Solution to God-Einstein-Oppenheimer Dice Puzzle

The only issue with these problems is this thing called google. When you want to take time to think, you realize you’re “competing” against people who are busily going through google searches for the answers. It’s almost like we need an honor code.

I can’t see that answer being correct.

Neither have an advantage.
Oppenheimer is asked to label the dice prior to being explained the rules of the game, and therefore, has no idea how to manipulate the dice in order to win the game.

-Is my response nitpicky…yes…

-Are 99% of the persons who responed to the question smarter than I am… Most likely…

-Is my answer correct… I think so

In post #285, I displayed my discovery of one of the 21/21/21 solutions, AND the solitary (!) 21/21/25 solution, which I uncovered by method #3 “fiddling with equals.” Took about half an hour of idle concentration.

I didn’t immediately assign values to each die; instead, I created a marker grid with three rows and 18 columns, with only one marker possible in each column and only six markers possible in each row. I followed the simple rule that every marker beats the markers one row below and all columns to the left. Once I found a pattern I liked, I assigned the markers the number value of the column it occupied.

I created a symmetrical arrangement of three die which all won 18/36 rolls. Then I fiddled till I got a 19/19/19, recorded this arrangement, then fiddled again till I got 20/20/20, and then 21/21/21. After this, I soon discovered it impossible to create an 22 triplet, but in the process, DID discover the 21/21/25.

I couldn’t mathematically PROVE those were the extreme results, but for half an hour’s work, I was satisfied … as I’m sure the Babylonians were with an answer of 25/8 for “pi.”

Wouldn’t you simply stack the discard die with 1-6, give one winner 18, plus the odd numbers 7-15, the other winner 17, plus the even numbers 8 – 16?

From a “game theory” perspective, 21/21/25 is no better than 21/21/21, especially if Einstein gets to choose which die he would use.

I read Dr. Mutalik’s dialogue and it gave me the shivers in a good sort of way. I would like my 17 year old grandson to read it.

However, I have one thought. Dr. Mutalik has God saying, “And so, without affecting anything that had gone on before, I redesigned the basis of reality…with Quantum Mechanics.” Wouldn’t Quantum Mechanics have been in the making of the universe from the very beginning? Or is this just poetic license?

Pat McKeage
English Major

I think the common fallacy many people were making (myself included), was comparing the expected values (which are transitive) of the possible dice, concluding that the player with the die that has the highest expected value wins. This is not the case since the expected value takes into account by how much a given die will win or lose every time. Meanwhile, we know that a win is a win regardless of by how much, and this is why it is possible to have non-transitivity of “winning” dice, as others have already given examples of.

Thank you, John, for awarding me the prize. The real prize was the excitement of solving the puzzle, and watching the solution grow richer with every one’s contributions. However, what I enjoyed most was writing the dialogue. Thanks again for this wonderful experience!

To those who’ve posted above:
Jonathan:
I took this point up with John, and his response was basically that it wasn’t a race. Although the first poster may get a mention, it’s not necessary that they will get the prize. And if you want to have the joy of solving the problem, why would you spoil it for yourself?

Len:
I’m amazed you were able to find some of the optimal solutions within half an hour. You obviously have a talent for playing with numbers. It might be worth analyzing your marker method to see if sets some sort of constrained limit on the movement of the markers for the three dice. If it does, it might yield a general method to yield all possible optimal solutions in the general case. On the other hand, there may not be any such methods. As I said, that’ll be a good math project.

Pat:
I’m glad you enjoyed the dialogue. Yes, I did use poetic license when I wrote the part that you quoted. However… if there is really someone up there who designed it all… something like this could have happened. Of course, quantum mechanics needs to have been in the making of the universe from the beginning. But several theorists, including the Nobel prize winner Gerald ‘t Hooft have suggested that there’s a “veiled reality” underlying QM that is deterministic, and gives rise to the probabilistic nature of QM. So conceivably, God could have “applied the veil” in 1920 say, and prevented humans from ever accessing it! This would not have changed anything that went before. Fascinating thought, eh?

