FiveThirtyEight Riddler

[albionmoonlight]

This board used to have a thing for puzzles. 538.com has a weekly puzzle column that seems to capture the style we like.

Here's this weeks version: How Many Cars Will Get Stuck In Traffic? | FiveThirtyEight (http://fivethirtyeight.com/features/how-many-cars-will-get-stuck-in-traffic/)

The answer might be on google. I am positing it here to see if we can figure it out. Right now, I am having trouble even figuring out a starting place. My gut says that the answer should be something like N/2. But I have no idea how to get there (or if my gut is anything close to right).

[cartman]

2/3rds

[QuikSand]

Cool, first I've heard of the feature there.



I know phrases like this can become unravelers:

Each preferred speed is chosen at random.

I think we spent some time here back in the day with a classic "two envelopes" puzzle that basically becomes an internal paradox due to something similar.

[QuikSand]

This board used to have a thing for puzzles.

Accurate.

[cartman]

I think it is like if you take a deck of cards, and use the values to equate to speed. So it will be N=number of cards and S=Speed. So the equation would be something like ((N-1)/S, (N-2)/S...)/N

But I failed Calculus three times in college, so take that for what it is worth.

[albionmoonlight]

One approach would be to start with small N and see what pattern develops. For n=1, you have 1 group. For n = 2, you average 1.5 groups. For n=3, you average 11/6 groups (~1.8).

So this might be a calculus problem with some answer as N approaches infinity.

[QuikSand]

that sounds promising

[QuikSand]

Just spitballing...

the slowest car creates a group, period (even with zero behind it)

the next slowest car creates a separate group unless it's immediately behind the slowest one (p=some weird but solvable function of n)

the third slowest car... etc etc etc


Probably amounts to the same limit sequence as you're setting up above, but just via intuition I think I'm getting to a similar place.

[albionmoonlight]

More spitballing:

the slowest car will be first 1/n times (one group)
it will be in second 1/n times (two groups)
the problem gets more complicated once the slowest car is in third or later.

But there does seem to be weight keeping the number lower than one might expect. 2/n times, you end up with one or two groups

[britrock88]

We have N cars with speeds assigned by a non-duplicative RNG.

You have a 1/N chance of 1 group if car 1 is the slowest and a 1/(N-1) chance of 2 groups is car 2 is the slowest. But starting with car 3, you have to reiterate the probability calculation for the multiple cars ahead of car n.

[Solecismic]

This is what I submitted:

1 + 1/2 + 1/3 + ... + 1/N

We have to assume a road of infinite length, and we have to assume cars may travel at an infinite number of different speeds to preserve independence. Otherwise, a collision would make it impossible to determine the true number of packs due to the interference of emergency vehicles and rubberneckers. The Twitter ramifications alone are enormous.

The answer is a series beginning with one pack. There's always at least the one pack. How do we get a second pack. That's if the first car isn't the slowest. Or 1 - 1/2. And a third pack? That's if the first car found after the first slower car isn't the slowest. Or 1 - 1/3 - 1/3. We keep adding the chances of finding a new pack. Keep in mind that the chance of finding N packs isn't 1/N in itself - it's 1/N!. But there's a (N-1)! "weight" attached to that event when calculating the average of our infinite number of iterations.

[EagleFan]

I say there is not enough information given to formulate an answer.

10 mile road... 20 cars... speed range 60 mph - 70 mph

gives you a different result than

5 mile road... 50 cars... speed range 25 mph - 60 mph

[rowech]

Fairly certain the answer is 2(1-.5^N). Ultimately, I think it's just a geometric series with r = .5 and I just used the summation formula for such a series. I fear there has to be more than this because that summation goes to 2 with an infinite number of cars and I think there has to be more than two groups that form. It just doesn't seem right.

[Solecismic]

Turns out the answer is 1 + 1/2 + 1/3 + ... + 1/N. But the explanation is a lot more complex than what I submitted. This week, it's ducks on a pond.

[QuikSand]

With the duck/dog, I think the logic must be obvious, but the calculation is trickier, to me.

Dog starts at 12 (on the clock) so the duck heads toward 6 (at least that seems obvious to me). Dog flips a coin and goes toward the 1. But as the dog's location changes, I presume the duck no longer chooses a direct route toward the 6 as his optimal route, he likely curves away from the dog's side...toward the 7.

