A math puzzle for Francois

The Coffee Shop
1

…or anyone else who is smart enough. From the NYTimes.com.

God does not throw dice, Albert Einstein famously declared, but suppose he was wrong. Suppose God decided to demonstrate otherwise by showing up one day at the Institute for Advanced Study. God announces that dice games are in fact wildly popular in heaven, and that the purpose of this visit it to teach a new game to Einstein and J. Robert Oppenheimer. God explains the rules:

There are three blank dice. First, Oppenheimer will take each of the six-sided dice and write the numbers from 1 to 18, in any order he likes, on the 18 faces of the three dice. Einstein will then examine the dice and select one of them as his own. Oppenheimer will then examine the remaining two dice and select one of them. (The third die will be discarded.) Oppenheimer and Einstein will then play repeated rounds of “Dice War” in which they roll the dice simultaneously, with a point being awarded each round to the player who rolls the higher number. The player with the most points wins.

Assume that Oppenheimer and Einstein employ the smartest possible strategies, and that the outcome will be determined by the laws of probability (meaning that God doesn’t skew the dice or the influence the rolls). Which player, if either, is favored to win?

2

Does Schrodinger get to view?

3

Or perhaps, more importantly, Heisenberg?

4

Hmmm…this is like the old pie-cutting game where the person who cuts gets the 2nd choice.

Seems like the best strategy for Oppie is to select the dice so that two of them have equal odds of rolling the highest number?

I think at least one partition that could do that is {7,8,9,10,11,12} {13,14,16,3,5,6} I think those have equal odds of rolling the highest number (18 each out of 36 possibilities}

Kind of a SWAG, there, but that’s my intuition…which probably means I just fell into a huge trap.

5

Seems like the best strategy for Oppie is to select the dice so that two of them have equal odds of rolling the highest number?

But that would only work if the 3rd die had lower odds of rolling the highest number, no?

(1,6,7,12,13,18) (2,5,8,11,14,17) (3,4,9,10,15,16) Would leave you will even odds on all 3, no? The other option would be to make one clearly the looser like (7,10,11,14,15,18) (8,9,12,13,16,17) & (1,2,3,4,5,6).

What I’m not understanding and probably because it’s way over my head or I’m missing the “Catch”, is that as far as I can tell the above leaves “Equal odds” which basically means there’s no winner if the number of rolls are infinite.

My guess is that I’m missing something about the particular “Mathematical beliefs” of Einstein and Oppenheimer.

~Matt

6

Seems like the best strategy for Oppie is to select the dice so that two of them have equal odds of rolling the highest number?

But that would only work if the 3rd die had lower odds of rolling the highest number, no?

(1,6,7,12,13,18) (2,5,8,11,14,17) (3,4,9,10,15,16) Would leave you will even odds on all 3, no? The other option would be to make one clearly the looser like (7,10,11,14,15,18) (8,9,12,13,16,17) & (1,2,3,4,5,6).

What I’m not understanding and probably because it’s way over my head or I’m missing the “Catch”, is that as far as I can tell the above leaves “Equal odds” which basically means there’s no winner if the number of rolls are infinite.

My guess is that I’m missing something about the particular “Mathematical beliefs” of Einstein and Oppenheimer.

~Matt

More:

Caution: Don’t assume there’s an immediately obvious or intuitive answer. When Mr. Sidhu first heard the puzzle, he promptly offered an answer only to be told by his professor, “You have only revealed that you know nothing about mathematicians.”

7

Assuming that both men know all the rules upfront,
Oppenheimer’s only chance for a tie is to have all 3 die have a total of 57 points 18+17+16… / 3
or if he stacks up two die, they have to have the same total ‘points’

This assumes the game occurs on earth - but if the Institute for Advanced Study is in Heaven, all bets are off. Heaven is a Socialist Utopia where everyone is a winner!!!

8

Assuming that both men know all the rules upfront,
Oppenheimer’s only chance for a tie is to have all 3 die have a total of 57 points 18+17+16… / 3
or if he stacks up two die, they have to have the same total ‘points’

One dice with 18/17/16/15/2/1 has 69 points. Another with 14/13/12/11/10/9 also has 69 points. The former will win 2/3 of the time, right?

9

So…we have 3 dice, A, B, C.

I’ll assume that Oppenheimer and Einstein have the same wits. I think that’s what you have in mind otherwise, you would have said Tibbs and Einstein, which favours Tibbs of course, who is a lot smarter (See his Facebook IQ as a proof).

The only chance for Oppenheimer to be favoured is for him to find a way to partition the 18 numbers in such a way that, no matter what die Einstein picks, Oppenheimer can pick a die that performs better. If such a partition does not exist, then Einstein is favoured to win because with equal wits, he will have picked the best die at the start. In this case, Oppenheimer would be better off choosing a neutral partition.

