Can You Take a "Risk"?
In the board game "Risk," you are dealt cards that come in one of three varieties. What are your chances of being dealt three cards of the same variety or all different varieties?
Welcome to Fiddler on the Proof! The Fiddler is the spiritual successor to FiveThirtyEight’s The Riddler column, which ran for eight years under the stewardship of myself and Ollie Roeder.
Each week, I present mathematical puzzles intended to both challenge and delight you. Puzzles come out Friday mornings (8 a.m. Eastern time). Most can be solved with careful thought, pencil and paper, and the aid of a calculator. The “Extra Credit” is where the analysis typically gets hairy, or where you might turn to a computer for assistance.
I’ll also give a shoutout to 🎻 one lucky winner 🎻 of the previous week’s puzzle, chosen randomly from among those who submit their solution before 11:59 p.m. the Monday after that puzzle was released. I’ll do my best to read through all the submissions and give additional shoutouts to creative approaches or awesome visualizations, the latter of which could receive 🎬 Best Picture Awards 🎬.
This Week’s Fiddler
Earlier this week, I was playing the board game Risk with one of my kids. At the end of each turn in the game in which you conquer at least one enemy territory on the board, you are dealt a card.
There are 42 territory cards in the deck—14 that depict an infantry unit, 14 that depict a cavalry unit, and 14 that depict an artillery unit. Once you have three cards that either (1) all depict the same kind of unit, or (2) all depict different kinds of units, you can trade them in at the beginning of your next turn in exchange for some bonus units to be placed on the board.
If you are randomly dealt three cards from the 42, what is the probability that you can trade them in?
This Week’s Extra Credit
The full deck of Risk cards also contains two wildcards, which can be used as any of the three types of cards (infantry, cavalry, and artillery) upon trading them in. Thus, the full deck consists of 44 cards.
You must have at least three cards to have any shot at trading them in. Meanwhile, having five cards guarantees that you have three you can trade in.
If you are randomly dealt cards from a complete deck of 44 one at a time, how many cards would you need, on average, until you can trade in three? (Your answer should be somewhere between three and five. And no, it’s not four.)
Making the ⌊Rounds⌉
There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. This week, I’m sharing an oldie but a goodie, relayed to me by reader Michael Amati. Apparently it’s making the rounds over on LinkedIn, but I immediately recognized it as a variant of Martin Gardner’s infamous “boy or girl paradox.”
If you haven’t seen this puzzle before, then boy (pun intended) are you in for a treat. Here’s how it was posed over on LinkedIn:
Mary has two children. One of them is a boy born on a Tuesday. What is the probability that the other child is a girl?
Now I should point out there’s some subtlety here. What were the criteria for selecting Mary? Was she chosen among all mothers of two children, and you’re simply being told the traits of the firstborn? Or was she chosen among all mothers of two children who possessed at least one boy born on a Tuesday?
These questions turn out to have different answers. The latter interpretation is the intended question in the paradox. The answer is, counterintuitively, not 50 percent, even with completely reasonable assumptions: A randomly selected child in the entire population is equally likely to be a boy or a girl, the gender of a child is independent of the genders of their siblings, etc.
The original poster then goes on to state that puzzles like this are “why people hate stats.” First off, this is probability, not statistics. Second, I know firsthand that this puzzle has the power to make people fall in love with probability, rather than hate it. So there!
Want to Submit a Puzzle Idea?
Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.
Standings
I’m tracking submissions from paid subscribers and compiling a leaderboard, which I’ll reset every quarter. All correct solutions to Fiddlers and Extra Credits are worth 1 point each. Solutions should be sent prior to 11:59 p.m. the Monday after puzzles are released. At the end of each quarter, I’ll 👑 crown 👑 the finest of Fiddlers. If you think you see a mistake in the standings, kindly let me know.
Last Week’s Fiddler
Congratulations to the (randomly selected) winner from last week: 🎻 Jerome Socolof 🎻 from Adams, Massachusetts. I received 55 timely submissions, of which 44 were correct—good for an 80 percent solve rate.
Last week, a group of people in a team building exercise were asked to quantify themselves based on questions like “What outdoor temperature do you prefer?” No one revealed their answer at first. Instead, each person placed a card with their name on an unmarked line, one at a time. In this example, folks who preferred higher temperatures placed their names farther right; folks who preferred lower temperatures placed their names farther left. However, the line was unmarked and didn’t have any units.
Once all the names were placed on the line, their values were revealed. For example, a group of six might have generated the following numbers, in order from left to right on the line: 60, 67, 65, 74, 70, 80.