I'm assuming there's some way to calculate the path/vector of the duck as a function of the location of the dog, and that it creates visually some sort of seashell like spiral. But the math in calculating all this is either non-obvious, or is simply beyond my knowledge/recollection.

Hmmm...i guess if the duck is perpetually just aiming for the spot diametrically opposed to the dog's location, that might not be impossible to sort out.

[Solecismic]

Yes, that's where I'd head with this question. The dog always aims for the spot on the shore closest to the duck and duck always aims for the spot furthest from the dog. That would create some sort of spiraling curve.

The initial relationship is 2π. So, what you have to prove, if the problem is relatively simple, is that by following this strategy, the dog is never in a worse situation than when the duck is in the middle if it can travel that fast.

[QuikSand]

So, I'm thinking about right triangles here... right, this ends up being a trig problem?

[ColtCrazy]

Sounds like there should be a math solution. The dog must travel half the circumference at least as fast as the duck can travel the radius. Wouldn't that mean the dog would have to travel pi times faster?

But that doesn't take in a sudden turn by the duck, which means the dog would have to travel even faster to make up for the turn.

[ColtCrazy]

So, I'm thinking about right triangles here... right, this ends up being a trig problem?

I think you are right. Assuming the duck turns at the last moment and moves diagonally, the duck should travel the radius and the hypotenuse of a right triangle to get to another quarter way around the lake. This means the dog would have to go 3/4 around the lake in the same time to catch the duck. I got that far.
I tried comparing those and got that the dog needs to be about 4 times faster (I got 3.9 and some change).

[QuikSand]

So, I'm thinking this is pretty tricky. If again, we assume that the duck continues to pursue the direction across from the dog at all times, then his amount of progress toward the perimeter will actually decrease over time, as the amount of lateral motion he needs increases linearly as he moves away from the center.

...and as I type this, I think I'm reaching a reductio ad absurdum with my premise there. Clearly the duck is gong to engage in some sideways evasion, but before too long he can't simply go in the "opposite" direction of the dog, who will be in the same side as him. So, his best path is some function, that I strongly suspects is of his distance from the center, but I don't have any way to express it.

[Solecismic]

Sounds like there should be a math solution. The dog must travel half the circumference at least as fast as the duck can travel the radius. Wouldn't that mean the dog would have to travel pi times faster?

But that doesn't take in a sudden turn by the duck, which means the dog would have to travel even faster to make up for the turn.

Yes, I counted r twice when I shouldn't have. The initial relationship is π. So, what is the relationship if the duck changes direction at some instant a after the simulation begins? Well, the dog is still πr away, but the duck is probably closer. What's the limit, though? Set a at 1/2r. The duck is now sqrt(5/4r) away from the shore along his new path. There's no way that's a good choice for the duck. So we know that the duck is best off if it either stays on its initial course or makes a change in the first half of its journey.

What if a is 1/9r? Where's the dog? He's at an angle of about 20 degrees from his starting point and he still has to travel πr. The duck is at .9r/.9397 or .9577r from his new destination. That's progress. So the simple case is ruled out. In fact, as long as the duck makes his change before the dog has traveled one sixth of the way around the half-circle, the duck has made progress and will reach shore safely. This is a complex problem as the dog will travel in a circle while the duck will travel in some form of what I guess is an Archimedian spiral.

The curvature of the spiral should be the maximum at which the dog never changes direction.

However, I've long since lost the geometry and calculus skills required to solve this problem.

[cuervo72]

However, I've long since lost the geometry and calculus skills required to solve this problem.

Yeah. These aren't just logic puzzles, they are math puzzles. I've looked at the solutions for this, and while I can grasp them there's no way I'd easily be able to calculate them on my own, not 20 years out from studying any of this.

[Toddzilla]

So the duck would necessarily have to always be traveling to the exact opposite part of the circle that the dog was positioned - since neither the dog nor the duck lose speed when changing directions, there is no advantage to turning suddenly.

Let's take this one "minute" at a time
So if the Dog is at 12 o'clock, the duck swims toward 6.
The dog picks a direction - say clockwise.
When the dog gets to 12:01, the duck will be heading towards 6:01.
When the dog gets to 12:02, the duck will be heading towards 6:02.
...
except when the dog get's off of 12:01, the duck has moved a distance of x from the center of the circle - in the direction of 6 - and now travels directly opposite of the dog which is no longer headed towards 6:02. This would be almost towards 6:02, but shorter (I'll show my work later)

so as each "minute" passes, the dog proceeds around the clock clockwise and for each minute the dog moves, the duck moves in a curve each time moving less than a minute and getting shorter each time.

i think.