Chuckre:
You are right, but aesthetically it is the best among equals out of the 15 optimal solutions. And you could always hope that Einstein would get dispirited facing certain defeat and make the wrong choice (Sorry, game theory police!)!

David:
The whole point is that you don’t know which will be the discard die till the end.

Rich:
Let’s pick nits. The problem clearly stated that God explained the rules, and there is no mention of the game having started before the explanation was done.

Einstein does not play dice.

Len: thank you for your description of your heuristic!

I am currently in India but immensely enjoyed John’s great puzzle, and the ensuing brilliant analysis from many of you. Of course I was delighted that Pradeep made such a useful contribution. I am as much a student of Philosophy as Pradeep is a scientist. I often quote him Will Durant that Science is conquered territory while philosophy is the relatively unorganized front trench in our conquest of truth! He says “Bah” to me. But he can’t reverse his genes! Returning to thoughts on QM there is a fascinating verse in Geeta (the kernel of Ancient Indian thought), that says God’s ways are inscrutable and says that he created the universe in a spasm of ecstasy and concealed the methodology from the humans so that they would not mess it up! You don’t have to believe this, to enjoy this as an interesting thought. So I am one of those who firmly believe that QM is deterministic!

John: Thanks for the kind words, but please don’t call it my puzzle — it was strictly your son’s, and your pride in him is entirely justified.

“Of course there are other ways to find solutions in modern times, with Google, Wikipedia and brute-force computer search. But that’s cheating, unless you solved the problem the old-fashioned way first!”

I take quite a bit of offense at having ‘brute-force’ computing equated with internet search. One is indeed cheating; it does not generate one’s own work. While inelegant to math purists, a Monte Carlo approach still requires quite a bit of problem setup, programming, and analysis. Those of us inclined to this method probably find as much joy in an elegant answer (which this problem indeed has, kudos to the submitter) as the purists do in an elegant process.

We simply are using all the tools God has put at our disposal. If you ask a question that can reasonably be solved using this type of approach, you shouldn’t be so disdainful of those that choose to utilize it.

-Flint Hasset

Flint:

You are right. In retrospect, I think I did come down too heavy-handedly on computer-based methods.

I enjoy programming algorithmic solutions as much as anyone, and the complete list of optimal solutions was indeed generated by such an effort, since, as I noted above, no analytical methods exist that I’m aware of. As you remarked, these methods do require hard work, analysis, skill, and I’d say, even imagination . The danger, however, is that once you have acquired facility in doing this, it gives you so much power, that it is tempting to use it as your first tool, when it should be at least your second. The amount that I learned about the nature of the problem and its solutions by solving it the old-fashioned way was invaluable. And it was nicely complemented by the computer stuff, and I might add, by the internet research.

If I had done it the other way around, I daresay it may not have been so insightful.

Best.

First, I don’t believe there are many people who actually would enjoy “programming algorithmic solutions” and secondly God realized that he omitted quantam mechanics in universe prime so God retroactivly redisigned the universe to suit purposes of this puzzle.

@Lawlerskates: I personally find both the thought-experiment and algorithmic methods satisfying and would even say there is an elegance in providing the simplest brute-force solution with as few lines of code as possible.

Re: #8, Pradeep, and #10, Keith

Thank you for your interest. Without repeating what I’ve already explained, here is the process I followed (my papers are not with me, so this is a cold start):

All win 18/36
Die 1: * * * * * *
Die 2: * * * * * *
Die 3: * * * * * *

The markers are place holders for numbers based on relative position, with numbers getting larger the farther to the right the marker is placed.

Looking at Die 1 as an example, three of its markers never best any of the markers for Die 2, while the other three always best all six of the markers for Die 2 … 3 x 6 = 18.

