## Don’t flip out

This week’s Fiddler is a probability question about a coin-flipping game.

Kyle and Julien are playing a game in which they each toss their own fair coins. On each turn of the game, both players flip their own coin once. If, at any point, Kyle’s most recent three flips are Tails, Tails, and Heads (i.e., TTH), then he wins. If, at any point, Julien’s most recent three flips are Tails, Tails, and Tails (i.e, TTT), then he wins.

However, both players can’t win at the same time. If Kyle gets TTH at the same time Julien gets TTT, then no one wins, and they continue flipping. They don’t start over completely or erase their history, mind you—they merely continue flipping, so that one of them could conceivably win in the next flip or two.

What is the probability that Kyle wins this game?

Extra Credit
Kyle and Julien write down all eight possible sequences for three coin flips (HHH, HHT, HTH, THH, HTT, THT, TTH, and TTT) on eight different slips of paper. They place these slips into a hat and shake it.

They will each randomly draw slips of paper out of the hat, at which point they will play the same game as previously described, but looking for the sequence specified on the slip of paper they each selected. Kyle draws first and looks at his slip of paper. After doing some calculations, he says: “Well, at this point, it’s about as fair a match as it could possible be.”

Which slip or slips of paper might Kyle have drawn? And what are his chances of winning at this point (i.e., before Julien selects his own slip of paper)?

My solution:
[Show Solution]

## Pinball probability

This week’s Fiddler is a challenging probability question.

You’re playing a game of pinball that includes four lanes, each of which is initially unlit. Every time you flip the pinball, it passes through exactly one of the four lanes (chosen at random) and toggles that lane’s state. So if that lane is unlit, it becomes lit after the ball passes through. But if the lane is lit, it becomes unlit after the ball passes through. On average, how many times will you have to flip the pinball until all four lanes are lit?

Extra credit: Instead of four lanes, now suppose your pinball game has $n$ lanes. And let’s say that $E(n)$ represents the average number of pinball flips it takes until all $n$ lanes are lit up. Now, each time you increase the number of lanes by one, you find that it takes you approximately twice as long to light up all the lanes. In other words, $E(n+1)$ seems to be about double $E(n)$. But upon closer examination, you find that it’s not quite double. Moreover, there’s a particular value of $n$ where the ratio $E(n+1)/E(n)$ is at a minimum. What is this value of $n$?

My solution:
[Show Solution]

## Catch the grasshopper

This week’s Riddler classic is a probability problem about a grasshopper!

You are trying to catch a grasshopper on a balance beam that is 1 meter long. Every time you try to catch it, it jumps to a random point along the interval between 20 centimeters left of its current position and 20 centimeters right of its current position. If the grasshopper is within 20 centimeters of one of the edges, it will not jump off the edge. For example, if it is 10 centimeters from the left edge of the beam, then it will randomly jump to anywhere within 30 centimeters of that edge with equal probability (meaning it will be twice as likely to jump right as it is to jump left). After many, many failed attempts to catch the grasshopper, where is it most likely to be on the beam? Where is it least likely? And what is the ratio between these respective probabilities?

My solution:
[Show Solution]

## Tetrahedral dice game

This week’s Riddler Classic is a game of four-sided dice:

You have four fair tetrahedral dice whose four sides are numbered 1 through 4.

You play a game in which you roll them all and divide them into two groups: those whose values are unique, and those which are duplicates. For example, if you roll a 1, 2, 2 and 4, then the 1 and 4 will go into the “unique” group, while the 2s will go into the “duplicate” group.

Next, you reroll all the dice in the duplicate pool and sort all the dice again. Continuing the previous example, that would mean you reroll the 2s. If the result happens to be 1 and 3, then the “unique” group will now consist of 3 and 4, while the “duplicate” group will have two 1s.

You continue rerolling the duplicate pool and sorting all the dice until all the dice are members of the same group. If all four dice are in the “unique” group, you win. If all four are in the “duplicate” group, you lose.

What is your probability of winning the game?

My solution:
[Show Solution]

## The luckiest coin

This week’s Riddler Classic is about finding the “luckiest” coin!

I have in my possession 1 million fair coins. I first flip all 1 million coins simultaneously, discarding any coins that come up tails. I flip all the coins that come up heads a second time, and I again discard any of these coins that come up tails. I repeat this process, over and over again. If at any point I am left with one coin, I declare that to be the “luckiest” coin.

But getting to one coin is no sure thing. For example, I might find myself with two coins, flip both of them and have both come up tails. Then I would have zero coins, never having had exactly one coin.

What is the probability that I will at some point have exactly one “luckiest” coin?

