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Friday Puzzler -- Tricky Dice

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Friday Puzzler -- Tricky Dice

#1

Friday Puzzler -- Tricky Dice

Alex Y

This week's puzzle is to design a special set of three dice. These dice have the unusual property that whichever die you opponent chooses, you can pick one of the remaining two that give you a better-than-even chance of beating your opponent in a roll of the dice.

1) Each face of each die has a single digit from 1 through 9 (and yes, you can distinguish between the 6 and the 9)

2) Opposing faces are marked with the same digit

Followup questions (that I do not have the answer to): What are the best odds you can give yourself in such a set of dice? If we remove the restriction that opposing faces are the same, can you design dice that give you a better chance of winning (assuming your opponent will choose the die that gives him the smallest chance of losing)?

Re: Friday Puzzler -- Tricky Dice

#2

Tricky Dice -- Some hints

Alex Y

You probably already have this, but with digits appearing in pairs on the dice, each die has only three outcomes on a toss, with equal probability. So when tossing two dice, we have to consider only 9 possible outcomes, not the usual 36.

Hint related to the two solutions I have found below:

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I have found two solutions, In both cases, the expected result on all three dice is the same.

Re: Friday Puzzler -- Tricky Dice

#3

Re: Tricky Dice -- Some hints

Larry Barrett

I don't understand the hint, but I think I have found 2 solutions.

In each case the digits 1 thru 9 are used only once (don't appear on more than one die).

If I write the digits used on the three dies in 3 rows and 3 columns, the sum of each row is 15, and the sum of the three columns is 6 15 24 for both solutions.

For each solution, if A selects die 1, B selects die 2; if A selects die 2, B selects die 3; if A selects die 3, B selects die 1.

Re: Friday Puzzler -- Tricky Dice

#4

Sounds like you got it

Alex Y

And if you found the same three solutions as I did, you have a 5/9 chance of winning in eery case.

My solutions: 942, 861, 753

and 951, 843, 762

I suspect, but don't see any argument more convincing than "I can't find one". that there are no solutions where you can always get better than 5/9 probability of winning

Re: Friday Puzzler -- Tricky Dice

#5

Re: Sounds like you got it

Larry Barrett

Those are my solutions, also.

I am thinking about the last part of your puzzle, but no ideas so far.

Re: Friday Puzzler -- Tricky Dice

#6

Re: Sounds like you got it

Alex Y

I think I have a proof that 5/9 is the best odds you can get if all 9 digits are used (no ties). Haven't worked on the possibility of ties yet.

👍 This page answered my questions

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