[Toddzilla]

https://pbs.twimg.com/tweet_video/CbMQQCOUAAA1OPS.mp4

sweet

[QuikSand]

So the duck would necessarily have to always be traveling to the exact opposite part of the circle that the dog was positioned - since neither the dog nor the duck lose speed when changing directions, there is no advantage to turning suddenly.

Let's take this one "minute" at a time
So if the Dog is at 12 o'clock, the duck swims toward 6.
The dog picks a direction - say clockwise.
When the dog gets to 12:01, the duck will be heading towards 6:01.
When the dog gets to 12:02, the duck will be heading towards 6:02.
...
except when the dog get's off of 12:01, the duck has moved a distance of x from the center of the circle - in the direction of 6 - and now travels directly opposite of the dog which is no longer headed towards 6:02. This would be almost towards 6:02, but shorter (I'll show my work later)

so as each "minute" passes, the dog proceeds around the clock clockwise and for each minute the dog moves, the duck moves in a curve each time moving less than a minute and getting shorter each time.

i think.

Problem is, the math here no longer works once the duck's distance from the center is of a greater ratio to the dog's distance than the ratio of their speeds. If duck is speed x and dog is speed 5x, say, then at some point the duck gets more than 1/5 of the way out from the circle, and he can no longer keep up with the concept of evading the dog and continuing to move outward. From that point on the dog is catching up.

I'm with the group above - I will read the eventual solution with interest, and I find the scripting and graphical representations amusing... but the math is beyond me, it seems.

[Toddzilla]

check the graphic, i think that explains my premise well

[albionmoonlight]

New puzzler is up. I've seen this one before, so I knew the answer. It did stump me when I originally saw it, though. This one you can get with logic. No need to resort to calculus.

[cuervo72]

My inclination is to say 50/50. There is a 1/100 chance that WPA sits in your seat. There is a 1/100 chance that WPA sits in WPA's seat, breaking the chain. The former means you don't get your seat, the latter means you do. If neither happens, we repeat the process. Each iteration equally adds to the chance of filling your seat or filling the WPA's seat -- down to two seats, where the second-to-last boarder has a 50/50 shot of grabbing either WPA's or yours if one hadn't been already.

I don't know, maybe there is some compounding of probability that I'm missing, but that would be my basic logic.

[QuikSand]

I think it's close to 50/50, but slightly less.

Two cases with the initial passenger -- he either sits in his seat (1/100) or not (99/100).

Assuming not, then there is always one person out of place on the plane... meaning than when passenger 99 sits, he's 50/50 to find his seat occupied. So, 99% of the time, it's a 50% chance your seat is taken by the second to last passenger.

Assuming the first passenger happens to land in his proper seat, then everyone gets his own seat, and your chances of being in your seat are 100%.

So... I make it at (0.5 * 0.99)+(0 * 0.01) = 0.495

[QuikSand]

hmmm... maybe have failed to account for the WPA sitting in your seat as a separate case...

[QuikSand]

So... I make it at:

WPA in his seat = 1% chance * if so then 0% chance yours is taken
WPA in your seat = 1% chance * if so then 100% chance yours is taken
WPA in other seat = 98% chance * if so then 50% chance yours is taken

And summing those gets right back to 50%. Yup.

[albionmoonlight]

50% is the answer I have seen before.

[Solecismic]

It has to be 1/2, no matter how many seats there are on the plane.

[QuikSand]

Looks like we missed a week, but here's a new one...

Can You Win This Hot New Game Show? | FiveThirtyEight (http://fivethirtyeight.com/features/can-you-win-this-hot-new-game-show/)

Two players go on a hot new game show called “Higher Number Wins.” The two go into separate booths, and each presses a button, and a random number between zero and one appears on a screen. (At this point, neither knows the other’s number, but they do know the numbers are chosen from a standard uniform distribution.) They can choose to keep that first number, or to press the button again to discard the first number and get a second random number, which they must keep. Then, they come out of their booths and see the final number for each player on the wall. The lavish grand prize — a case full of gold bullion — is awarded to the player who kept the higher number. Which number is the optimal cutoff for players to discard their first number and choose another? Put another way, within which range should they choose to keep the first number, and within which range should they reject it and try their luck with a second number?

