Paradoxical Dice
The daily problem for Feb 27 asks which of two special dice (1,1,3,3,5,5) and (2,2,2,2,5,5) has a better chance of winning if both dice are rolled simultaneously and the higher roll wins (In case of a tie, you roll both dice again.)
We'll say die A beats die B if, when A and B are rolled simultaneously, the probability that A rolls a higher number than B is bigger than 1/2.
A, B and C are six-sided dice whose faces have numbers in some set S . Is is possible to construct dice A, B and C such that A beats B, B beats C and C beats A?
Notation: for a set S , ∣ S ∣ denotes the number of elements in the set.
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1 solution
For the dice
( i , j , k , l , m ) = ( 1 , 2 , 3 , 4 , 5 ) ⇒ ⎩ ⎪ ⎨ ⎪ ⎧ A = ( 4 , 4 , 4 , 4 , 1 , 1 ) B = ( 3 , 3 , 3 , 3 , 3 , 3 ) C = ( 5 , 5 , 2 , 2 , 2 , 2 )
their expected values are also all the same (3).
I think you made a typo in the comparison of A and C: "and C rolls i ". I think it should be j .
There also exists a set of 5 such non-transitive dice ( Grime dice ).
Red: 4 4 4 4 4 9
Yellow: 3 3 3 3 8 8
Blue: 2 2 2 7 7 7
Magenta: 1 1 6 6 6 6
Olive: 0 5 5 5 5 5
Sorted alphabetically, each die beats the next die in a circle (Yellow also beats Blue). Sorted by word length, each die also beats the next die.
Any four of these dice also form a set of non-transitive dice.
How do we prove that | S | = 5 is the only answer possible ??
You can't prove it because it is not true.
The choice given in the question does not claim that ∣ S ∣ = 5 the only possible way to get non-transitive dice. (That's why it says 'if', not 'if and only if'.)
As to why none of the other given choices are correct, again, it is because of the logic. Both of the other choices ( S = { 1 , 2 , 3 , 4 , 5 } and ∣ S ∣ ≥ 9 ) say 'only if', which is wrong because of the example given in the solution.
Let S = { i , j , k , ℓ , m } , with i < j < k < ℓ < m . Then the dice A = ( ℓ , ℓ , ℓ , ℓ , i , i ) , B = ( k , k , k , k , k , k ) , C = ( m , m , j , j , j , j ) have the desired property:
The probability that A rolls higher than B is 2/3, so A beats B
The probability that B rolls higher than C is 2/3, so B beats C
To compare A and C , note that A rolls higher than C only when A rolls ℓ and C rolls j , which happens with probability (2/3)(2/3) = 4/9 <1/2. Hence C beats A .