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Welcome to The Riddler. Each week I propose problems related to the things that are dear to us here: math, logic, and probability. Two puzzles are featured each week: the Riddler Express for those of you who want something small and the Riddler Classic for those of you in the slow motion of the puzzle. Submit a correct answer for either, and you might get a shout in the next column. Please wait until Monday to share your responses publicly! If you need a clue or have a favorite puzzle to collect dust in your attic, find me on twitter.

## Riddler Express

From Michael Fienberg comes a tower of not so much terror:

My condo complex has a single elevator that serves four floors: the garage (G), the first floor (1), the second floor (2) and the third floor (3). Unfortunately, the elevator is malfunctioning and stops on every floor no matter what. The elevator always goes G, 1, 2, 3, 2, 1, G, 1, 2, etc.

I want to get on the elevator to a random floor (all four floors are also likely). As I turn the corner to approach the elevator, I hear that its doors have closed, but I have no further information on which floor it is on or whether the elevator is going up or down. The doors may have just closed on my floor, as far as I know.

On average, how many stops will the elevator make until it opens to my floor (including your floor stop)? For example, if I wait on the second floor and hear the doors close at the garage level, then the elevator will open to my floor in two stops.

*Additional credit: *Instead of four floors, suppose my condo has *NOT* floors. On average, how many stops will the elevator make until it opens to my floor?

## Classic Riddler

You are the coach of Riddler Fencing Academy, where your three students face a neighboring team. Each of your students has a different probability of winning a given point in a match. The strongest shooter has a 75% chance of winning each point. The weakest has only a 25 percent chance of winning each point. The remaining shooter has a 50 percent chance of winning each point.

The match will be a relay. First, one of your students will face an opponent. As soon as one of them reaches a score of 15, they are both traded. Then another of your students takes on another opponent, continuing where the score left off. When a team reaches 30 (not necessarily from the same team that reached 15 first), the two shooters are switched. The remaining two shooters continue the relay until a team reaches 45 points.

As a coach, you can choose the order in which your three students occupy the three relay positions: first, second or third. How are you going to order them? So what will your team’s chances be of winning the relay?

## Solution to last week’s Riddler Express

Congratulations to ðŸ‘ Steve Schaefer ðŸ‘ from Carlsbad, Calif., Winner of last week’s Riddler Express.

Last week was Hanukkah, which meant it was time for some Menorah math!

I had a very special menorah. Like most menorahs, there were nine candles in total – one central candle, called the shamash, four to the left of the shamash, and four to the right. But unlike most menorahs, the eight candles on either side of the shamash were numbered. The two adjacent shamash candles were both “1”, the next two shamash candles were “2”, the next pair was “3” and the outermost pair was “4”.

The shamash was still on. How many ways were there to light the remaining eight candles so that the sums on either side of the menorah were “balanced”? For example, one of these methods was to light candles 1 and 4 on one side and candles 2 and 3 on the other. In this case, the sums on both sides were 5, so the menorah was balanced.

Solver Madeline Argent approached this puzzle by first noting that the sum of each side could be any integer from 0 (when no candles were lit) to 10 (when all four candles were lit) . However, many sums were achievable in several ways, such as the aforementioned 5, which was both 1 + 4 and 2 + 3.

Here are all the possible sums, as well as the number of ways they could be reached:

- There was a way to have a sum of 0 (no candles lit).
- There was a way to get a sum of 1 (just candle 1 was lit).
- There was a way to have a sum of 2 (just the 2 candle).
- There were two ways to get a sum of 3 (just candle 3, or candles 1 and 2).
- There were two ways to get a sum of 4 (just candle 4, or candles 1 and 3).
- There were two ways to get a sum of 5 (candles 1 and 4, or candles 2 and 3).
- There were two ways to get a sum of 6 (candles 2 and 4, or candles 1, 2 and 3).
- There were two ways to get a sum of 7 (candles 3 and 4, or candles 1, 2 and 4).
- There was a way to have a sum of 8 (candles 1, 3 and 4).
- There was a way to have a sum of 9 (candles 2, 3 and 4).
- There was a way to have a sum of 10 (candles 1, 2, 3 and 4).