Now, to think about that last comment…if such a partition exists then Oppenheimer chooses it and he wins. If this partition does not exist, then Oppenheimer will choose a neutral partition (for instance {1,2,3,16,17,18}, {4,5,6,13,14,15} and {7,8,9,10,11,12}) and will at least not lose. So overall, because Oppenheimer can select the partition, he is favoured…

10

Ooops I guess I know little of mathematics as well.

Ignoring all the red herrings in the story, it would appear that the trick comes down to: Is there a configuration of the three die where all three die have equal potency, or, 2 of the 3 die have equal potency and the third is less potent.

Oppie has to play for a tie

Unless there is a way Oppie can setup the die so that die #1 can beat
die #2, and die #2 can beat die #3, and die #3 can beat die #1 - - then
no matter what die Albert picks he can pick one that can beat it.

11

Right. The sum of the numbers on the die isn’t important. What matters is the number of winning combinations out of all possible combinations of the two die. Getting 18 vs. 1 is just as much a win as 2 vs. 1.

12

I would say that it will be a toss up. There is most likley a combination of numbers that would be employed on the two dice, that gives a 50/50 chance it each player. Oppie would know this, and also know that if he did not do this, Einstein would figure it out and take the advantaged dice…Thats my take anyway…

13

Unless there is a way Oppie can setup the die so that die #1 can beat die #2, and die #2 can beat die #3, and die #3 can beat die #1 - - then no matter what die Albert picks he can pick one that can beat it.
Rock, paper, scissors.

14

“Seems like the best strategy for Oppie is to select the dice so that two of them have equal odds of rolling the highest number?”

Unless there happens to be a way of numbering the dice to set up a scissors/paper/rock (non-transitive) situation. Then if Einstein chose the “scissors” die, Oppenheimer would choose the “rock” die, etc., so that the game would be to Oppenheimer’s advantage. Without further analysis, I’m not sure whether there exists such a numbering scheme or not.

EDIT: Oops, Lorenzo and Ken beat me to it.

15

The point of the problem was to see if someone is favoured. Since there is a neutral partition, the worst that happens to Oppenheimer is a draw. Hence, he is favoured. I don’t even care if there is a non-transitive relation.

16

This assumes the game occurs on earth - but if the Institute for Advanced Study is in Heaven, all bets are off.
I assure you, NJ is nobody’s idea of heaven.

17

Die 1: 18,13,10,8,5,3 1>2 in 19/36 combos
Die 2: 17,15,12,7,4,2 2>3 in 19/36 combos
Die 3: 16,14,11,9,9,1 3>1 in 19/36 combos

So,
pick 1, then 3 wins
pick 2, then 1 wins
pick 3, then 2 wins
.

18

Here’s a nontransitive relation:

A = {3,4,11,12,13,14}
B = {1,2,9,10,17.18}
C = {5,6,7,8,15,16}

If Oppenheimer numbers the dice as above, A wins over B 20/36 of the time, B wins over C 20/36 of the time, and C wins over A 20/36 of the time. Therefore Oppenheimer is favored.

If there did not exist a non-transitive relation, then there would be no way that Oppenheimer could number the dice and then draw a die without allowing Einstein to reestablish parity by choosing his die correctly. Therefore, while it would be true, as you say, that “the worst that happens to Oppenheimer is a draw,” it would also be true that the best that could happen to him is a draw. Hence you are mistaken in not caring about the non-transitive relation: it is crucial to establish such such a relation exists in order to solve the problem.

EDIT TO ADD: BTW, non-transitive relations also happen in voting patterns and are therefore important in political analysis, which is why I tend to tune in on them.

19

No, I’m not. See Nash’s notion of equilibrium for games. Had Ken asked for a strict favouring of one player, then yes, I would need to find a strictly winning strategy.

20

“Had Ken asked for a strict favouring of one player, then yes, I would need to find a strictly winning strategy.”

His question was: “Which player, if either, is favored to win?” How can you answer that without figuring out whether there or not there exists a winning strategy for Oppenheimer?

EDIT TO ADD: This would really only be a game-theoretic type of situation if, at the same time that Oppenheimer was numbering his dice, Einstein was making some kind of decision in the game, with neither player privy to the strategy of the other. But the way the problem was defined, Oppenheimer makes a decision (how to number the dice), then Einstein makes a decision (which die to choose for himself), and then Oppenheimer makes a decision (which die to choose for himself). Just as in chess, at no point is either player “in the dark.” I never heard anyone suggest that Nash’s theory would be helpful in chess.