To change the advantage for Die 1 to 19/36, all I need to do is shift one of its “total loser” markers into a postion where it bests one marker for Die 2. Like this:

Die 1: * * * * * *
Die 2: * * * * * *
Die 3: * * * * * *

3 x 6 + 1 x 1 = 19

The same can then be done for the relationship between Die 2 and Die 3:

Die 1: * * * * * *
Die 2: * * * * * *
Die 3: * * * * * *

Next, however, moving a marker for Die 3 to best Die 1 alters the relationship for Die 2 and Die 3, so that Die 2 only beats Die 3 by 16/36:

Die 1: * * * * * *
Die 2: * * * * * *
Die 3: * * * * * *

Fortunately, because of the flexibility for moving markers in the center of the field, only one more adjustment to the marker for Die 2 is necessary to give all three die a 19/36 superiority.

Die 1: * * * * * *
Die 2: * * * * * *
Die 3: * * * * * *

The shifts are much more straight-forward for the next step:

All win 20/36
Die 1: * * * * * *
Die 2: * * * * * *
Die 3: * * * * * *

Then, a double-iteration sweep is needed again:

All win 21/36
Die 1: * * * * * *
Die 2: * * * * * *
Die 3: * * * * * *

And finally, to achieve 25/36 for one die requires only one position change on Die 2:
Die 1: * * * * * *
Die 2: * * * * * *
Die 3: * * * * * *

Hope that was clear enough.

Arrgh, the markers didn’t space properly!!

Re: #8, Pradeep, and #10, Keith

Thank you for your interest. Without repeating what I’ve already explained, here is the process I followed (my papers are not with me, so this is a cold start):

All win 18/36
Die 1: * * *———————— * * *
Die 2: ——* * *————-* * *
Die 3: ————* * * * * *

The markers are place holders for numbers based on relative position, with numbers getting larger the farther to the right the marker is placed.

Looking at Die 1 as an example, three of its markers never best any of the markers for Die 2, while the other three always best all six of the markers for Die 2 … 3 x 6 = 18.

To change the advantage for Die 1 to 19/36, all I need to do is shift one of its “total loser” markers into a postion where it bests one marker for Die 2. Like this:

Die 1: * * –*———————- * * *
Die 2: —-*—* *———— * * *
Die 3: ————* * * * * *

3 x 6 + 1 x 1 = 19

The same can then be done for the relationship between Die 2 and Die 3:

Die 1: * * –*———————- * * *
Die 2: —-*– *– *———- * * *
Die 3: ———-*– * * * * *

Next, however, moving a marker for Die 3 to best Die 1 alters the relationship for Die 2 and Die 3, so that Die 2 only beats Die 3 by 16/36:

Die 1: * * –*——————– *– * *
Die 2: —-* –*– *——– * * *
Die 3: ———-* –* * * *——– *

Fortunately, because of the flexibility for moving markers in the center of the field, only one more adjustment to the marker for Die 2 is necessary to give all three die a 19/36 superiority.

Die 1: * * –*——————– *– * *
Die 2: —-*– *——– *– * * *
Die 3: ———-* * * *– *——– *

The shifts are much more straight-forward for the next step:

All win 20/36
Die 1: * * —–*—————— * *– *
Die 2: —–* *———— * * * *
Die 3: ———–* * * * *———— *

Then, a double-iteration sweep is needed again:

All win 21/36
Die 1: –* * *——————– * * *
Die 2: *——– *——– * * * *
Die 3: ——–* –* * * *————– *

And finally, to achieve 25/36 for one die requires only one position change on Die 2:
Die 1: –* * *——————— * * *
Die 2: * —————–* * * * *
Die 3: ———* * * * *—————- *

Hope that was clear enough.

Does God play dice where it really counts?

Obviously, none of Vegas casinos are going to let God play at their craps tables.

They know a no win situation when they see one.

But, I’m sure they’d give him their best VID (very important deity) suite!

FWIW, I worked on a simplified problem by visualizing triangles with 3 sides instead of dice with 6 and numbers 1-9. (I guess I’d stick a pin in the center of each triangle to make it a spinner.) That made the trial and error much easier.

