Math Puzzles (from FiveThirtyEight - new puzzle every Friday) (1 Viewer)

[Ignoratio Elenchi]

But Player 2's goal isn't to beat Player 1's average value, right? It's to beat his "naive strategy" of staying on anything above 0.5 (which we already know yields a 0.625 AV).

What's the difference?  We don't know what Player 1's actual score is, but we know if he employs the strategy of using a 0.5 cutoff, his expected score is 0.625.

If Player 2 draws, say, 0.6, he knows that against P1's naive strategy he has a 40% chance of winning if he stands pat (he'll beat P1 60% of the 50% of the time that P1's first number was <0.5, plus 100% of the 10% of the time it was between 0.5 and 0.6). Why would he throw that away for a random new number that has only a 37.5% chance of winning?

This seems like a compelling argument but I think the issue might be that you're confusing two different concepts.  In your last sentence, you say that the random new number has only a 37.5% chance of winning (implying Player 1 "actually has" a score of 0.625).  But if that's the case, then Player 2's score of 0.6 doesn't win 40% of the time, it wins 0% of the time.  Or perhaps a better way to look at it: you're right that staying on 0.6 gives you a 40% chance of winning - so why would you stay on 0.6?  By symmetry you and the other player should have an equal shot at the money, so why would you settle for a strategy that gives you only a 40% chance of winning and a 60% chance of losing?

I haven't had a chance to work on it since earlier this morning, but I'm thinking the crux of the problem might hinge on the fact that each individual number is drawn from a uniform distribution, but each player's eventual number is NOT uniformly distributed.  The distributions are skewed, depending on where you place the cutoff.  

 

[Mr. Irrelevant]

Or perhaps a better way to look at it: you're right that staying on 0.6 gives you a 40% chance of winning - so why would you stay on 0.6?  By symmetry you and the other player should have an equal shot at the money, so why would you settle for a strategy that gives you only a 40% chance of winning and a 60% chance of losing?

Your "equal shot at the money" is only the case at the start of the game, before you draw your first number. Once that number is known, all you can do is maximize your chances of winning given that number - which chances may be (in fact, usually will be) more or less than 50%.

It's like hitting a 15 against a dealer's 10 in blackjack. Your EV of doing so is negative, but it's still your optimal play because it's less negative than the alternative of standing pat.

 

[CBusAlex]

I think this will be the key to figuring out the answer mathematically. Whatever the optimum number is, you should have a 50% chance of beating it if you respin on anything below.

Let X be the optimal number, and P be the percent chance of beating that number with one spin (1-X). Your chances of beating X if you respin on everything below X are

P + (1-P)P = 0.5

P^2 - 2P + 0.5 = 0

Solving with the quadratic equation

P = 0.2928932188134524

Which means X is 0.7071067811865476

Well, I can't help but notice that IE's experimental solution of .617 is extremely close to what you get if you solve

P + (1-P)P = (1-P)

-P^2 + 3P - 1 = 0

P = 0.3819660112501051

X = 0.6180339887498949

Which means that it's probably the correct equation to use, although I can't quite wrap my mind around why that is yet.

 

[Ignoratio Elenchi]

Your "equal shot at the money" is only the case at the start of the game, before you draw your first number. Once that number is known, all you can do is maximize your chances of winning given that number - which chances may be (in fact, usually will be) more or less than 50%.

The same exact thing is true for the other guy in the other booth.  You're both playing the same exact game, free to employ any strategy you want.  You start with a 50/50 shot of winning and by symmetry you should never employ a strategy that would drop that number below 50%. 

It's not a matter of seeing the first number and then deciding what your strategy should be based on what that number is.  The strategy itself should already be decided ahead of time.  If you think standing pat on anything greater than or equal to 0.6 is a winning strategy, then that's what you'd employ.  

And actually, I think the real issue is that you can't just stop there.  It's not that a 0.6 cutoff isn't a winning strategy against a 0.5 cutoff.  It very well might be.  But we don't know if Player 1 is going to use a 0.5 cutoff.  He's got the same freedom as you and nothing in the problem suggests he's any less intelligent or strategic than Player 2, so he almost certainly won't use a 0.5 cutoff.  So now you have to recalibrate.  

 

[Ignoratio Elenchi]

X = 0.6180339887498949

It would be pretty dope if that's the answer, because then it's just (phi-1).

 

[CBusAlex]

It would be pretty dope if that's the answer, because then it's just (phi-1).

Well great, now I just have to figure out how the #### the golden ratio applies here.