[QuikSand]

So...instinct surely starts you with a number of 0.5, right? Higher than that and the EV of your second value is lower than what you have in hand.

Making it a two-person game, I presume, has some recursive issue, then. Because the EV of another person's final value of 1-2 numbers isn't normally distributed, then you no longer simply shoot for the highest possible value.

I don't yet have a clear idea how to calculate this... but I assume that to maximize your chances of winning (rather than just maximizing the value of your number) then you have to turn the range into a binary setup = where you're looking at P (your score > EV of other player's score).

And somehow, since both players have the same information and motivation, I assume this has to converge for both players.












Aww, hell, how about 2/3.

[albionmoonlight]

So if it were just me trying to get the largest number, I would keep anything over .5 and redo anything under .5 (no idea what I would do with .5 on the nose, but let's leave that out of it).

That means, I think, that I would keep half of my rolls. Because they are all above .5 distributed randomly, they would average .75.

I would redo half my rolls, which would average .5.

So my final average would be (.5 * .75) + (.5 * .5). Which would equal .625. Not quite 2/3 :(

Now, my opponent could also get to .625 using the same simple strategy.

So what's the next step? Or is .625 the best either of us can do?

[QuikSand]

So, let's say I assume you're going with the .625 strategy.

Sketching through whether I do better by tinkering based on that. What if I set my bar a shade higher, say at 0.6 (I keep everything below that and reroll above it)?

My EV is easy:
low rolls .6 x .5
high rolls .4 x .8
=total EV of .620

But mapping "who wins" is different, right?

Assuming there's some calculus that can solve the likelihood of winning each of those four quadrants (set this up like a number theory matrix)...then it seems it must be possible to pick a strategy that "beats" your .625 strategy. Right? Otherwise why publish the puzzle at all?

I'm still stuck that this is some sort of recursive thing...you move to .625, I beat that my adjusting a little, then you adjust downward, I adjust upward, and we both meet at some equilibrium in the middle. I'm lacking the math, but comfortably accepting guidance from the human context here, and assuming there's some perfect balance that represents optimal strategy by the player who knows his opponent will also employ optimal strategy.

[QuikSand]

So, I guess the "who wins" between two random number between 0 and p1 and between 0 and p2 looks like this (where p2>p1, and both are expressed as decimals between 0 and 1):

((p2-p1)/p2) = situations where the p2 always wins
(p1/p2) = situations where the wins are evenly split

(and there's an inverse for the values between a certain value and 1)

Not as bad as I had thought initially. So, in my example above... I think the math looks like this:

p1 = 0.5 (albion chooses to re-roll everything below 0.5)
p2 = 0.6 (Quik choses to re-roll everything below 0.6)

Grid out the four outcomes, using the above:
P1 keeps, P2 keeps
P1 keeps, P2 rerolls
P1 rerolls, P2 keeps
P1 rerolls, p2 rerolls

...and I'm guessing that despite a lower overall EV, that the 0.6 strategy wins more than it loses. (I'll fill in the math above with brute force if I can get my kids in line sometime today)

Then you figure out what function guides this. Presumably it could be done fairly well with brute force and Excel, but it also likely sorts out to one big formula, I suppose.

[Radii]

Otherwise why publish the puzzle at all?


This is where I ended up, after not really being able to figure out why the answer isn't .5 as your cutoff point to re-rolling. If the answer was to keep anything above 0.5 and roll anything below, surely they wouldn't have bothered to pose the question. I look forward to the explanation.

[QuikSand]

Here's what I come up with:

Grid out the four outcomes, using the above:
P1 keeps, P2 keeps: likelihood = 0.2, p(P2 wins) = 0.6, aggregate p(P2 wins) = 0.120
P1 keeps, P2 rerolls: likelihood = 0.3, p(P2 wins) = 0.25, aggregate p(P2 wins) = 0.075
P1 rerolls, P2 keeps: likelihood = 0.2, p(P2 wins) = 0.8, aggregate p(P2 wins) = 0.160
P1 rerolls, p2 rerolls: likelihood = 0.3, p(P2 wins) = 0.5, aggrebate p(P2 wins) = 0.150
TOTAL p(P2 wins) = 0.505

So...in the event I have the math right above, choosing 0.6 as your breakeven point is a FAVORITE to beat a player who picks 0.5 as his, even though it doesn't maximize the expected value of the outcome.

[QuikSand]

So, break all this down into one formula, it has a maximum point (derivative zero?) at some value. .6 is better than .5, I'm guessing.

