3 ms·

Randall Munroe's Math Puzzle

8 points by BIackSwan 15y ago

He posted it on his G+ wall. (https://plus.google.com/111588569124648292310/posts) I pick two real numbers through some process unknown to you. It might be random and it might not. Maybe I always pick "3" and "100". Maybe I roll two dice. Maybe I write C code by mashing a keyboard until it compiles and prints two numbers (or produces Windows ME). Maybe I always use 0 for the first number, and for the second I call my aunt and ask her for a negative real number, which I multiply by the estimated number of protons in the universe. (At this point, my aunt is used to that kind of call from me.) I put these two numbers on slips of paper and put them in two envelopes. I thoroughly shuffle the envelopes, and then you choose one via a fair coin toss. You open it and look at the number. You are now given the option (as in the infamous but very different Monty Hall problem) of switching to the other envelope. Your goal is to pick the envelope with the higher number. Can you come up with a strategy that guarantees you a better-than-even chance of winning? It has to do better than 50% no matter how I picked the numbers--if your strategy includes the rule "switch if the number is a 2", it's wrong, because I could always be picking 2 and negative 100, and in that scenario your strategy will fail at least 50% of the time. This means your strategy must work even if I have guessed what your strategy is and am cherry-picking numbers specifically to defeat it. I cannot emphasize enough that you do not know anything about my process. Not only do you not know the numbers, you don't know which random distribution I am picking them from—or whether it's even a random distribution at all—and there is no deductive basis for estimating the odds that I'm using any particular method. (This gets at what is sometimes called the difference between risk and uncertainty.) I'll post again tomorrow with the correct answer and a bit more about the puzzle.

ColinWright 15y ago
Firstly, you don't have to shuffle the envelopes, and you don't have to specify that I must use a fair coin to choose one. The problem as stated is unnecessarily over-specified. Secondly, you say: > Can you come up with a strategy that guarantees > you a better-than-even chance of winning? It should be pointed out that this should be a fixed strategy that will be used on every occasion we play this game. The challenge is to find a strategy which will, independently and on every occasion we play, have, on that occasion, a chance of winning that is strictly greater than 50%. Finally, this is well-known (in some circles) and controversial. I remember getting into a protracted discussion (probably around 100 emails) with someone claiming that the stated strategy that achieves better than 50% was wrong, and that it was in fact still exactly 50%. It took a very, very long time to convince them that they were calcualting the wrong thing. So for your sake, I hope you get the explanation crystal clear, and avoid all the pitfalls. Otherwise you will be hounded. Very brave. Added in edit: I've just gone and looked at the comment thread there, and now I'm depressed. I should know better than to look at people trying to talk about math and probability, even a sub-group such as comment on XKCD blag posts.
- BIackSwan 15y ago
  Lets see the solution he puts up.
  - ColinWright 15y ago
    I'm pretty sure there's only one solution (I have no proof of that) and I'm pretty sure he'll put up the solution I know. There will then be a deluge of replies from people who either don't understand the problem, think it's a different problem, or don't understand the solution. Already in the comments we see the solution, followed by many many posts from people who really just don't understand it. The cynic in me says it's a great way to generate traffic, but it's a really poor way to create understanding. It's almost a math equivalent to FUD. If you want to know my solution, email me. Address is in my profile. Although the lack of upvotes suggests that no one is that interested, I won't post it here and spoil the fun in case someone wants to think about it first. In case you're interested, I've uploaded a statement of the problem (slightly altered to make it a little more general) along with a solution. The MD5 of the file is 0a600b1f7f26c44854b8a8118f5491a1 - so if you ask later and download the file you can check that I've most likely not changed it.
- impendia 15y ago
  You don't have to both shuffle the envelopes and flip a fair coin, but you have to do at least one of these or something equivalent. The first player chooses two numbers by any means whatsoever, but must pick one of them genuinely randomly. If the first player simply chooses two numbers and then tells you one of them arbitrarily, he/she could always pick the higher number -- or the lower number -- and you can't defeat both of these strategies independently of knowing which one he/she picked. BTW, I cheated and looked at the (official XKCD) solution -- and I recommend no one else do the same! This puzzle is awesome.
SamReidHughes 15y ago
Well obviously you just keep playing until you've won more than you've lost, and then you stop playing.
gujk 15y ago
Randall Monroe posted the solution ages ago on his blog. Is he syndicating himself on G+?
gujk 15y ago
For the record, the solution assumes the existence of an infinitesimal real number larger than zero, or a finite bound on the adversaries' range of choices. The larger his range, the smaller your chances of beating 50-50, with no limit on how small your chances may be.

8 points by BIackSwan 15y ago

ColinWright 15y ago

Firstly, you don't have to shuffle the envelopes, and you don't have to specify that I must use a fair coin to choose one. The problem as stated is unnecessarily over-specified. Secondly, you say: > Can you come up with a strategy that guarantees > you a better-than-even chance of winning? It should be pointed out that this should be a fixed strategy that will be used on every occasion we play this game. The challenge is to find a strategy which will, independently and on every occasion we play, have, on that occasion, a chance of winning that is strictly greater than 50%. Finally, this is well-known (in some circles) and controversial. I remember getting into a protracted discussion (probably around 100 emails) with someone claiming that the stated strategy that achieves better than 50% was wrong, and that it was in fact still exactly 50%. It took a very, very long time to convince them that they were calcualting the wrong thing. So for your sake, I hope you get the explanation crystal clear, and avoid all the pitfalls. Otherwise you will be hounded. Very brave. Added in edit: I've just gone and looked at the comment thread there, and now I'm depressed. I should know better than to look at people trying to talk about math and probability, even a sub-group such as comment on XKCD blag posts.

BIackSwan 15y ago
Lets see the solution he puts up.
- ColinWright 15y ago
  I'm pretty sure there's only one solution (I have no proof of that) and I'm pretty sure he'll put up the solution I know. There will then be a deluge of replies from people who either don't understand the problem, think it's a different problem, or don't understand the solution. Already in the comments we see the solution, followed by many many posts from people who really just don't understand it. The cynic in me says it's a great way to generate traffic, but it's a really poor way to create understanding. It's almost a math equivalent to FUD. If you want to know my solution, email me. Address is in my profile. Although the lack of upvotes suggests that no one is that interested, I won't post it here and spoil the fun in case someone wants to think about it first. In case you're interested, I've uploaded a statement of the problem (slightly altered to make it a little more general) along with a solution. The MD5 of the file is 0a600b1f7f26c44854b8a8118f5491a1 - so if you ask later and download the file you can check that I've most likely not changed it.
impendia 15y ago
You don't have to both shuffle the envelopes and flip a fair coin, but you have to do at least one of these or something equivalent. The first player chooses two numbers by any means whatsoever, but must pick one of them genuinely randomly. If the first player simply chooses two numbers and then tells you one of them arbitrarily, he/she could always pick the higher number -- or the lower number -- and you can't defeat both of these strategies independently of knowing which one he/she picked. BTW, I cheated and looked at the (official XKCD) solution -- and I recommend no one else do the same! This puzzle is awesome.

SamReidHughes 15y ago

Well obviously you just keep playing until you've won more than you've lost, and then you stop playing.

gujk 15y ago

Randall Monroe posted the solution ages ago on his blog. Is he syndicating himself on G+?

gujk 15y ago

For the record, the solution assumes the existence of an infinitesimal real number larger than zero, or a finite bound on the adversaries' range of choices. The larger his range, the smaller your chances of beating 50-50, with no limit on how small your chances may be.