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        riddles >> medium >> Custom dice game
        
(Message started by: towr on Apr 29^th, 2009, 6:38am)

Title: Custom dice game
Post by towr on Apr 29^th, 2009, 6:38am

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?

Title: Re: Custom dice game
Post by SMQ on Apr 29^th, 2009, 7:26am

Hmm, given specific values for the various faces it's a fairly straightforward probability calculation: [hide]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[/hide], but with no a priori knowledge of the values I'm not seeing a clear (or clever) way to proceed.

--SMQ

Title: Re: Custom dice game
Post by towr on Apr 29^th, 2009, 7:41am

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.

Title: Re: Custom dice game
Post by Grimbal on Apr 29^th, 2009, 9:39am

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.

Title: Re: Custom dice game
Post by towr on Apr 29^th, 2009, 10:21am

on 04/29/09 at 09:39:15, 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.

?!?
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.

Title: Re: Custom dice game
Post by towr on Apr 29^th, 2009, 10:27am

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]

Title: Re: Custom dice game
Post by Grimbal on Apr 30^th, 2009, 12:24am

on 04/29/09 at 10:21:30, 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.

Uh. Indeed. For some reason I assumed the earning was proportional to the difference in points. :-/

Title: Re: Custom dice game
Post by SMQ on Apr 30^th, 2009, 4:51am

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?

--SMQ

Title: Re: Custom dice game
Post by towr on Apr 30^th, 2009, 5:33am

on 04/30/09 at 04:51:55, SMQ wrote:

Wait, which of us is player 2 again?

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?

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.