When the sums on each side were 0, 1, 2, 8, 9 or 10, there was only one way to balance the menorah, since each of these sums was only possible in one direction. But when the sums on each side were 3, 4, 5, 6 or 7, there was *four* ways to balance the menorah. For example, if the sums were 5, then both sides of the menorah could have been 1 + 4, both sides could have been 2 + 3, the left side could have been 1 + 4 and the right side could have been 2 + 3, or the left side could have been 2 + 3 and the right side 1 + 4.

In all, there were six sums with balanced lighting, and another five sums with four balanced lights. It meant that there was **26** balanced ways of lighting the menorah.

Maybe Chanukah should last 26 nights, instead of eight. Personally, I wouldn’t complain.

By the way, if you didn’t count the case where both sums were 0, which meant your answer was 25, I still considered that correct. Who wants an empty menorah? (You look, Hewitt school students!)

## Solution to last week’s Riddler Classic

Congratulations to ðŸ‘ Jake Gacuan ðŸ‘ from Manila, Philippines, winner of the Riddler Classic last week.

Last week, Ice Master and Official Delivery Guy Kristoff and his loyal pal Sven were making ice cubes that would be sent to the hot springs, where Kristoff’s extended family lived.

They were studying a cube whose sides were 1 meter, which meant that its area to volume ratio was 6. (Kristoff wasn’t particularly concerned with the units here, but knew they were reverse meters.) Sven was. so concerned that the ice would melt before it reached the hot springs, so he suggested that Kristoff cut the ice along a single plane to minimize the area to volume ratio of the resulting larger chunk.

What plan should Kristoff have cut the ice on? And what was the resulting surface / volume ratio?

First, several solvers noted that any cut, regardless of the resulting surface / volume ratio, would have melted the ice faster. To this, I say that Sven is just a reindeer and that his mathematical prowess exceeds his knowledge of thermodynamics.

Back to ice – for any given area (or volume), the 3D solid with the minimum area to volume ratio is the sphere. And so a reasonable approach was to make the ice cube a bit more “spherical” (sort of).

To convince yourself that it was even possible, you could have cut a small corner of the cube along a plane cutting three edges at a distance. *X* from a vertex, as shown below.

In this case, you were cutting a tetrahedron with a volume *X*^{3}/ 6. You also deleted an area consisting of three right triangles with a combined area of â€‹â€‹3*X*^{2}/ 2. At the same time, you were creating a *New* equilateral surface with side length *X*âˆš2, and therefore the area (*X*^{2}3) / 2. Putting it all together, the area / volume ratio was (6âˆ’ (3 âˆ’ âˆš3)*X*^{2}/ 2) / (1âˆ’*X*^{3}/ 6). For small values â€‹â€‹of *X*, the denominator was very close to 1, while the numerator was slightly less than 6. In other words, it was quite possible that the ratio was less than 6!

Minimize the above rational function gave you an optimal ratio of around 5.962 when x was around 0.4254. And this is the response that several readers have submitted. But was it possible for Kristoff to do even better?

Another way to make the cube slightly more “spherical” was to cut along a plane parallel to an edge:

If the distance between the cut corners and the intersection of the plane with the edges was again *X*, the volume deleted this time was *X*^{2}/ 2, while the removed surface was *X*^{2}+2*X*. During this time, the newly created surface was *X*2. This time, the resulting ratio was (6âˆ’ (2 âˆ’ âˆš2)*X*–*X*^{2}) / (1âˆ’*X*^{2}/ 2). Minimize *this* rational function gave you a ratio of about **5.957** when *X* was about 0.1481. It turned out that cutting parallel to an edge was slightly better than cutting a corner.

This week’s winner Jake extended the puzzle by looking at what would happen if Kristoff did several cuts, cutting several corners or edges. According to Jake, with eight cuts, Kristoff could achieve an area-to-volume ratio as low as 5.66. And if you want to see how solver Tyler Barron got this week’s answer in real time, check out his work on Tic!

## Want more puzzles?

Well, are you unlucky? There is a whole book full of the best puzzles from this column and puzzles never seen before. It’s called “The Riddler”, and it’s in store now!

## Would you like to submit a puzzle?

Send an email to Zach Wissner-Gross at [email protected]

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