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Colored cards at Morty's


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31 replies to this topic

#11 HoustonHokie

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Posted 01 August 2008 - 04:51 AM

I may have missed something, but I read "you can bet any amount".

What if you bet zero and count cards, until the last card?

It's not that you can bet any amount - you can bet "any fraction of your current worth". So if Davey doesn't make any bets until the last card, he'll double his money to $2, but Alex will take $1 and leave Davey with $1, which is exactly what he started with - no gain, no loss.
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#12 HoustonHokie

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Posted 01 August 2008 - 05:32 AM

Spoiler for how much he'll win that way...

I just realized that I messed up the probability math here. This is better:

Spoiler for how much he'll really win that way...

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#13 d3k3

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Posted 01 August 2008 - 06:12 AM

I just realized that I messed up the probability math here. This is better:

Spoiler for how much he'll really win that way...


I didn't quite follow that last bit, but I believe the probability of the last two cards being of the same color is roughly 0.5.

Anyway, I haven't worked it all out yet, but I think there may be a way to increase your winnings, if you bet a portion of your stack proportional to the odds of the next card turning up a particular color. For instance, if you don't bet on the first card and it turns up black, the odds of the next card being red is 26/51. So you bet $0.50 on red. If I understand the rules correctly, you keep the $0.50 when you win, only paying the initial $1 stake at the end of the deck. You therefore expect to win about $0.01 on this bet ((26-25)/51*0.5). If you lose this bet, your stack is cut in half, but your odds of winning the next bet increases to 26/50, so you now bet $0.25. Your expectation is now exactly $0.01 ((26-24)/50*0.25). If you had won your first bet, you pass on the second bet, and your odds of winning on the third card is 25/49 (the largest proportion of your stack smaller than your odds of winning). Either way, as you progress through the deck, your odds of winning each bet increases whenever there is not an equal number of cards of each color remaining. Since you are only betting when the odds favor you, you expect to have more than $1 to bet on the last card, or on the last cards that are known to be of a particular color.

[edited: grammar]

Edited by d3k3, 01 August 2008 - 06:14 AM.

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#14 d3k3

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Posted 01 August 2008 - 06:52 AM

I didn't quite follow that last bit, but I believe the probability of the last two cards being of the same color is roughly 0.5.

Anyway, I haven't worked it all out yet, but I think there may be a way to increase your winnings, if you bet a portion of your stack proportional to the odds of the next card turning up a particular color. For instance, if you don't bet on the first card and it turns up black, the odds of the next card being red is 26/51. So you bet $0.50 on red. If I understand the rules correctly, you keep the $0.50 when you win, only paying the initial $1 stake at the end of the deck. You therefore expect to win about $0.01 on this bet ((26-25)/51*0.5). If you lose this bet, your stack is cut in half, but your odds of winning the next bet increases to 26/50, so you now bet $0.25. Your expectation is now exactly $0.01 ((26-24)/50*0.25). If you had won your first bet, you pass on the second bet, and your odds of winning on the third card is 25/49 (the largest proportion of your stack smaller than your odds of winning). Either way, as you progress through the deck, your odds of winning each bet increases whenever there is not an equal number of cards of each color remaining. Since you are only betting when the odds favor you, you expect to have more than $1 to bet on the last card, or on the last cards that are known to be of a particular color.

[edited: grammar]


Er, oops. "(the largest...odds of winning)" should come after "... so you now bet $0.25."
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#15 bonanova

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Posted 01 August 2008 - 07:53 AM

Overheard at Morty's this afternoon:

Did you hear?
Davey took Alex to the cleaners last night!
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#16 taliesin

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Posted 01 August 2008 - 12:28 PM

Overheard at Morty's this afternoon:

Did you hear?
Davey took Alex to the cleaners last night!


dont give any clues yet, im working on this one
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#17 taliesin

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Posted 01 August 2008 - 12:47 PM

dont give any clues yet, im working on this one

Spoiler for Worst case I can statically earn a profit of
Spoiler for and, best case

Edited by bonanova, 01 August 2008 - 05:43 PM.
Bonanova added spoilers for taliesin's results

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#18 Chuck Rampart

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Posted 01 August 2008 - 01:39 PM

Spoiler for Keeping your $1 stake until only one color is left


You can get a higher expected value by betting money earlier in the deck, but I'm not sure what the optimum strategy would be in that case. That also would allow the possibility of a loss of money (but you only started with $1, so I guess it's not that big a deal).
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#19 foolonthehill

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Posted 01 August 2008 - 02:33 PM

Spoiler for still thinking, but my two cents...

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#20 bonanova

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Posted 01 August 2008 - 05:42 PM

Spoiler for Worst case I can statically earn a profit of
Spoiler for and, best case

No clues for a while. Several are still working.

Care to put your worst case strategy into a spoiler?
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell




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