 

[Ignoratio Elenchi]

Well, I can't help but notice that IE's experimental solution of .617 is extremely close to what you get if you solve

P + (1-P)P = (1-P)

-P^2 + 3P - 1 = 0

P = 0.3819660112501051

X = 0.6180339887498949

Which means that it's probably the correct equation to use, although I can't quite wrap my mind around why that is yet.

Also still trying to wrap my head around the mechanics of it, but I think it's like:

Both players start with an equal shot at the money.  If there was some strategy one could employ to increase their chances, the other would do the same.  So basically it seems the solution is for both players to settle on the same cutoff, which seems obvious since they're symmetrical. 

So the iterative strategy is to choose a cutoff equal to your opponent's expected value.  And the end result is that you end up with the same cutoff.  So you want to find the precise cutoff point that gives you an expected value of that same cutoff you chose, which I assume is what you're calculating above.  

 

[Mr. Irrelevant]

So the iterative strategy is to choose a cutoff equal to your opponent's expected value.  And the end result is that you end up with the same cutoff.  So you want to find the precise cutoff point that gives you an expected value of that same cutoff you chose, which I assume is what you're calculating above. 

I want that to be the answer; it certainly seems more elegant than 0.5.

But I keep coming back to the fact that a property of co-optimal strategies is that neither player can improve his outcome unilaterally. In this case, if I know that my opponent is going to re-draw on anything below 0.618, and his average will thus be 0.618, why would I not revert to drawing on anything < 0.5, which yields an average of 0.625?

 

[Ignoratio Elenchi]

I want that to be the answer; it certainly seems more elegant than 0.5.

But I keep coming back to the fact that a property of co-optimal strategies is that neither player can improve his outcome unilaterally. In this case, if I know that my opponent is going to re-draw on anything below 0.618, and his average will thus be 0.618, why would I not revert to drawing on anything < 0.5, which yields an average of 0.625?

Going back to an earlier part of the discussion, the reason is because the outcomes aren't uniformly distributed.  Going with a 0.5 cutoff, you'll have a higher expected value but you'll win fewer times.  

 

[Ignoratio Elenchi]

So you want to find the precise cutoff point that gives you an expected value of that same cutoff you chose, which I assume is what you're calculating above.  

Just to confirm, let C be the cutoff.  When your first spin is below the cutoff (which will happen C% of the time) you'll spin again with expected value of 5.  And when it's above the cutoff, you'll stand pat with expected value (1+C)/2.  You want your overall expected value to equal the cutoff itself, so you end up with:

C/2 + (1-C)(1+C)/2 = C

which simplifies to CBusAlex's equation above.

 

[Ignoratio Elenchi]

This one seems too easy.  I like when puzzles seem too easy but end up having a surprising solution, but I have a bad feeling this one is as easy as it seems.  

So glad I was wrong about this. 

 

[Ignoratio Elenchi]

David Sklansky agrees with our answer. 

 

[GregR]

On the one hand, it would then seem that using any other cutoff would be suboptimal.  However, if Player 2 assumes Player 1 is using this simple strategy, then he knows on average he's going to have to beat a score of 0.625 to win.  So perhaps he should use that as a cutoff instead (like on the Price is Right, it normally wouldn't make sense to spin the big wheel again if your first spin was $0.90, but if the guy before you already got $0.95 you have to spin again).  

Good stuff here, Vizzini.

But unfortunately for you, your opponent has spent years building up an immunity to iocaine powder...

 

[Ignoratio Elenchi]

New puzzle and solution to last week's.

Hark! The NCAA Tournament starts next week, and the “granny shot” has reappeared as a free throw technique. Its proponents claim that it improves accuracy because there are fewer moving parts — the elbows and wrists are held more stable, for example, and the move is symmetric because one’s arms are, more or less, equal length. Let’s find out how effective the granny shot really is.

Consider the following simplified model of free throws. Imagine the rim to be a circle (which we’ll call C) that has a radius of 1, and is centered at the origin (the point (0,0)). Let V be a random point in the plane, with coordinates X and Y, and where X and Y are independent normal random variables, with means equal to zero and each having equal variance — think of this as the point where your free throw winds up, in the rim’s plane. If V is in the circle, your shot goes in. Finally, suppose that the variance is chosen such that the probability that V is in C is exactly 75 percent (roughly the NBA free-throw average).

But suppose you switch it up, and go granny-style, which in this universe eliminates any possible left-right error in your free throws. What’s the probability you make your shot now? (Put another way, calculate the probability that |Y| < 1.)