[Solecismic]

You're not trying to maximize your expected value; you're trying to maximize your chances of beating your opponent's expected value.

[Solecismic]

Against the .5 opponent, 7/12 is your best play, based on taking that derivative. This gets into calculus I've long since forgotten if that hypothesis doesn't hold true against any opponent. My guess is that the answer is close to 7/12, though, because these curves are very flat.

[Maple Leafs]

This riddle made me feel dumb. Then I went to the 538 site and read the comments from people who were completely perplexed by the concept of "between 0 and 1" and I felt smart again.

[albionmoonlight]

But if .6 beats .5, then don't both players then use .6? Which then means that .6 may not be the best strategy.

Which, of course, is another way of saying Quik's equilibrium point above.

[Maple Leafs]

In searching for the airplane riddle you guys were mentioning, I found this site that tries to solve the game show puzzle with code.

The result is... well, probably not especially satisfying. Spoiler warnings, obviously.

The 538 Riddler: Hot New Game Show – Some Disagree (http://somedisagree.com/2016/03/06/the-538-riddler-hot-new-game-show/)

[Vince, Pt. II]

This week's version is the first one I've gotten intuitively from the get go.

[Vince, Pt. II]

Apparently only the first half though...I find that I am unable to determine whether or not she would know if he says no. Which tells me that perhaps I'm wrong on the first half of things as well?

My logic:
The only information Pete has is that there are two hidden numbers from 1 to 9 (inclusive), and he has their product on a piece of paper. We are working with three different sets here:

1. Legal combinations of numbers.
2. The products of these combinations.
3. The sums of these combinations.

The only possibly way he could know what the two numbers were off the bat is if the number on his paper was a product that appeared uniquely in the set of all products. Since he doesn't know this, we can intuitively eliminate all number combinations that produce a unique product.

Susan is in the same scenario, only she now knows that any combination that produced a unique product is eliminated. Since she also answers 'no,' we can eliminate all unique sums from this new set of combinations, further reducing the set of possible combinations.

Repeating this recursively, when you get to the first time either of them says 'yes,' the answer can be any combination of numbers left in our set of legal combinations which produces a unique sum/product, depending upon who is doing the answering. If there is only one unique sum/product, you can determine exactly which combination of numbers were chosen.

It seems as if determining whether someone will say 'yes' or 'no' to a following question is only definitive if there are ZERO or ALL unique answers left; if there are some unique answers and some duplicate answers, the person may or may not know.

[albionmoonlight]

New Riddle (http://fivethirtyeight.com/features/can-you-solve-this-napoleonic-puzzle/)

Complete this series:

10, 11, 12, 13, 14, 15, 16, 17, 21, 23, 30, 33, …

[albionmoonlight]

dola:

Five minutes of staring at it has given me nothing

[britrock88]

New Riddle (http://fivethirtyeight.com/features/can-you-solve-this-napoleonic-puzzle/)

Complete this series:

10, 11, 12, 13, 14, 15, 16, 17, 21, 23, 30, 33, …

My first thoughts when looking at series have to do with basic arithmetic functions (*, /, +, -) and primes.

This series catches primes 11, 13, 17, and 23, while missing 19 and 29.

I also noticed a relatively high number of multiples of 3--12, 15, 21, 30, and 33, though 18, 24, and 27 are omitted.

Then my eye caught something interesting. No trailing digit in the series is higher than 7. What if this series were affected by the fact that the numbers are represented in base 8? That would reset this series to represent 8, 9, 10, 11, 12, 13, 14, 15, 17, 19, 24, and 27. Primes and multiples of 3 come up again, but not without omissions.

[QuikSand]

I also thought of base-8, but that stagger from 17-21-23-30-33 is just hard to pin down to anything rational. Most series speed up (or slow down) but to make that big of a jump (from 23 to 30) only to slow down (30 to 33) is tough to figure out.

[britrock88]

What do you think "Can you solve this Napoleonic puzzle?" means?

That sentence yields words of lengths 3, 3, 4, 5, 6, and 10 letters. You can add combinations of those digits to get to every object in the series set <30 (in addition to some other positive integers not in the series).

[cartman]

I was trying to get the Napoleonic thing to fit in, besides their blurb about short and deadly. I tried to match it up to known figures from things like the march on Moscow and back, but couldn't get anything to line up.

