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towr
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Custom dice game
« on: Apr 29th, 2009, 6:38am » |
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I've got a set of five custom (6-sided) dice. They're fair dice, in that each face has an equal chance of 1/6th of coming up. But I changed the values on the faces. You get to pick two of the dice, and then I pick two from the remaining ones. Whoever throws the highest sum wins (and, heck, if it's a tie I'll even give you the win).
If we were to take a gamble on the result, what do do you think would be fair odds?
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| « Last Edit: Apr 29th, 2009, 6:38am by towr » |
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SMQ
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Re: Custom dice game
« Reply #1 on: Apr 29th, 2009, 7:26am » |
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Hmm, given specific values for the various faces it's a fairly straightforward probability calculation: assume I will choose dice to my greatest advantage and that you will do likewise: of the 1296 possible (and possibly equivalent) outcomes, how many are a win for me, but with no a priori knowledge of the values I'm not seeing a clear (or clever) way to proceed.
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towr
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Re: Custom dice game
« Reply #2 on: Apr 29th, 2009, 7:41am » |
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Well, the most obvious way to proceed would to make such a set of dice and try to find the largest minimum advantage you can give to player 2.
I'm not quite certain I found the best set of dice either, though; but it gives an decent advantage. Looking up non-transitive dice may give you a few ideas.
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| « Last Edit: Apr 29th, 2009, 7:42am by towr » |
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Grimbal
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Re: Custom dice game
« Reply #3 on: Apr 29th, 2009, 9:39am » |
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If there is any non-zero fair value to the gamble, then after doubling all facet values the fair value doubles as well.
So either the fair value is zero or it must be expressed in terms of the actual values on the facets. Or maybe the facet values must be restricted in some way, for instance to the integers 1..6.
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towr
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Re: Custom dice game
« Reply #4 on: Apr 29th, 2009, 10:21am » |
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on Apr 29th, 2009, 9:39am, Grimbal wrote:| If there is any non-zero fair value to the gamble, then after doubling all facet values the fair value doubles as well. |
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?!?
The chance of winning remains the same whatever multiple of the facet values you put on the dice. Instead of sum1 vs sum2 you'll just be comparing sum1*X vs sum2*X.
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| « Last Edit: Apr 29th, 2009, 10:21am by towr » |
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towr
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Here's an example, we have dice A,B,C,D,E:
A: 0,5,5,5,5,5
B: 1,1,6,6,6,6
C: 2,2,2,7,7,7
D: 3,3,3,3,8,8
E: 4,4,4,4,4,9
The odds of winning are (at least) 55 to 53 for the second player.
But you can do much better at picking the dice.
[edit]Attachment shows which pair of dice to choose given the first players choice, and what number of times out of 108 you'll win[/edit]
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| « Last Edit: Apr 29th, 2009, 11:13am by towr » |
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Grimbal
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Re: Custom dice game
« Reply #6 on: Apr 30th, 2009, 12:24am » |
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on Apr 29th, 2009, 10:21am, towr wrote:
?!?
The chance of winning remains the same whatever multiple of the facet values you put on the dice. Instead of sum1 vs sum2 you'll just be comparing sum1*X vs sum2*X. |
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Uh. Indeed. For some reason I assumed the earning was proportional to the difference in points. :-/
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SMQ
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Re: Custom dice game
« Reply #7 on: Apr 30th, 2009, 4:51am » |
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Wait, which of us is player 2 again? In any case, in order to answer the original question, I'm to assume that in designing the dice you have attempted to choose values to your greatest advantage, and now need to decide what odds I would accept in order to play, yes?
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towr
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Re: Custom dice game
« Reply #8 on: Apr 30th, 2009, 5:33am » |
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on Apr 30th, 2009, 4:51am, SMQ wrote:| Wait, which of us is player 2 again? |
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The one that chooses a pair of dice last (and can therefore select dice that on average beat the pair player one picked).
Quote:| In any case, in order to answer the original question, I'm to assume that in designing the dice you have attempted to choose values to your greatest advantage, and now need to decide what odds I would accept in order to play, yes? |
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Yeah.
And of course that's just a roundabout way of saying: find the* best set of dice where the second player can always pick a pair of dice that beats whichever pair the first player picked.
*) or rather 'a', since there's usually more than one even if we disregard addition and multiplication.
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| « Last Edit: Apr 30th, 2009, 5:33am by towr » |
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