 

[heckmanm]

First thought is:

Spoiler

• Given: P(|X| < 1) * P(|Y} < 1) = 0.75
  
• Assume in a "normal" shot, misses are random in both axes, so P(|X| < 1) = P(|Y| < 1)
  
• Thus, P(|X| < 1) = P(|Y| < 1) = SQRT(0.75) = ~0.866
  
• With a granny-style shot, P(|X| < 1) = 1.0
  
• Therefore P(granny shot) = P(|Y| < 1) = ~0.866
  

 

[Ignoratio Elenchi]

This week's puzzle feels too mathy and too much like work.  I'm going to work on it but I have the feeling that unless you're pretty fluent with probability distributions and whatnot (which I'm not) it'll be a grind.  

 

[Ignoratio Elenchi]

First thought is:

Hidden Content

I think the third bullet point is incorrect. The probability that you'll make a "normal" shot isn't just P(X<1) * P(Y<1).  For example, when X and Y both equal 0.99, that point is clearly not in the circle.  What you need is for sqrt(X2 + Y2) < 1, so that the distance from the origin is less than one, not just that each individual component is less than 1.

 

[heckmanm]

I think the third bullet point is incorrect. The probability that you'll make a "normal" shot isn't just P(X<1) * P(Y<1).  For example, when X and Y both equal 0.99, that point is clearly not in the circle.  What you need is for sqrt(X2 + Y2) < 1, so that the distance from the origin is less than one, not just that each individual component is less than 1.

 

[heckmanm]

OK, visually, I think of a circle inside a set of lines like

| |o| |

For a normal shot, P(ball in circle) = 0.75. Some of the other 0.25 (call this a) will fall in the "center lane" either above or below the circle, and the rest (call it b) in the "outer lanes" (which are actually of infinite width, but whatever).

So, P(a) + P(b) = 0.25, but we need to figure out the individual probabilities.

Then, for a granny shot, P(b) = 0, and P(make) = 0.75 / (0.75 + P(a))

 

[CBusAlex]

Is this the same free throw shooter as the earlier question? Because that's going to introduce a lot of additional math.

 

[JayMan]

Not much time to look it up right now... but it’s seems like a classic bivariate normal distribution – that can be solve with the joint probability density function (set at 75%), express only along the Y axis... more later.

 

[heckmanm]

Or, could we consider a circle inside a square, where the area of the circle is 75% of the area of the square, so that any given point has a 75% probability of being inside the circle? (This assumes misses in all directions are equally likely). If so, then the area of the square is 4/3 area of the circle, which would looks like this:

The circle has a radius of 1, and thus an area of  [SIZE=10pt]π [/SIZE]

The outermost lines (the yellow line at the bottom, not the black one) define a square with an area of 4/3 [SIZE=10pt]π, so the circle is 75% of the square[/SIZE]

[SIZE=10pt]The vertical lines just inside the square are at x = +/- 1.  This defines the "miss" areas to the left and right that are eliminated on the "granny shot".  The remainder of the area that lies outside the circle are the potential "miss" areas that remain.[/SIZE]

 

[heckmanm]

Based on the above, I came up with about 77%, but I could easily be making faulty assumptions.

 

[JayMan]

Or, could we consider a circle inside a square, where the area of the circle is 75% of the area of the square, so that any given point has a 75% probability of being inside the circle? (This assumes misses in all directions are equally likely). If so, then the area of the square is 4/3 area of the circle, which would looks like this:

The circle has a radius of 1, and thus an area of  [SIZE=10pt]π [/SIZE]

The outermost lines (the yellow line at the bottom, not the black one) define a square with an area of 4/3 [SIZE=10pt]π, so the circle is 75% of the square[/SIZE]

[SIZE=10pt]The vertical lines just inside the square are at x = +/- 1.  This defines the "miss" areas to the left and right that are eliminated on the "granny shot".  The remainder of the area that lies outside the circle are the potential "miss" areas that remain.[/SIZE]

Given that both x/y have normal distribution around the (0,0) point - there are a lot more shots that will be concentrated around that point, with decreasing density (moving away from (0,0)) so we can't just look at the overall area circle/square to deduce probabilities

 

[heckmanm]

Based on the above, I came up with about 77%, but I could easily be making faulty assumptions.

Like I said...

 

[Ignoratio Elenchi]

Yeah, I don't know if there's an accessible way to solve this one.  Might require double integrals and ####, which would be disappointing as I don't think that's really in the spirit of these types of puzzles.  My initial assumptions about these puzzles have frequently been wrong though, and hopefully that's the case again and there's a fun and not too strenuous way to figure this one out. 

 

[Ignoratio Elenchi]

Before cracking open the old math books I ran some simulations and am getting:

Spoiler

~90% chance to make the granny shot.