Here is my solution:
[Show Solution]

## Vehicular trouble

This week’s Riddler Classic is about steady-state mixing of fluids. Here is the paraphrased problem.

Your old van holds 12 quarts of transmission fluid. At the moment, all 12 quarts are “old.” But changing all 12 quarts at once carries a risk of transmission failure. Instead, you decide to replace the fluid a little bit at a time. Each month, you remove one quart of old fluid, add one quart of fresh fluid and then drive the van to thoroughly mix up the fluid. Unfortunately, after precisely one year of use, what was once fresh transmission fluid officially turns “old.” You keep up this process for many, many years. One day, immediately after replacing a quart of fluid, you decide to check your transmission. What percent of the fluid is old?

Here is my solution:
[Show Solution]

## Can you eat all the chocolates?

This week’s Riddler Classic is a neat puzzle about eating chocolates.

I have 10 chocolates in a bag: Two are milk chocolate, while the other eight are dark chocolate. One at a time, I randomly pull chocolates from the bag and eat them — that is, until I pick a chocolate of the other kind. When I get to the other type of chocolate, I put it back in the bag and start drawing again with the remaining chocolates. I keep going until I have eaten all 10 chocolates.

For example, if I first pull out a dark chocolate, I will eat it. (I’ll always eat the first chocolate I pull out.) If I pull out a second dark chocolate, I will eat that as well. If the third one is milk chocolate, I will not eat it (yet), and instead place it back in the bag. Then I will start again, eating the first chocolate I pull out.

What are the chances that the last chocolate I eat is milk chocolate?

Here is my original solution:
[Show Solution]

And here is a far more elegant solution, courtesy of @rahmdphd on Twitter.
[Show Solution]

## Delirious ducks

This week’s Riddler Classic is about random walks on a lattice:

Two delirious ducks are having a difficult time finding each other in their pond. The pond happens to contain a 3×3 grid of rocks.

Every minute, each duck randomly swims, independently of the other duck, from one rock to a neighboring rock in the 3×3 grid — up, down, left or right, but not diagonally. So if a duck is at the middle rock, it will next swim to one of the four side rocks with probability 1/4. From a side rock, it will swim to one of the two adjacent corner rocks or back to the middle rock, each with probability 1/3. And from a corner rock, it will swim to one of the two adjacent side rocks with probability 1/2.

If the ducks both start at the middle rock, then on average, how long will it take until they’re at the same rock again? (Of course, there’s a 1/4 chance that they’ll swim in the same direction after the first minute, in which case it would only take one minute for them to be at the same rock again. But it could take much longer, if they happen to keep missing each other.)

Extra credit: What if there are three or more ducks? If they all start in the middle rock, on average, how long will it take until they are all at the same rock again?

Here is my solution:
[Show Solution]

## Beer pong

This interesting twist on the game of Beer Pong appeared on the Riddler blog. Here it goes:

The balls are numbered 1 through N. There is also a group of N cups, labeled 1 through N, each of which can hold an unlimited number of ping-pong balls. The game is played in rounds. A round is composed of two phases: throwing and pruning.

During the throwing phase, the player takes balls randomly, one at a time, from the infinite supply and tosses them at the cups. The throwing phase is over when every cup contains at least one ping-pong ball. Next comes the pruning phase. During this phase the player goes through all the balls in each cup and removes any ball whose number does not match the containing cup. Every ball drawn has a uniformly random number, every ball lands in a uniformly random cup, and every throw lands in some cup. The game is over when, after a round is completed, there are no empty cups.

How many rounds would you expect to need to play to finish this game? How many balls would you expect to need to draw and throw to finish this game?

Here is my solution:
[Show Solution]

## Is this bathroom occupied?

After a brief hiatus from Riddling, I’m back! This Riddler problem is about probability and bathroom vacancy.

There is a bathroom in your office building that has only one toilet. There is a small sign stuck to the outside of the door that you can slide from “Vacant” to “Occupied” so that no one else will try the door handle (theoretically) when you are inside. Unfortunately, people often forget to slide the sign to “Occupied” when entering, and they often forget to slide it to “Vacant” when exiting.

Assume that 1/3 of bathroom users don’t notice the sign upon entering or exiting. Therefore, whatever the sign reads before their visit, it still reads the same thing during and after their visit. Another 1/3 of the users notice the sign upon entering and make sure that it says “Occupied” as they enter. However, they forget to slide it to “Vacant” when they exit. The remaining 1/3 of the users are very conscientious: They make sure the sign reads “Occupied” when they enter, and then they slide it to “Vacant” when they exit. Finally, assume that the bathroom is occupied exactly half of the time, all day, every day.