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Topic: Quick Draw Aces (Read 2469 times) |
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fieldazed
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Quick Draw Aces
« on: Apr 26th, 2012, 6:39am » |
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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. Not until after worringly asking for the third time if he had had enough did Coop 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?
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rmsgrey
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Re: Quick Draw Aces
« Reply #1 on: Apr 26th, 2012, 9:33am » |
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Flash your red ace every time - Coop pays you $5 (the two aces shown are different) and you pay him $2 if he manages to find the other red ace...
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SMQ
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Re: Quick Draw Aces
« Reply #2 on: Apr 26th, 2012, 9:46am » |
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What Coop forgot is that you don't have to play black or red with equal probability! Let's say you play black with probability p and Coop plays black with probability q, then the expected value of the game for Coop, EV = 10pq - 5p(1 - q) - 5(1 - p)q + 2 = 22pq - 7p - 7q + 2. Let EVp be the best expected value for Coop at a given p. Then EVp = 2 - 7p for p < 7/22, -5/22 for p = 7/22, and 15p - 5 for p > 7/22. Furthermore, dEV/dq = 22p - 7, so at the magic value of p = 7/22, Coop's strategy becomes irrelevant (dEV/dq = 0 so changing q has no effect on EV) and Coop can expect to lose about 23c per game. I'm just surprised he kept it up for long enough (probably around 440 games) for you to win $100 from him!
--SMQ
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--SMQ
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fieldazed
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Re: Quick Draw Aces
« Reply #3 on: Apr 27th, 2012, 10:10am » |
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Yes indeed, once again, SMQ.
Am a little more impressed then surprised with Coop's virility tho  . What I find more surprising ishis opponent's ability to randomly select Black seven out of twenty two times. Is there a decent strategy in that regard? And is there an issue if Coop decides he's had enough after only a few games? So is there a better strategy with that in mind? And maybe the only decent extension to this puzzle - at what amount of winnings/losses should Coop bow out?
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towr
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Some people are average, some are just mean.
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Re: Quick Draw Aces
« Reply #4 on: Apr 27th, 2012, 11:43am » |
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on Apr 27th, 2012, 10:10am, fieldazed wrote:| What I find more surprising ishis opponent's ability to randomly select Black seven out of twenty two times. Is there a decent strategy in that regard? |
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Can you use props? Say a coin, a pin and a piece of paper with parallel lines a pin's distance from each other? Drop the pin on the paper, if it crosses a line and you throw heads, bet black, otherwise red. That should happen 1/pi times, which is about 7/22
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Wikipedia, Google, Mathworld, Integer sequence DB
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fieldazed
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Re: Quick Draw Aces
« Reply #5 on: May 1st, 2012, 6:28am » |
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ah yes - the ole Buffon's needle experiment - very clever, towr!
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