A Counterintuitive Probability Game

https://www.evanward.org/posts/a-counterintuitive-probability-game

by EvanWard97 • 7 years ago

60 65 7 years ago

dsukhin2• 7 years ago

This is a super facinating observation. At first it does seem very counterintuitive but after a little examination, the resulting probabilities make a lot of sense.

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

EvanWard975• 7 years ago

I know, it was totally counterintuitive until I coded it. Same kind of deal with the Monte Hall problem.

esrauch• 7 years ago

I think it can be more intuitively explained without getting into math: if you know someone wrote down two random numbers between 1 and 10. You look at one and have to guess if the other one is higher or lower.

Say you flip over a 2: what do you think is the right guess?

ChristianBundy• 7 years ago

Of course, but I'm having trouble understanding why this works on an infinite number line. Say you flip over a 2,387,290,723,013,172,348,238,987: what do you think is the right guess?

godelski2• 7 years ago

I think this distracts from the actual problem. They have a more generalized case, (-inf, inf).

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.

amalcon• 7 years ago

The reason it's counterintuitive is because, in the game, you don't know that the two numbers are random. In fact, they are chosen by an adversary. You flip a 2: do you think this is most likely the lower number, or do you think the other player picked 1 and 2 just to mess with you?

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.

BeetleB4• 7 years ago

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.

mannykannot2• 7 years ago

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.)

tomas789• 7 years ago

I’m struggling with choosing a random number t from (-inf, inf) where P(t) > 0.

szemet2• 7 years ago

Any distribution is good with these properties, e.g. just use standard normal distribution, it is available in every statistics libraries, but easy to generate from uniform random reals, for example:

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).

kangnkodos• 7 years ago

Almost every number chosen randomly between negative infinity and infinity would have so many digits that it would not fit on any computer currently in existence. It would not fit on a piece of paper. In the original problem, a number is written down on one piece of paper. That limits the possible numbers involved considerably.

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.

munificent• 7 years ago

Me too. Wouldn't choosing a random number on the entire real line require you to generate an infinite amount of information?

gentaro2• 7 years ago

I may be missing something here, but I'm not sure this is quite right.

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.

EvanWard975• 7 years ago

The idea is that by picking a random real number C, there is a chance it will fall between A and B over time, no matter how the other person picks their numbers A and B. It's the infinite range C can be drawn from which gives you a strictly >50% of guessing the higher of A or B no matter how the other player tries to game the system.

gentaro2• 7 years ago

> The idea is that by picking a random real number C, there is a chance it will fall between A and B over time, no matter how the other person picks their numbers A and B.

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).

dsukhin2• 7 years ago

If you have a range (-inf, inf) and you split it at some random C, even though there are still infinite values on either side, the probability you choose another number on the left side of C is (p) and on the right side is (1-p) regardless of how you choose the next number. That's what makes these probabilities finite and makes the math work out.

OskarS• 7 years ago

But... but... you can't "pick a number uniformly distributed in (-inf, inf)". Like, that's not a thing that is possible [0], so saying anything about the probabilities once you've done so is not reasonable. You can pick a random number in infinite range non-uniformly (e.g. to pick a number in the range [1,inf), pick a number X in (0,1] and use 1/X), but then the probabilities are no longer so neat and tidy.

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".

[0] https://math.stackexchange.com/a/14169/641751

_v7gu• 7 years ago

Pick a random C from R. A and B have equal probabilities of being higher or lower than C. Now with the restriction A>B, there are only three equally likely options: A>B>C, A>C>B, C>A>B. You can guess half the time correctly for A>B>C and C>A>B and all the time for A>C>B, which gives you 66% which is shown with the random trials.

BeetleB4• 7 years ago

The paper attached describes a scenario where the player A chooses a number between -infinite to infinite.

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.

bko• 7 years ago

From the paper, the only conditions of the first player (the person picking the initial two random numbers) is that they are distinct. They don't have to be random and (as the implementation has) bound by the same range as the second player. If you use a different bound range for the two players, you get about 50/50 odds.

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)

deckar012• 7 years ago

If you have ever played a card game where you have to guess if the next card is going to be higher or lower than another card, this should not be a surprising result. You should always assume the next card will be the average of all possible cards. What makes these kinds of games fun is having to recalculate the average card in you head based on the cards that you can see.

lonelappde5• 7 years ago

This is a different problem. Cards are finite. The OP problem is infinite.

deckar012• 7 years ago

The set of real numbers used in the simulation is finite, but it doesn't prevent the result from being approximately the same. Using a small number of cards is no different. The best solution should be obvious, but was not presented in the cited paper, and was glossed over in this example. The more interesting result to me is that there is no difference between choosing the larger of two finite sets and choosing the more probable of two infinite sets.

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.

nabdab• 7 years ago

I’m really struggling to see how this is counter intuitive? You pick one of the numbers, and if it’s positive you say “this is likely the largest” and if negative “this is likely the smallest”. Since there’s a 50/50 chance for each sign, and all positive numbers are larger than all negative numbers. That already gives you the minimum 1/2 probability they are asking for. But you’ve got an improvement because not only does the other number have to match the sign for you to lose, it also has to be larger in magnitude, and that happens in half the cases where they have the same sign, so that gives you 1/4 extra. You end up being correct 75% of the time. The thing with a random number in between is just an odd way of reducing your probability of winning by adding randomness to the split, what that’s supposed to show I’m really confused by.