[britrock88]

The author retweeted this... https://twitter.com/xaqwg/status/719202348858380289

[britrock88]

If 22 were also included in the series, the series would represent all the digits (10 less than x less than 99) that could be represented by more than one of base 8, base 6, and base 4.

(Anyone know how to use the less than symbol without breaking things?)

EDIT: hang on, that's not true for 16/17. My bad.

[cartman]

(Anyone know how to use the less than symbol without breaking things?)

You have to use an escape notation.

& lt ; for less than, & gt ; for greater than (remove the spaces after the & and the t)

quote this to see: &lt; example &gt;

[cuervo72]

I googled the answer, and the Napoleonic thing doesn't mean anything. We've trained ourselves to look for any hidden clues but that's really not one, at least not one that I understand after looking at the answer.

[QuikSand]

Well, I actually spent a little time noodling that, but didn't land on it. Feel disappointed, it was very "gettable."

[cuervo72]

Yeah, I was disappointed in myself for not trying harder before looking for a spoiler. But with how some of the other ones were, I didn't know if it was actually gettable or not and just wanted an answer before I gave up and forgot about it.

[cartman]

Here's a new one from them:

While traveling in the Kingdom of Arbitraria, you are accused of a heinous crime. Arbitraria decides who’s guilty or innocent not through a court system, but a board game. It’s played on a simple board: a track with sequential spaces numbered from 0 to 1,000. The zero space is marked “start,” and your token is placed on it. You are handed a fair six-sided die and three coins. You are allowed to place the coins on three different (nonzero) spaces. Once placed, the coins may not be moved.

After placing the three coins, you roll the die and move your token forward the appropriate number of spaces. If, after moving the token, it lands on a space with a coin on it, you are freed. If not, you roll again and continue moving forward. If your token passes all three coins without landing on one, you are executed. On which three spaces should you place the coins to maximize your chances of survival?

Extra credit: Suppose there’s an additional rule that you cannot place the coins on adjacent spaces. What is the ideal placement now? What about the worst squares — where should you place your coins if you’re making a play for martyrdom?

[cuervo72]

I would probably go 4-5-6. 1/2 shot on the first roll. If you fail, you should have a 1/2 shot on the second roll. Probably some permutation that squeezes more percentage points out for you though.

Will have to think about the other one. Gut says 4-6-8, but there's more math involved there I think.

[cartman]

I think this is a case for using a Monte Carlo Simulation, but not sure of the approach.

[cartman]

or take the simpler approach of using two rolls, kind of like craps. 6,7,8 are the most common combinations, 16 out of 36 ways (44% chance) to get those totals with two rolls. If you go with the non-consecutive numbers, then 5,7,9 gets you 14 out of 36 ways (39% chance). Anything more than that might be noise and blind luck landing on a number as you approach 1000. But there is probably something I'm overlooking.

edit: there is. You could possibly get a 6 (or 5 for nonconsecutive) on your initial roll.

[CrescentMoonie]

I was thinking 994, 995, 996, or something to that effect. Push the probability to the end and hopefully that would help account for luck.

[Maple Leafs]

I would probably go 4-5-6. 1/2 shot on the first roll. If you fail, you should have a 1/2 shot on the second roll. Probably some permutation that squeezes more percentage points out for you though.

Yeah, this gives you a slightly better than 75% chance (since there's also a small possibility that the first two rolls end up totalling 2 or 3 and giving you another 50-50 roll). But given how these puzzles go, that seems like it's too easy.

[cuervo72]

Yeah, the "too easy" is what I was figuring.

[Maple Leafs]

Saw this one recently, don't think it was a riddler but maybe one of the sites he links to. I'll put it on here only because I tried to figure it out, then decided there must be some trick to it and looked up the answer and now I feel dumb, and hopefully I can make someone else sad too.

Using the standard operators +, -, * and /, and each of the numbers 2, 4, 6 and 8 once each, try to get something that equals 25.

[cuervo72]

Well at least concatenation isn't an option.

[nol]

4 * 6 2/8? Doesn't seem to be one where you can multiply to get something bigger and then divide or subtract to get 25, at least.

[cuervo72]

I'd accept that. Nice.

[Maple Leafs]

Here's a new one from them:

While traveling in the Kingdom of Arbitraria, you are accused of a heinous crime. Arbitraria decides who’s guilty or innocent not through a court system, but a board game. It’s played on a simple board: a track with sequential spaces numbered from 0 to 1,000. The zero space is marked “start,” and your token is placed on it. You are handed a fair six-sided die and three coins. You are allowed to place the coins on three different (nonzero) spaces. Once placed, the coins may not be moved.

