Probability Puzzler 5 - Strange dice
Posted Friday, September 23 (solutions by evening of Wednesday, September 28)
The puzzle: I show you four unusual six-sided dice. They are unusual, because they do not have the numbers 1 through 6 on their six faces. Instead, here's what the dice look like (what I'm going to do is list the six numbers on the faces, in increasing order):
- Dice A: 0, 0, 4, 4, 4, 4
- Dice B: 3, 3, 3, 3, 3, 3
- Dice C: 2, 2, 2, 2, 6, 6
- Dice D: 1, 1, 1, 5, 5, 5
I play the following game with you: you choose one of the dice, then I choose another. Having made our choices, we both roll our respective dice. You win the game if you roll a higher number than me (notice that ties are impossible).
Which dice do you choose, in order to maximize your probability of winning this game?
A solution: It turns out that, by choosing first, you are dooming yourself to (2/3) certain loss!
- If you pick dice B, then I pick A, and my probability of winning is 2/3 (if I roll a 4 I win; if I roll a 0, I lose)
- If you pick dice C, then I pick B, and my probability of winning is still 2/3 (if you roll a 2 I win; if you roll a 6, I lose)
- If you pick dice D, then I pick C, and my probability of winning is still 2/3 (if I roll a 6 (which I do 1/3 of the time) I win; if I roll a 2 (which I do 2/3 of the time), I win as long as you roll a 1 (which you do 1/2 of the 2/3 of the time))
- Finally, if you pick dice A, then I pick D, and my probability of winning is, bizzarely, still 2/3 (if I roll a 5 (which I do 1/2 of the time) I win; if I roll a 1 (which I do 1/2 of the time), I win as long as you roll a 0 (which you do 1/3 of the 1/2 of the time))
This is an example of a phenomenon that we do not encounter too often: non-transitive. If A is better than B, and B is better than C, and C is better than D, then it makes sense that A would be better than D; but not in this case!
The four dice of this problem are called Effron's dice, after their inventor. For more on non-transitive dice, see the wikipedia page Nontransitive dice.
Solvers: (in no particular order)
- Dane Krzyszowski (Ace)
- Sean Meehan (2)
- Ankur Chawla (3)
- Maria Corsaro (4) (winner)
- Gabe Schepergerdes (5)
- Mike Wiederecht (6)
- Elliot Pearce (7)
- Kevin Park (8)
- Joseph Fallon (9)
- Kathryn Murphy (10)
Winner was decided by a draw from a deck of cards with picture cards removed.