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.

EvanWard975• 7 years ago

They don't have to pick numbers equally from below and above zero. They could try to make your odds as difficult as possible by picking 10^20 and 10^20-1. The power of picking a random C is that there is a nonzero probability it will fall between their two numbers.

com2kid• 7 years ago

It is confusing (and I am still confused) because my understanding is:

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?

MauranKilom2• 7 years ago

The important part in the original is that every choice of splitting point has nonzero probability. This means you will _eventually_ land on a splitting number between the two where you will then have 100% chance to be right (instead of 50% chance for all the other times), pushing your average above 50%.

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).

BeetleB4• 7 years ago

But you’ve got an improvement because not only does the other number have to match the sign for you to lose, it also has to be larger in magnitude, and that happens in half the cases where they have the same sign, so that gives you 1/4 extra.

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:

https://news.ycombinator.com/item?id=21160811

greenbay202• 7 years ago

Note that the paper claims that no matter how player A chooses the two numbers to write down, you can always guarantee >50% chance in guessing which one is the higher one after only looking at one of the numbers. The reason this seems counter-intuitive is that the only information you're given is one of the numbers. You are not told anything about how the first player picked her numbers. Now, the reason why OP's code does not seem counter-intuitive to most is that it shows a quite different result. The OP assumed a specific distribution (which is a strong assumption, one the paper does not make) and found a strategy that yields > 50%.

EvanWard975• 7 years ago

I did not assume they chose A or B from a specific distribution, especially one centered around zero. It's just how I decided to finish coding it. You'll get >50% as long as C can fall between A and B. I guess I should have selected weird distributions for A and B to be chosen from to show that the principle holds.

EvanWard975• 7 years ago

I updated the code to highlight the principle a bit better.

cortesoft2• 7 years ago

Does this hold true if Player 1 is adversarial?

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).

nimih• 7 years ago

The result holds even if Player 1 knows Player 2's strategy and attempts to foil it, although as you correctly observe, the paper only claims >50%, and Player 1 can minimize that margin by picking the two numbers to be right next to each other (and trying to choose them such that they occur in a low-density area of Player 2's distribution f(t) ).

mraudiobook_com• 7 years ago

Exactly, came to say this. The TL,DR should be, it actually is intuitive if you make specific assumptions.

KwisatzHaderack• 7 years ago

Very interesting, thanks for sharing.

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.

ryandrake• 7 years ago

The code linked in the article demonstrates this.

laegooose• 7 years ago

Strictly speaking, the _probability_ that Player 2 did correct choice is not defined unless the random distribution used by Player 1 is defined.

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!

dfelix• 7 years ago

Player 1 is not constrained to choose numbers randomly from a distribution. Player 1's choices are considered to be arbitrary. Player 2's probability of winning can then be calculated from Player 1's choices and Player 2's privately chosen density function.

godelski2• 7 years ago

From reading the comments, it seems the counter-intuitive part is that a bound like (-inf, inf) is different than a bound like [0,10].

kgwgk7• 7 years ago

the benefit of using the entire Real number line as the possible range for C is to ensure that you at least sometimes

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

lonelappde5• 7 years ago

It's not probability 0, it's a number greater than 0 *but it's impossible to prove a lower bound greater than 0".

So, 0 in the real (ha!) world.

kgwgk7• 7 years ago

https://rationalwiki.org/wiki/Probability#Zero_probability

markisus2• 7 years ago

I think I drew a picture that could explain the effect. In the picture, player 2's lower number is A, and their higher number is B.

https://imgur.com/a/DKCtH81/

Red is the region where you lose, and Green is the region where you win.

kgwgk7• 7 years ago

Now imagine zooming out to show the whole real line and it’s clear why the “guaranteed win” doesn’t ever happen :-)

markisus2• 7 years ago

And it can get even worse because your adversary can choose A and B as close together as they want.

lonelappde5• 7 years ago

The "solution" relies on the fact that positive numbers arbitrarily close to 0 are still technically positive. If you assume that the range of numbers is finite, the solution is much more intuitive. If you assume that the range of numbers is actually infinite, and don't use a misleading python program, a win ratio of "0.5 + unknowably arbitrarily small episilon" is the least impressive "strictly greater than 0.5" you can imagine.

You can't win real money gambling in this game if you actually have infinite range of numbers.

dTal• 7 years ago

The improvement in guessing accuracy over 50% is a measure of the overlap of the density functions used by the two players (specifically, I believe it is the area under both curves). The result notes that if Player 2 uses a density function +/-inf, they can guarantee at least epsilon overlap.

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.

zxcmx• 7 years ago

This still makes no sense to me. Picking a random number C adds no information, so I can't see how it can influence the accuracy of your decision. Just from a naive information theoretic point of view.

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.

Ragib_Zaman• 7 years ago

Great problem OP.

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!

kgwgk7• 7 years ago

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.

A Counterintuitive Probability Game

Related Stories