After placing the three coins, you roll the die and move your token forward the appropriate number of spaces. If, after moving the token, it lands on a space with a coin on it, you are freed. If not, you roll again and continue moving forward. If your token passes all three coins without landing on one, you are executed. On which three spaces should you place the coins to maximize your chances of survival?

Extra credit: Suppose there’s an additional rule that you cannot place the coins on adjacent spaces. What is the ideal placement now? What about the worst squares — where should you place your coins if you’re making a play for martyrdom?

And here's the answer they posted:


To give yourself the best shot at staying alive, place your coins on spaces 4, 5 and 6. You’ll survive about 79.4 percent of the time.

A little intuition here goes a long way. For starters, you’ll definitely want to place a coin on square 6. It’s the best square because it maximizes the number of rolls that could land on it. You could roll a 6 and hit it in one shot, for example. If you don’t get there in your first roll, you’re guaranteed to have at least one more shot at landing there. Square 5 is another great candidate for the same reason. You could hit it on your first roll, of course, but if you roll something less than 5 you’ll get another chance to save your life on that square.

You can confirm this intuition with math, of course. Once you’ve got your coins on squares 5 and 6, let’s say you’re struggling with whether to place your final coin on square 4 or on square 7. Here’s how you can decide, as calculated by Aaron Cote.

For 4-5-6, here are the probabilities you win and you…

Roll a 4, 5 or 6: 0.5

Roll a 3: 1/6*0.5 = 0.083

Roll a 2: 1/6*(0.5+0.083) = 0.097

Roll a 1: 1/6*(1/2+0.083+0.097) = 0.113

For a total probability of about 0.794.

For 5-6-7, here are the probabilities you win if you…

Roll a 5 or 6: 0.333

Roll a 4: 1/6*1/2 = 0.083

Roll a 3: 1/6*(1/2+0.083) = 0.097

Roll a 2: 1/6*(1/2+1/6*1/2+1/6*0.097) = 0.113

Roll a 1: 1/6*0.794 = .132

For a total probability of about 0.758. So 4-5-6 is better.

[cuervo72]

Yay little intuitions.

[britrock88]

I'm not really satisfied with that explanation. They didn't take these calculations out any farther down the number line? I was doing some brute-force math and there seemed to be a second peak around 10/11.

[Maple Leafs]

No, they went into more detail, including a probability chart. I just didn't post the whole thing. You can find it here:
These Challenges Will Boggle Your Mind | FiveThirtyEight (http://fivethirtyeight.com/features/this-challenge-will-boggle-your-mind/)

[britrock88]

No, they went into more detail, including a probability chart. I just didn't post the whole thing. You can find it here:
These Challenges Will Boggle Your Mind | FiveThirtyEight (http://fivethirtyeight.com/features/this-challenge-will-boggle-your-mind/)

Thanks for linking! Pointing out the dependence of 11 on not-6 is a strong point.

[QuikSand]

This week's renewal is chock full of goodness:

https://fivethirtyeight.com/features/can-you-rule-riddler-nation

[Vince, Pt. II]

Excellent.

[digamma]

This week's renewal is chock full of goodness:

https://fivethirtyeight.com/features/can-you-rule-riddler-nation

They had you at 2/3.

[albionmoonlight]

They had you at 2/3.

:D

[Umbrella]

Both questions are going to be tricky. I am wondering if I will even be in the right ballpark.

[QuikSand]

They had you at 2/3.

It moved

[QuikSand]

I'm a dummy and neglected to enter on time for the latter puzzle. Had two theories, was leaning toward the quirkier of the two. One was just a log-based scale and investing in each castle:

1
2
3
4
6
8
11
15
21
29

..the other was to put in only token soldiers in most places (I opted against zero, and then decided to opt against one, going with two) and then load up on enough to get to 28 points if we won them all.

Something like this:

1
1
7
2
13
17
23
30
3
3

[Umbrella]

I'm 99.9% sure I got the express right this week, although I did have to brush up on how to do derivatives of square roots.

[Maple Leafs]

Thought this one seemed up FOFC's alley...

