FiveThirtyEight Riddler

[britrock88]

Originally Posted by albionmoonlight

New Riddle

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]

Originally Posted by britrock88

(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: < example >

[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]

Originally Posted by 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.

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]

Originally Posted by 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?

And here's the answer they posted:

Spoiler

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

[britrock88]

Originally Posted by 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

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...riddler-nation

[Vince, Pt. II]

Excellent.

[digamma]

Originally Posted by QuikSand

This week's renewal is chock full of goodness:

https://fivethirtyeight.com/features...riddler-nation

They had you at 2/3.

[albionmoonlight]

Originally Posted by digamma

They had you at 2/3.

[Umbrella]

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

[QuikSand]

Originally Posted by digamma

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]

Originally Posted by 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"?

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