Desert escape

This week’s Riddler classic is about geometry and probability, and desert escape! Here is the (paraphrased) problem:

There are $n$ travelers who are trapped on a thin and narrow oasis. They each independently pick a random location in the oasis from which to start and a random direction in which to travel. What is the probability that none of their paths will intersect, in terms of $n$?

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]

Frustrating elevator

This weeks Riddler Express is a problem about a frustrating elevator! Here it goes:

You are on the 10th floor of a tower and want to exit on the first floor. You get into the elevator and hit 1. However, this elevator is malfunctioning in a specific way. When you hit 1, it correctly registers the request to descend, but it randomly selects some floor below your current floor (including the first floor). The car then stops at that floor. If it’s not the first floor, you again hit 1 and the process repeats.

Assuming you are the only passenger on the elevator, how many floors on average will it stop at (including your final stop, the first floor) until you exit?

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]

Squid game

This week’s Riddler Classic is Squid Game-themed!

There are 16 competitors who must cross a bridge made up of 18 pairs of separated glass squares. Here is what the bridge looks like from above:

To cross the bridge, each competitor jumps from one pair of squares to the next. However, they must choose one of the two squares in a pair to land on. Within each pair, one square is made of tempered glass, while the other is made of normal glass. If you jump onto tempered glass, all is well, and you can continue on to the next pair of squares. But if you jump onto normal glass, it will break, and you will be eliminated from the competition.

The competitors have no knowledge of which square within each pair is made of tempered glass. The only way to figure it out is to take a leap of faith and jump onto a square. Once a pair is revealed — either when someone lands on a tempered square or a normal square — all remaining competitors take notice and will choose the tempered glass when they arrive at that pair.

On average, how many of the 16 competitors will make it across the bridge?

Here is my solution.
[Show Solution]

And here is a much better 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]

Flawless war

his week’s Riddler Classic has to do with the card game “War”. Here is the problem, paraphrased:

War is a two-player game in which a standard deck of cards is first shuffled and then divided into two piles with 26 cards each; one pile for each player. In every turn of the game, both players flip over and reveal the top card of their deck. The player whose card has a higher rank wins the turn and places both cards on the bottom of their pile. Assuming a deck is randomly shuffled before every game, how many games of War would you expect to play until you had a game that lasted just 26 turns (with no ties; a flawless victory)?

Here is my solution:
[Show Solution]

Cutting a ruler into pieces

This week’s Riddler Classic is a paradoxical question about cutting a ruler into smaller pieces.

Recently, there was an issue with the production of foot-long rulers. It seems that each ruler was accidentally sliced at three random points along the ruler, resulting in four pieces. Looking on the bright side, that means there are now four times as many rulers — they just happen to have different lengths. On average, how long are the pieces that contain the 6-inch mark?

With four cuts, each piece will be on average 3 inches long, but that can’t be the answer, can it?

Here is my solution:
[Show Solution]

Dungeons & Dragons

This week’s Riddler Classic is a probability problem about the game Dungeons & Dragons. Here it goes:

When you roll a die “with advantage,” you roll the die twice and keep the higher result. Rolling “with disadvantage” is similar, except you keep the lower result instead. The rules further specify that when a player rolls with both advantage and disadvantage, they cancel out, and the player rolls a single die. Yawn!

There are two other, more mathematically interesting ways that advantage and disadvantage could be combined. First, you could have “advantage of disadvantage,” meaning you roll twice with disadvantage and then keep the higher result. Or, you could have “disadvantage of advantage,” meaning you roll twice with advantage and then keep the lower result. With a fair 20-sided die, which situation produces the highest expected roll: advantage of disadvantage, disadvantage of advantage or rolling a single die?

Extra Credit: Instead of maximizing your expected roll, suppose you need to roll N or better with your 20-sided die. For each value of N, is it better to use advantage of disadvantage, disadvantage of advantage or rolling a single die?

Here is a detailed derivation of the relevant probabilities:
[Show Solution]

And here are the results:
[Show Solution]

Flip to freedom

This week’s Riddler Classic is a problem about coin flipping. The text of the original problem is quite long, so I will paraphrase it here:

There are $n$ prisoners, each with access to a random number generator (generates uniform random numbers in $[0,1]$) and a fair coin. Each prisoner is given the opportunity to flip their coin once if they so choose. If all of the flipped coins come up Heads, all prisoners are released. But if any of the flipped coins come up Tails, or if no coins are flipped at all, the prisoners are not released. If the prisoners cannot communicate in any way and do not know who is flipping their coin or not, how can they maximize their chances of being released?

Here is my solution:
[Show Solution]