For sake of easy math, pick C, your threshold, to be fixed at 0, which is halfway between -inf and inf.
Now there are 4 scenarios. The numbers player 1 chooses, A and B, can be both above or both below C with probability 1/4 each. In both those cases, you have a 50/50 chance of being correct on which number is larger depending on which number you chose to see.
Then, with the other 1/2 of the times, one number will be above and the other will always be below C. In those cases, the selected strategy will have you always choose the correct highest number.
All together you are right 1/2 + 2 * 1/4 * 1/2 = 3/4 = 75% of the time. Simple probability but totally counterintuitive from the onset.
Now what if C!=0 or numbers are not being selected uniformly by your opponent? You can replace the 1/2's with (p) and (1-p) in the right spots without making any distributional assumptions and it seems (without doing the math out fully) that things cancel nicely and show that you always have a strictly greater than 1/2 chance of guessing right no matter what. Exercise left for the reader :P
Well, frankly, you're merely restating the problem :-)
Intuitively, if my f(t) is extremely heavily weighted in the interval (-100, -90), and my splitting value is usually in this range, and if Player 1 only picks positive values (randomly or otherwise), then ...
Never mind - I get it now :-)
To complete the thought, then almost every time I'll have a 50% chance of being right. But since f(t)>0 everywhere, every once in a while, even if it happens once in a billion years, I will pick a splitting value in between A and B, and this will bump the overall probability to be over 50% just slightly. Furthermore this doesn't matter whether player 1 is even utilizing randomness - it will work if he is always picking 10 and 20.
So you were on the right track, but you do not replace 1/2 with p and 1-p. Assuming Player 2 randomly picks one slip of paper (stated in problem), he will be right 50% of the time when the splitting value is not between the two numbers (almost always in my scenario). On the rare occasions your splitting number is in between the two, you will be right 100% of the time.
I prefer this explanation as the 75% one has a lot of assumptions (uniform randomness, splitting at 0, etc), and there are quite a few comments asserting 75% when in reality it can be any probability over 50%. I can relax these constraints to drastically reduce the probability to 50 + epsilon, but epsilon > 0 always.
The thing that continues to bother me: How does one get a splitting value? In classical probability, if you have a continuous distribution (which f(t) is), then using it to pick a point is impossible/meaningless. If I try to come up with an "approximation" scheme that discretizes f(t), then I can always come up with a strategy for Player A to defeat that scheme.
Player A does not know what your scheme is, and you are free to change it - in fact, you can change it for each round, and thus defeat any attempt by A to deduce it. In other words, don't have a scheme, forget about f(t), and just pick a number. Any number you pick is, ipso facto, from any number of distributions that are nonzero everywhere (and also, as it happens, from any number that are not), regardless of whether or not you have any of them in mind.
The distribution f(t) does not appear in the explanation of the outcome, and I think it's true to say that the only reason f(t) is mentioned at all is that if you do instead choose from a distribution from which some ranges of numbers are excluded, then the puzzle-poser can no longer say that your odds are strictly greater than 50-50, as player A might always pick from an excluded range (for example, if player B stubbornly insists on only picking positive numbers, and player A is determined on only playing negative numbers.) Without the requirement for f(t) to be nonzero everywhere, I think then either one would have to drop the 'strictly' claim, which makes the result look far less paradoxical (B can sometimes do better than 50-50), or one would have to put some constraints on how A chooses the number-pairs (and, furthermore, those constraints would be dependent on which particular distribution B was using.)
Say you flip over a 2: what do you think is the right guess?
That's the point of picking a random pivot: it reduces your opponent's ability to influence the situation to only placing their numbers as close to each other as possible.
In your case, if we don't allow double selecting, 2 is the lowest number between 1 and 10. So you have a 100% chance of guessing right. You're using the bound to help you. Being unbounded you still have an infinite set of numbers smaller than the number you see and an infinite number higher. To be more analogous, that would be like flipping over 5 every single time (on [0,10]). You have {1,2,3,4} below and {6,7,8,9} above. EVERY TIME.
The paper attached describes a scenario where the player A chooses a number between -infinite to infinite. The simulation uses ranges -1000000 to 1000000, it's not a surprise that the strategy works in this case.