 

[heckmanm]

Given that both x/y have normal distribution around the (0,0) point - there are a lot more shots that will be concentrated around that point, with decreasing density (moving away from (0,0)) so we can't just look at the overall area circle/square to deduce probabilities

So is it more of a 3D bell curve like this, with 75% of the shots falling inside the circle within +/- 1 on each axis?

 

[JayMan]

So is it more of a 3D bell curve like this, with 75% of the shots falling inside the circle within +/- 1 on each axis?

This is exactly that:

1. Find the standard deviation of the bivariate normal distribution (both with mean:0 and variance:1) that has 75% of the area inside it... since both are independant (correlation null) - it'll be the same standard deviation for x and y;

2. Given that with granny shot there is no variance in x... with the standard deviation found in 1. - find the probability of being within that range with a normal distribution of mean:0 and variance:1... that is your answer.

 

[heckmanm]

One more question.  

Does " with granny shot there is no variance in x " mean that x is always zero (i.e., perfectly dead-center)? Or just that it never falls outside +/- 1?

The former just makes it a 2D bell curve; the latter leaves it 3D but slices off the sides

 

[JayMan]

One more question.  

Does " with granny shot there is no variance in x " mean that x is always zero (i.e., perfectly dead-center)? Or just that it never falls outside +/- 1?

The former just makes it a 2D bell curve; the latter leaves it 3D but slices off the sides

From the problem: "But suppose you switch it up, and go granny-style, which in this universe eliminates any possible left-right error in your free throws."... perfectly dead-center... 2D bell curve with the probability of making your shots falling within the range of the same standard deviation (as the 3D belle curve with 75% in)

 

[heckmanm]

From the problem: "But suppose you switch it up, and go granny-style, which in this universe eliminates any possible left-right error in your free throws."... perfectly dead-center... 2D bell curve with the probability of making your shots falling within the range of the same standard deviation (as the 3D belle curve with 75% in)

Intuitively, it would seem that the "dome" of the 3D curve would be flatter than the 2d (thinking of it as the volume under the 3D curve vs the area under the 2D curve) in order to carve out 75% within +/- 1 of 0,0. Meaning a higher standard deviation 3D, which when applied to the 2D curve, gives you MORE than 75%..

I was unable to find a z-score for a bivariate normal distribution before getting distracted by other shiny internet objects, but it's 0.674 for 2D 75th percentile

 

[James Bond]

New puzzle and solution to last week's.

If your name is Shaq, the probability is clearly zero.

 

[JayMan]

Intuitively, it would seem that the "dome" of the 3D curve would be flatter than the 2d (thinking of it as the volume under the 3D curve vs the area under the 2D curve) in order to carve out 75% within +/- 1 of 0,0. Meaning a higher standard deviation 3D, which when applied to the 2D curve, gives you MORE than 75%..

I was unable to find a z-score for a bivariate normal distribution before getting distracted by other shiny internet objects, but it's 0.674 for 2D 75th percentile

Bolded part is correct - and intuitively also... since the granny style gives you perfect left-right alignment on every shot (and that you front-back distribution is normal with same mean and variance as it was with the regular shooting style)... you'll hit more than 75%

This is exactly what we need to find, the stdev for the bivariate (3D) that fits 75% and apply it to the normal (2D) to deduce the %.

 

[heckmanm]

Bolded part is correct - and intuitively also... since the granny style gives you perfect left-right alignment on every shot (and that you front-back distribution is normal with same mean and variance as it was with the regular shooting style)... you'll hit more than 75%

This is exactly what we need to find, the stdev for the bivariate (3D) that fits 75% and apply it to the normal (2D) to deduce the %.

As a rough approximation, I compared the volume of a cone vs the area of a triangle of the same dimensions.  For a cone, the 75th percentile of volume occurs at roughly the 91st percentile of height. (i.e., the top 91% of the cone contains 75% of the volume).  This appears to be constant regardless of the height/base ratio.

Since the area of a triangle is linear with its height, the cross-section at that point contains ~91% of the area of the full triangle.

I realize a normal distribution is not actually triangular, but I expect the actual answer to be somewhere around this figure.

 

[Ignoratio Elenchi]

New puzzle and solution to last week's.

A man in a trench coat approaches you and pulls an envelope from his pocket. He tells you that it contains a sum of money in bills, anywhere from $1 up to $1,000. He says that if you can guess the exact amount, you can keep the money. After each of your guesses he will tell you if your guess is too high, or too low. But! You only get nine tries. What should your first guess be to maximize your expected winnings?