A giant troll captures 10 dwarves and locks them up in his cave. That night, he tells them that in the morning he will decide their fate according to the following rules:

- The 10 dwarves will be lined up from shortest to tallest so each dwarf can see all the shorter dwarves in front of him, but cannot see the taller dwarves behind him.
- A white or black dot will be randomly put on top of each dwarf’s head so that no dwarf can see his own dot but they can all see the tops of the heads of all the shorter dwarves.
- Starting with the tallest, each dwarf will be asked the color of his dot.
- If the dwarf answers incorrectly, the troll will kill the dwarf.
- If the dwarf answers correctly, he will be magically, instantly transported to his home far away.
- Each dwarf present can hear the previous answers, but cannot hear whether a dwarf is killed or magically freed.

The dwarves have the night to plan how best to answer. What strategy should be used so the fewest dwarves die, and what is the maximum number of dwarves that can be saved with this strategy?

[Radii]

If you can use some sort of cadence here you can save 9. I assume that's not the spirit of the question (or is it?)

The tallest dwarf is screwed. His answer will be the color of the dot of the dwarf in front of him and he's got a 50/50 shot. That dwarf will answer with his own color which he now knows. He will speak softly if that color matches the dwarf in front of him, and he will yell if it does not. So dwarf #2 is saved and has signaled dwarf #3 what his color is based on whether he answered loudly or quietly. Continuing on like that every dwarf will know their own color and can signal the dwarf in front of him.


Is that "cheating"?

[Radii]

If the spirit of the question is that you just know the answer of previous dwarves but can't gain any additional information than that, then the baseline has to be 5 guarenteed saved.

Odd numbered dwarves answer with the color of the one in front of them and have a 50/50 shot themselves. Even numbered answer with the color they were given by the previous dwarf.

So dwarf 1 doesn't know his own, he answers with the color for dwarf 2. Dwarf 2 knows his so he saves himself. Dwarf 3 doesn't know his, so he answers with the color for dwarf 4. Repeat.

That will save all even numbered dwarves.

I immediately feel like there is a better answer and this is just the starting point.

[Maple Leafs]

I like the yell/whisper answer, but I'm guessing they want something else. If I'm a dwarf, I take this to the troll appeals court, though.

With these puzzles, it's usually best to start with a smaller number. I can figure it out up until three, then I get lost.

One dwarf: Agree with Radii, it's just 50/50 for the first guy. No way around that.

Two dwarves: First guy names the second guy's color. (Where "first" refers to the order they speak, i.e. first is the tallest.) That guarantees he'll be right.

Three dwarves: First guy looks at the two colors in front of him. If they match, he says "white". If not, he says "black". That tells the second guy what to say, based on the color he can see. And third guy says white if both answer match, or black if they're different. So we guarantee two of the three.

So, uh, only seven more to go and we're done.

[QuikSand]

If you can use some sort of cadence here you can save 9. I assume that's not the spirit of the question (or is it?)

The tallest dwarf is screwed. His answer will be the color of the dot of the dwarf in front of him and he's got a 50/50 shot. That dwarf will answer with his own color which he now knows. He will speak softly if that color matches the dwarf in front of him, and he will yell if it does not. So dwarf #2 is saved and has signaled dwarf #3 what his color is based on whether he answered loudly or quietly. Continuing on like that every dwarf will know their own color and can signal the dwarf in front of him.


Is that "cheating"?

The perpetual angle-shooter in me LOVES this answer.

[Maple Leafs]

The answer to the troll riddle, which ends up being a lot simpler than we probably thought.

Nine of the 10 dwarves can be saved for sure and, with a little luck, all 10 will escape the troll’s clutches. How? The dwarves agree on the following plan: The first, tallest dwarf will risk life and limb to save the others. Since he has no information to go on to determine his own dot’s color, he can use his guess to inform the others. The dwarves agree that if the number of white dots the tallest dwarf sees is even, he should say “white,” and if it’s odd, he should say “black.”

That first dwarf only has a 50-50 chance of survival, but all of his compatriots will now survive for sure because they know why he said the color he said. Suppose the first dwarf says “white,” meaning he sees an even number of white dots. Then it’s the second dwarf’s turn. If he also sees an even number of white dots, then he knows for sure that his dot is black. If, instead, he sees an odd number of white dots, then he knows for sure that his dot is white. Based on the responses of the first two dwarves, the third can then also determine the “evenness” or “oddness” of the remaining white dots. If what he sees matches that, his must be black, if not, white, and so on.

[bhlloy]

That's brilliant, although I'd probably be the dwarf that fucked it up and got everyone else killed