A minor conundrum courtesy of the NYT, that is very much relevant to L2...enjoy!
The God-Einstein-Oppenheimer Dice Puzzle
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 is 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 influence the rolls). Which player, if either, is favored to win?
'god' does not play a role in this problem at all... so why even include the idea of 'god'.
As for the actual problem at hand... well it all depends on how Oppenheimer decided to put the numbers on the dice. But!! It really all depends if Einstein picks the number with the higher quantity of 'higher numbers'. Then the person who took the other dice might not have as many high numbers which lowers his odds of winning.
Then again, Im not sure if the same rules of probability would apply with these die like they would a dice with 1-6 on it. Then it would be 83.33% of NOT rolling a 6 and 16.66% of rolling a 6 (given just one dice).
Then again, 'god' does not exist in my beliefs so technically this problem could not and will never exist :D
The "god" reference in the problem is just an attempt at a physics joke. Oppenheimer is clearly at least 50/50 by marking the dice as follows:
die 1 : 1,2,3,4,5,6
die 2: 7,10,12,14,16,17
die 3: 8,9,11,13,15,18
Die 1 will always lose, 2 and 3 are equivalent, they each have a 50% chance of beating the other, so Einstein has to pick one of them and then neither player is favored. Still thinking about if there's a sequence such that 1 > 2, 2 > 3 and 3 > 1 where Oppenheimer is always favored to win.
The "god" reference in the problem is just an attempt at a physics joke. Oppenheimer is clearly at least 50/50 by marking the dice as follows:
die 1 : 1,2,3,4,5,6
die 2: 7,10,12,14,16,17
die 3: 8,9,11,13,15,18
Die 1 will always lose, 2 and 3 are equivalent, they each have a 50% chance of beating the other, so Einstein has to pick one of them and then neither player is favored. Still thinking about if there's a sequence such that 1 > 2, 2 > 3 and 3 > 1 where Oppenheimer is always favored to win.
You're on the right track with that last line. That's the L2 connection...think: Rock, Paper, Scissors.
Ok, found it. Intuitively, that they're non-transitive has to be the answer because it's quite trivial to make them equal and that wouldn't be much of a puzzle :) Just takes a minute when you're trying to actually find the combinations by hand, or at least it did for me.
die 1: 1,2,9,10,17,18
die 2: 5,6,7,8,15,16
die 3: 3,4,11,12,13,14
Probably many other equivalent solutions. Anyways, Oppenheimer wins as expected.
Yes, Oppenheimer wins, as long as he numbers the dice such that each dice essentially becomes the proverbial rock, paper or scissor. Then all he has to do is pick the dice that trumps whichever dice Einstein chooses. God removes the last dice, and voila, Oppenheimer wins.
Of course we all know that Einstein would never allow himself (or his principles) to be abused again -- not after what Oppenheimer did with E=MC2. :D
Of course we all know that Einstein would never allow himself (or his principles) to be abused again -- not after what Oppenheimer did with E=MC2. :D
Hmm, I thought Einstein always said his greatest mistake was the cosmological constant, not his contributions to the atomic bomb (though he did try to talk Truman out of using it).
There is no possible way to answer this question, its a clearly hypothetical and circumstancial assessment of the situation. How can you give a right or wrong answer to a question about a transient relationship without knowing the terms of the 2 players' dice?
It IS a mathematical question, a very simple one actually with many correct answers depending on which variable and mathematical equation you want to employ to determine the average (mean) of any number of random rolls over a period of time.
However, this is not a proper question with an answer because there is not enough given information to let us know which dice had what, and which dice was chosen by either participant.
We could postulate that player A labelled the dice in a numerical order, 1 face on each dice at a time, from 1 - 18; i.e.
In which case we can say that on average, over time, dice 3 will difinitively win.
But thats only a circumstancial answer, based on how player A decided to label the dice, and based on player B being the one who took dice 3.
This really isn't a transient relationship without any more information about any of the circumstances involved in the question. Any answer could be right or wrong as to does player A win, or player B win, or neither.. Its all in the extrapolation of the conditions of the non-mathematical forces working on the variables of the equation.
Assume that Oppenheimer and Einstein employ the smartest possible strategies. Which player, if either, is favored to win?
Jayy...you may want to read the quoted part of the problem again. We are to assume they both play perfectly.
Oppenheimer has a dominant strategy which results in his always being favored to win. If we assume he plays perfectly, then he chooses this strategy. It's a fairly standard question in game theory, to ask what the result will be given optimal play from all sides.
If I asked you, who wins in a standard game of 3x3 tic-tac-toe assuming both players use the best possible strategy, you wouldn't be able to answer that? Clearly it's a tie. The same in checkers, assuming perfect play from both sides, it's a tie.