Damn, stop me at \$100." Shortly thereafter, Coop handed me what I hoped was the same \$100 bill he flashed last week as enticement into Was I just lucky? If not, what was my strategy?

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Never mind, I misread the problem completely

Edited by CaptainEd
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Clever, but are you considering alternate strategy for Coop?

It is obvious that you change odds in your favor by never playing Black, thus no \$10 payout. This gives you a win 2/3's of the the time while only 1/3 for Coop, but Coop should know this and thus opt for red. If you predict this for his logic, then you should play black but then he may figure you would figure this and play black. This logic could be extended so it comes down to who can read the tells best, especially after observing patterns of play. You can only guarantee winning if Coop picks randomly.

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At this weeks friendly game I thought I'd try to get back a little of my money that was still in Coop's chip stack. Here was my proposed set-up: "I'll give you one red and one black ace and I'll take the other two aces. We'll both simultaneously show one ace of our chosing. If both shown aces are black, I'll give you \$10; if the two shown aces are different, you pay me \$5; and if both shown aces are red, I'll give you \$1. Rob, our most philosophical member (read that least mathematical) asked simply, Why?" Coop now had to chime in, "You see there are four possible combinations, black-black, black-red, red-black, and red-red. So 25% of the time I should win \$10, 50% of the time I lose \$5, and 25% of the time I should win \$1. Out of four games then, I should win \$1 on average." I liked the way Coop was already speaking in the first person; it was time to reel him in. "Okay Coop, I'll make that \$2 I'll pay out for every red-red combination." Coop couldn't resist at this point, "You're on but I get to decide when we quit. Huck can keep the tally." Vern just chuckled in amusement. I gave Coop the two pointed aces and the game was on. At least three times I asked Coop in a worried tone if he had had enough but not until we must have played 300 hands or more did he ask Huck what the tally was. When Huck told him he was down \$72 Coop exclaimed, "You're drunk!" Though that very usually is the case, the results were accurate. "Damn, stop me at \$100." Shortly thereafter, Coop handed me what I hoped was the same \$100 bill he flashed last week as enticement into his little prop bet. Was I just lucky? If not, what was my strategy?

This is a really nice puzzle.

At every game, choose Red with probability 15/22, and choose black with probability 7/22. You have an expected wining of 5/22 dollar/hand regardless of what Coop does.

Edited by bushindo
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very tricky problem.

if both players play randomly, coop is right he should win 2 dollars with four draws.

1/4 *10 -1/2*5 +1/4*2 = .5

that is with every draw, coop should win on average 50 cents.

however, both players can choose whether they play a black or a red card.

if you for example play black 25% of the time, then coop is forced to do the same, as he needs to match red for red to win.

expected winnings for black at 25%

1/16*10 -6/16*5 +9/16*2 = -0.125

that is, out of 8 hands you win 1 dollar.

now, how to calculate the best play?

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x^2 *10 -(1-((1-x)^2 +x^2)*5 +(1-x)^2 *2 = min

22x^2 -14x +2 = min

44x -14 = 0

x = 14/44

or

x = 7/22.

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I agree with Bushindo's solution. Here's a loose point to keep going on this - How do you implement it?

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Thanks for the compliment bushindo. Nice solve. And phillip 1882, as well. My approach was really quite simple. Not requiring any derivatives and pretty neat I think. Wondering some on how bushindo went about it. As for DeeGee's extension, have one idea as to where he may be going but not sure how specific of an implementation strategy he may have in mind. Will ponder further on that.

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Thanks for the compliment bushindo. Nice solve. And phillip 1882, as well. My approach was really quite simple. Not requiring any derivatives and pretty neat I think. Wondering some on how bushindo went about it. As for DeeGee's extension, have one idea as to where he may be going but not sure how specific of an implementation strategy he may have in mind. Will ponder further on that.

This is what I did

Let x be the probability that we draw a Red card at every turn, and let (1-x) be the probability that we draw a Black card. Let y_R( x ) be a function of our expected winning given x and given that Coop plays a Red card. Let y_B( x ) be our expected winning function if Coop plays a Black card. The explicit forms for the functions are

y_R( x ) = -2 * x + 5 * (1-x)

y_B( x ) = 5*x - 10( 1- x )

To find the optimal probability x, simply solve the equation -2 * x + 5 * (1-x) = 5*x - 10( 1- x ).

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