 

[Ignoratio Elenchi]

My first thought, not sure if it's this simple:

Spoiler

With 9 guesses you can only guarantee to find the number in a range of 511 numbers.  We want to maximize the expected value so we'll look for the highest 511 numbers, that is anything from $490 to $1000.  Thus if the amount in the envelope is >= $490, we'll find it; if it's less than that, we won't.  This gives an expected value of 0.511 * 745 ~= 380.7.

 

[matttyl]

My first thought, not sure if it's this simple:

Hidden Content

How does one hide content?

 

[Ignoratio Elenchi]

Put it in spoiler tags:

[spoiler]
Stuff you want to hide
[/spoiler]

 

[Thorpe]

I got the same number, IE.  You are incorrect about your expected value, because you can still guess the correct number a percentage of time for the lower amounts.  You just aren't guaranteed to guess it.

 

[heckmanm]

The question is "what should your first guess be?" - so is it

Spoiler

Guess 490, because if it's higher than that, we know we'll find it. And if it's lower, we just d!ckpunch the guy and walk away because we're FBGs and we don't waste our time on small change.

 

[matttyl]

The question is "what should your first guess be?" - so is it

Hidden Content

Incorrect.  Cause then we'd only have 8 guesses left.

 

[heckmanm]

Incorrect.  Cause then we'd only have 8 guesses left.

Good point.  Should be halfway between that number and 1000.

 

[matttyl]

Spoiler

Spoiler
Since we know with 9 guesses that we can guess any whole number in a range of 511, and we want to maximize our winnings - shouldn't the first guess be at the midway point of the highest 511 possible outcomes (halfway from 490 to 1000, or 745)?

 

[Ignoratio Elenchi]

I got the same number, IE.  You are incorrect about your expected value, because you can still guess the correct number a percentage of time for the lower amounts.  You just aren't guaranteed to guess it.

No, you'll never guess a number less than 490 with this strategy.  You're basically assuming that the value is between 490 and 1000 and guessing accordingly.  If it's less than that you'll never hit it.  If the strategy allowed you to guess numbers less than 490, then there would necessarily be numbers greater than 490 that you'd miss, which is obviously a suboptimal tradeoff. 

 

[Thorpe]

No, you'll never guess a number less than 490 with this strategy.  You're basically assuming that the value is between 490 and 1000 and guessing accordingly.  If it's less than that you'll never hit it.  If the strategy allowed you to guess numbers less than 490, then there would necessarily be numbers greater than 490 that you'd miss, which is obviously a suboptimal tradeoff. 

I stand corrected.  

 

[Ignoratio Elenchi]

I'm assuming we don't need the spoiler tags at this point... 

Yes, matttyl, that's correct.  You'd start at 745 and keep guessing halfway between the remaining range, which will guarantee you'll land on the right number within 9 guesses if it's greater than or equal to 490.  

 

[JayMan]

No, you'll never guess a number less than 490 with this strategy.  You're basically assuming that the value is between 490 and 1000 and guessing accordingly.  If it's less than that you'll never hit it.  If the strategy allowed you to guess numbers less than 490, then there would necessarily be numbers greater than 490 that you'd miss, which is obviously a suboptimal tradeoff. 

You are both correct with the 511 range (and the initial guess at 745 as others have pointed out also) to maximize winning... but IE, I think Thorpe's point on expected value (which isn't a part of the 538 question) is that you could still get it right if the number isn't in the 490-1000 range with an initial guess of 745 - but aren't guaranteed of getting it... thus an expected value above the one stated.

 

[Ignoratio Elenchi]

You are both correct with the 511 range (and the initial guess at 745 as others have pointed out also) to maximize winning... but IE, I think Thorpe's point on expected value (which isn't a part of the 538 question) is that you could still get it right if the number isn't in the 490-1000 range with an initial guess of 745 - but aren't guaranteed of getting it... thus an expected value above the one stated.

No you couldn't because the strategy dictates that you'll never guess a number below 490.  For example, if your first guess is 745 and the man says "Lower", your next guess isn't the midpoint between 1 and 745, it's the midpoint between 490 and 745.  You want to make sure that if it's greater than or equal to 490, you'll definitely land on it, at the expense of guaranteeing that you'll never get it right if it's less than 490.  

(Granted, I'm not 100% sure this is the optimal strategy, there might be some better approach, but this is what I came up with at first and it seems like it would be best.)

 

[JayMan]

You want to make sure that if it's greater than or equal to 490, you'll definitely land on it, at the expense of guaranteeing that you'll never get it right if it's less than 490.  

Good point - I didn't think of the strategy as locking-in the 490-1000 range, at the expense of all other lower numbers.

(fun twist to the problem would be that the man with the coat knows this strategy, and that the guesser knows the man with the coat knows it also... the usual back and forth we saw with a few problems lately)