I think if the ranges were truly -infinite to infinite, this strategy would fall apart. No matter where you set C, there are infinite values above and below it.
The game only makes sense if you attach limits to the numbers the other person is allowed to pick, and then the only "split" that works is in the middle of the range. At which point the game is sort-of "obviously true".
But for any two finite numbers, A and B, the chance that C will fall between them if selected from an infinite range is 1/infinity (or 0).
No. It doesn't even require Player A to pick them at random. The strategy for Player B still works over 50% of the time.
If I was the Player 1 in this case, and was trying to prevent Player 2 from guessing with a greater than 50% probability, I would use this method:
Pick a random 100 digit number N (using a random number generator to pick 100 digits in order).
Flip a coin. Heads, the second number is N - 1. Tails, it is N + 1.
I would imagine you could slightly beat 50%, but the percentage would approach 50% as you added more digits to your initial number. It would probably be close to unmeasurable at 100 digits (i.e. you couldn't tell if you were doing better than 50% or not).
It feels like the “puzzlement” you are supposed to feel that you can beat 50% comes from people ignoring the fact that you can look at the number before making your call.
This assumes that Player 1 allows for negative numbers. What if Player 1 always writes down positive numbers?
Let's take it a step further. What if Player 1 only writes A and B whenever he plays the game (but we keep changing Player 2 so they never realize this)?
The claim in the paper is that even then they'll guess correct over 50% of the time.
I'm probably wrong about the statement below, but will state it anyway:
I suspect (but cannot quickly prove in my mind) that the other flaw in your argument is utilizing rules like "If the distribution is uniform, you have a 50% chance of picking a positive number, and 50% chance of picking a negative number". The reason I have trouble disproving it is that classical probability theory doesn't allow for such statements: The probability of picking a single point in a continuous distribution is 0.
Edit: I figured out why it works. See this comment and its parent:
1. Person A picks 2 random #s 2. Person B picks a random split #. 3. Person B looks at one of the random #s. Based on its relationship to the arbitrary split # he choose, he then decides it the other random # is larger or smaller than the one he has in hand.
I am confused as to how choosing a random split # has any impact on the probability of the other paper in hand. I'd normally think that each random # choice is an independent event and that my "split #" has no impact on the system as a whole.
Like, you choose the numbers -500 and 250.
I choose to split at 1000. Or 200. Or ten billion.
Obviously the math works, but why does my picking another random # make a difference in the system as a whole? The relationship of all the #s to each other is still presumably completely random.
Would this work if I am instead handed two numbers, and I have to guess if a third # is greater than the first number, and I make a choice based on the relationship of the 1st number to the 2nd number? Since the 2nd number is random, it shouldn't matter who provides it, right?
So stating it that way,
"Here is 200, 750, and some third number, is the third number higher or lower than 200?"
Does that still work?
If the splitting point is given to you (by an adversary, not an actual random process), this nonzero probability for any given number is not necessarily fulfilled (at least you didn't indicate so).
https://en.wikipedia.org/wiki/Box%E2%80%93Muller_transform
Also wikipedia has an easy to understand summary too, in the two envelope problem article:
https://en.wikipedia.org/wiki/Two_envelopes_problem#Randomiz...
Here it explains with numbers larger than 0, and uses exponential distribution (even easier to generate from uniform random variable).
If instead, the game was played between two supernatural beings with access to an inifinite piece of paper, and infinite time to write the numbers down, that would be a very different game.
This demonstrates that a sequence of coin flips which could be carried out in the lifetime of a human, or by a computer while the universe exists, converted to a number between 0 and 1, and then mapped to the whole number line is not random between negative infinity and positive infinity. Because of the small number of coin flips, you will never end up with any of the really big numbers. And the really big numbers is where it's at.
Red is the region where you lose, and Green is the region where you win.
This may be a probability zero event unless you make some additional assumptions about how the numbers are selected. The whole real line is quite big and the probability of two finite intervals overlapping is null.
> choose a value in between the ranges the other player is selecting numbers from
So, 0 in the real (ha!) world.
