[EventHorizon]

I just stumbled on this puzzle at http://tierneylab.blogs.nytimes.com/2009/03/30/the-god-einstein-oppenheimer-dice-puzzle/

I thought it was pretty good, so I'm copying it here to let braindenners have a try:

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?

Instead of simply asking who is favored to win, my question for you is this: What are the best odds for the dice war game that Oppenheimer can get given that Einstein will minimize it by his die choice?

In other words, I want to know how much the winning player will win by, and not just who wins. Good luck...

[Guest]

I just stumbled on this puzzle at http://tierneylab.blogs.nytimes.com/2009/03/30/the-god-einstein-oppenheimer-dice-puzzle/

I thought it was pretty good, so I'm copying it here to let braindenners have a try:

Instead of simply asking who is favored to win, my question for you is this: What are the best odds for the dice war game that Oppenheimer can get given that Einstein will minimize it by his die choice?

In other words, I want to know how much the winning player will win by, and not just who wins. Good luck...

they both have a 50-50. I can't come up with any scenario where O could trick E into an inferior die.

[EventHorizon]

they both have a 50-50. I can't come up with any scenario where O could trick E into an inferior die.

one player does have an advantage...

[bushindo]

one player does have an advantage...

If only I get paid for working on puzzles...

It is possible to construct 3 dice so that dice A beats dice B, B beats C, and C beats A. Then the one who gets to construct the die always win. Here's a smaller example from 1-9.

Dice A: 9,3,2

Dice B: 8,7,1

Dice C: 6,5,4

We can extend this to the case of 1-18 easily. The optimality case will have to wait. I'm in a meeting right now, so I have to pretend to be working.

[Guest]

one player does have an advantage...

O would have a 19:17 advantage

[Guest]

O can indeed mark the dice such that for whichever dice E chooses, O would have 19:17 odds of winning.

O can mark the dice as below to get these odds:

Dice A: 1,3,8,12,15,18

Dice B: 2,6,7,11,14,17

Dice C: 4,5,9,10,13,16

A beats B by 19:17

B beats C by 19:17

C beats A by 19:17

[Guest]

I got the same odds as DeeGee with slightly different dice can anything better be achieved?

Die 1 die 2 die 3

2 3 1

5 6 4

9 7 8

12 10 11

13 14 15

16 17 18

[bushindo]

I got the same odds as DeeGee with slightly different dice can anything better be achieved?

Die 1 die 2 die 3

2 3 1

5 6 4

9 7 8

12 10 11

13 14 15

16 17 18

I can get 20:16 odds, but even better odds might be possible.

This is an extension but my earlier smaller example from 1-9.

Dice A: 9,3,2

Dice B: 8,7,1

Dice C: 6,5,4

We can see from above that the smallest edge is 5:4 (Dice A over dice B). We can extend this to 1-18 by letting the numbers in the smaller example be consective numbers in the larger example (i.e. smaller example number 1 = larger example number 1 and 2, smaller example number 8 = larger example number 16 and 15). So this configuration would have the same 20:16 odds as the 1-9 example

Dice A: 18, 17, 6, 5, 4, 3

Dice B: 16, 15, 14, 13, 2, 1

Dice C: 12, 11, 10, 9, 8, 7

But better odds might be posssible by taking advantage of the finer granularity of the dice. Still working on this, but what I would really like is a conclusive proof of optimality, as opposed to just an example though.

[Guest]

every time you win you get 1 point, it's not based on how much higher the number is, only whether or not it is higher.

bushido, with your 3 dice, dice A will beat dice B 2/3 of the time, and B beat dice C 2/3 of the time, and dice C will beat dice A 2/3 of the time.

Edited August 28, 2009 by phillip1882

[Guest]

okay wait. with bushido's dice,

for A vs B: A has 1/3 chance of guaranteed win, and 2/3 chance of win 1/3 of a time. (.555)

for B vs C: 2/3 chance of guaranteed win. (.666)

for C vs A: C wins 2/3 of the time. (.666)

[Guest]

every time you win you get 1 point, it's not based on how much higher the number is, only whether or not it is higher.

bushido, with your 3 dice, dice A will beat dice B 2/3 of the time, and B beat dice C 2/3 of the time, and dice C will beat dice A 2/3 of the time.

bushindo gets 20:16 A vs B, 24:12 B vs C, and 24:12 C vs A. am i missing something? perhaps a tweak could bring A:B at the same odds as the others

Edited August 28, 2009 by ljb

[bushindo]

bushindo gets 20:16 A vs B, 24:12 B vs C, and 24:12 C vs A. am i missing something? perhaps a tweak could bring A:B at the same odds as the others

A conjugate descent algorithm allows me to improve the odd a bit further, now it's 21:36. It doesn't seem possible to get the lowest odd up much further.

     [,1] [,2] [,3] [,4] [,5] [,6]

[1,]    3   16    2   15    4   17

[2,]    6   18    5   10    9    7

[3,]    1   13   12   11   14    8

Edited August 28, 2009 by bushindo

[superprismatic]

I just exhausted over all 2,858,856 ways of numbering the dice and I get odds of 7:5 that Oppenheimer wins (probability of 7/12) with the dice:

Die 1: 1,2,9,14,15,16
Die 2: 6,7,8,11,12,13
Die 3: 3,4,5,10,17,18

1 beats 2 with probability 7/12;

2 beats 3 with probability 7/12;

3 beats 1 with probability 7/12.

Edited August 28, 2009 by superprismatic