The following is the quickest way I’ve found of generating the fifteen solutions in which Oppenheimer is guaranteed a 21/36 advantage:

Each way of numbering the dice (disregarding relative positions on the individual dice) can be represented as an 18-letter word containing exactly 6 instances each of ‘a’, ‘b’, and ‘c'; the numerical position of each letter in the word represents a number on the corresponding die A, B, C.

Start with the arbitrary arrangement
i. aaaaaabbbbbbcccccc
Obviously this is a poor choice for Oppenheimer, as die C will always win. So balance things out:
ii. cccaaaaaabbbbbbccc
C now breaks even against A and B, while B always beats A. Still not satisfactory. B is already beating A; to complete the cycle, C needs to beat B, and A to beat C. Thus we must move at least one ‘a’ past at least one terminal ‘c’ (with the greatest advantage accruing to A if we move the ‘a’ to the very end) and at least one ‘b’ ahead of at least one initial ‘c’ (again, with the best result if the ‘b’ goes to the very beginning):
1. bcccaaaaabbbbbccca
Lo and behold: The optimal solution. A beats C 21 of 36, C beats B 21 of 36, and B still beats A 25 of 36. Moving additional ‘a’s or ‘b’s out of the center to increase A’s advantage over C or C’s over B doesn’t work, because B’s advantage over A would drop below 21 of 36. Since B has the extra margin over A, however, we can swap letters in the center section (thus having no impact on the interaction with C) to obtain slightly less optimal results:
2. bcccaaaababbbbccca (B beats A 24 of 36)
3. bcccaaabaabbbbccca (B beats A 23 of 36)
4. bcccaabaaabbbbccca (B beats A 22 of 36)
5. bcccabaaaabbbbccca (B beats A 21 of 36)
6. bcccaabaababbbccca (B beats A 21 of 36)
7. bcccaaabbaabbbccca (B beats A 21 of 36)
8. bcccaaababbabbccca (B beats A 21 of 36)
9. bcccaaaabbbbabccca (B beats A 21 of 36)
10. bcccaaaabbbabbccca (B beats A 22 of 36)
11. bcccaaaabbabbbccca (B beats A 23 of 36)
12. bcccaaabababbbccca (B beats A 22 of 36)
Swapping ‘ba’ for ‘ab’ reduces B’s wins by 1; swapping ‘ab’ for ‘ba’ increases B’s wins by 1; doing both simultaneously has no impact on the results. The order of presentation here was arbitrarily chosen in order that each solution derives from the immediately previous one either by a single swap or a pair of offsetting swaps.

Note that every one of these both begins and ends with the string ‘bccca’. In three cases, there is an identical string both following the initial ‘bccca’ and preceding the terminal one: ‘b’ in #5, ‘aabb’ in #7, and ‘a’ in #9. Swapping these strings simultaneously at both ends does not change the overall results (the effect at one end being exactly offset by that at the other), yielding the three remaining solutions
13. bbcccaaaaabbbcccab (from #5: ‘b’ ‘bccca’ ‘aaaabb’ ‘bccca’ ‘b’)
14. aabbbcccabcccaaabb (from #7: ‘aabb’ ‘bccca’ ‘bccca’ ‘aabb’)
15. abcccaaabbbbbcccaa (from #9: ‘a’ ‘bccca’ ‘aabbbb’ ‘bccca’ ‘a’)

Obviously it is possible to obtain other solutions by this method if we permit swaps that reduce one or more of the win numbers below 21. I haven’t tried to determine whether it’s possible to obtain _all_ the solutions in which C beats B beats A beats C through a chain of swaps from the ‘bcccaaaaabbbbbccca’ master solution without going through any non-solutions along the way, but I wouldn’t be surprised if it were. Cutting B’s wins over A to 20 permits us to reach the positions
bcccbaaaaabbbbccca and bcccaaaabbbbbaccca
from which the ‘bcccb’ and ‘accca’ strings give a lot of maneuvering room.