The team’s score was the length of the “longest increasing subsequence.” In other words, it was the maximum number of elements in the list you could keep such that they formed an increasing subsequence. In the example above, you could remove the 67 and the 74 to get the following increasing subsequence with four numbers: 60, 65, 70, 80. There were a few other ways to get an increasing subsequence of length four, but there was no way to get a sequence of length five or more, so the team’s score was four.
Suppose a total of four people were participating in this team building exercise. They all wrote down different numbers, and then independently placed their names at random positions on the line.
On average, what did you expect the team’s score to be?
A team size of four was small enough that you could work out the answer by hand (thank goodness!). But before we get to far, let’s look at even smaller team sizes to get the lay of the land. For simplicity, let’s assume that a team size of N produced the whole numbers from 1 to N.
With a team size of one, there wasn’t really much of a “team.” Nevertheless, the longest increasing subsequence was always 1, which meant the average score was 1.
With a team size of two, there were two possible ways for the team to order their numbers on the line. Remember, since the positions on the line were random, each ordering was equally likely.
1, 2. The longest increasing subsequence was 1, 2, which had length 2.
2, 1. The longest increasing subsequence was 2 (or 1), which had length 1.
The average score was (2 + 1)/2, or 1.5.
With a team size of three, there were six (i.e., 3!) possible ways for the team to order their numbers on the line.
1, 2, 3. The longest increasing subsequence was 1, 2, 3, which had length 3.
1, 3, 2. The longest increasing subsequence was 1, 3 (or 1, 2), which had length 2.
2, 1, 3. The longest increasing subsequence was 2, 3 (or 1, 3), which had length 2.
2, 3, 1. The longest increasing subsequence was 2, 3, which had length 2.
3, 1, 2. The longest increasing subsequence was 1, 2, which had length 2.
3, 2, 1. The longest increasing subsequence was 3 (or 2, or 1), which had length 1.
The average from these six cases was 2. So far, for N = 1, 2, and 3, a formula for the longest increasing subsequence was (N + 1)/2. But would that hold up when N was 4? If so, then the average longest increasing subsequence would have length 2.5. Let’s find out if that was right!
With a team size of four, there were 24 (i.e., 4!) possible ways for the team to order their numbers on the line.
1, 2, 3, 4. The longest increasing subsequence had length 4.
1, 2, 4, 3. The longest increasing subsequence had length 3.
1, 3, 2, 4. The longest increasing subsequence had length 3.
1, 3, 4, 2. The longest increasing subsequence had length 3.
1, 4, 2, 3. The longest increasing subsequence had length 3.
1, 4, 3, 2. The longest increasing subsequence had length 2.
2, 1, 3, 4. The longest increasing subsequence had length 3.
2, 1, 4, 3. The longest increasing subsequence had length 2.
2, 3, 1, 4. The longest increasing subsequence had length 3.
2, 3, 4, 1. The longest increasing subsequence had length 3.
2, 4, 1, 3. The longest increasing subsequence had length 2.
2, 4, 3, 1. The longest increasing subsequence had length 2.
3, 1, 2, 4. The longest increasing subsequence had length 3.
3, 1, 4, 2. The longest increasing subsequence had length 2.
3, 2, 1, 4. The longest increasing subsequence had length 2.
3, 2, 4, 1. The longest increasing subsequence had length 2.
3, 4, 1, 2. The longest increasing subsequence had length 2.
3, 4, 2, 1. The longest increasing subsequence had length 2.
4, 1, 2, 3. The longest increasing subsequence had length 3.
4, 1, 3, 2. The longest increasing subsequence had length 2.
4, 2, 1, 3. The longest increasing subsequence had length 2.
4, 2, 3, 1. The longest increasing subsequence had length 2.
4, 3, 1, 2. The longest increasing subsequence had length 2.
4, 3, 2, 1. The longest increasing subsequence had length 1.
The average from these 24 cases was 29/12, or approximately 2.417. That was the answer to last week’s Fiddler, and it fell slightly short of our hypothesis of 2.5 from a moment ago.
What happened with larger values of N? For that, let’s turn to Extra Credit.
Last Week’s Extra Credit
Congratulations to the (randomly selected) winner from last week: 🎻 Nate McIntosh 🎻 from Toledo, Ohio. I received 33 timely submissions, of which 29 were correct—good for an 88 percent solve rate.
Instead of four people, now there were 10 people participating in the team building exercise. As before, they all wrote down different numbers, and then independently placed their names at random positions on the line.
On average, what did you expect the team’s score to be?