I think the whole pick a random number and modulate the choice based on it is rather trivial. Knowing that you are starting with a 75% success rate with the optimal strategy, you should be able to produce a strategy with an arbitrary success rate P=[0.25, 0.75] with one very simple extra rule: X% of the time, choose the non-optimal option. P = 0.75 - 0.5*X. If you want to construct a 66% strategy, choose sub-optimally 0.166% of the time.
Consider a paradox: Player 1 picks a random number X, writes it on a slip of paper, and 2X on another slip, and puts them randomly in front of Player 2 as "left" and "right". Player 2 wins amount of money written on a slip he chose.
Let's say he is about to pick left one (but didn't see it yet). Let's say left has number Y. The right one has either Y/2 or 2Y, with 50/50 probability. Which means right one is more profitable to pick, because it has 1.25Y on average!
You can't win real money gambling in this game if you actually have infinite range of numbers.
However, it's interesting to note that, practically speaking, if you played this game in a pub, Player 1's density function would likely be highly predictable -especially if they came up with the numbers mentally. As long as Player 2 chooses a reasonably similar density function they can likely do quite significantly better than 50%. It would take some patter, and you might lose a friend, but I could see it being possible to win "real" money this way.
If C is chosen from the exact same distribution as A and B, I wonder if the strategy works even better than if C were any other random distribution. My intuition says yes.
1 - As some people have already clarified for other commenters, it indeed makes no difference how Player 1 picks their numbers. They can pick them from some distribution of their own, or in an adversarial manner. The probability of winning by following the strategy in the paper is still strictly greater than 1/2.
2 - In fact, even if Player 1 can read Player 2's mind and knows their strategy and even the exact distribution they will sample from (but can't see into the future to see the sample from the distribution), the probability is still strictly greater than 1/2.
3 - Since it isn't actually included in the paper or any of the comments, for the sake of completeness I'll write down the computation.
Let P(E) denote the probability of an event E, and W be the event that Player 2 wins by following the strategy suggested in the paper. Let a, b be the smaller, larger number respectively. A is the event that Player 2 picked a, B is the event that Player 2 picked b. Then summing over disjoint events,
P(W) = P(A and W) + P(B and W) = P(W | A)P(A) + P(W | B)P(B)
We have P(A) = P(B) = 1/2. Now let x be the result of Player 2 sampling from their distribution. Given that they picked A, they win if and only if a <= x, so P(W | A) = P(a <= x). Given that they picked B, they win if and only if x < b, so P(W | B) = P(x < b). Therefore,
P(W) = (1/2) [P(a <= x) + P(x < b)] = (1/2) [1 + P( x in [a,b) )]
4 - If I were to show this problem to someone else, I may try to emphasize the potentially adversarial nature of Player 1 and the odds seemingly being stacked against Player 2 by phrasing it like this (although this may be _over_ exaggerated):
* Player 1 gets to write down any two distinct real numbers on two pieces of paper, and then flips a coin. Player 2 gets to see the number on the left if the coin lands on Heads, the number on the right if the coin lands on Tails. After seeing the number, Player 2 must declare whether they are seeing the smaller or larger number. If Player 2 guesses correctly, they win $1 from Player 1. Otherwise, Player 2 pays $1 to Player 1.
Further, now knowing the rules of the game, both players can choose any particular strategy of playing the game. However, whatever Player 2 chooses as their strategy, they must inform Player 1 of their strategy and not deviate from it when the game is played. Player 1 is allowed to adjust their strategy after hearing Player 2s strategy. Would you prefer to be Player 1 or Player 2? *
I think worded like that, many peoples first guess might even be Player 1. Then their next answer would be that it doesn't matter, and then they're in for a treat when they see that they should choose to be Player 2!
But player 1 can make the edge of player 2 as low as he wishes so it really doesn't matter.
The part I changed is scaling down the first players range by a factor of 100
a=random.randint(-R//100,R//100)
b=random.randint(-R//100,R//100)
If it's possible to get some epsilon of advantage by choosing a C once, then... can you do it more? Can you combine or average more of them? Is 1 optimal? Why?
I fee like there's some shenanigans here regarding assumptions about the distribution of A, B and C but I can't really put my finger on it.
