Welcome to The Riddler. Each week, I offer problems related to the things we hold dear here: math, logic, and probability. Two puzzles are featured each week: the Riddler Express for those of you who want something bite-sized, and the Riddler Classic for those of you into the slow puzzle movement. Submit a correct answer for either one and you may receive an acknowledgment in the next column. Please wait until Monday to publicly share your answers! If you need a clue or have a favorite puzzle gathering dust in your attic, find me on twitter or send me an email.

## riddle express

As a citizen of the Riddler Nation, you are visiting the United States. Upon landing at an American airport, you would like to exchange your 100 Riddlerian rupees for some American currency. Fortunately, you see a currency exchange station where it might be possible to make a profit.

The dollar is known to be more valuable than the rupee. Now this station says they’ll give you *D.* dollars for each rupee, where *D.* is a decimal less than 1 that goes to the hundredths place. So *D.* it can be 0.99, 0.50, or 0.37, but not values like 0.117 or 1/𝜋. And when exchanging dollars for rupees, the station uses an exchange rate of *R.*where *R.* is equal to 1/*D.* *rounded to the nearest hundredth*. (Yes, that last part is very important.)

For example, suppose *D.* is 0.53. In this case, when you exchange 100 rupees, you will receive $53. When exchanging the $53 dollars, the station uses an exchange rate of 1/0.53, or 1.88679…, which rounds to 1.89. And so, paying back the $53 gives you Rs 100.17, a net profit!

what value of *D.* will you get the most profit for your 100 rupees? (Remember, *D.* is a decimal to the hundredths place and is less than 1.)

send your answer

## classic riddle

From Dave Moran comes a handy parking puzzle:

There is a parking lot behind Dave’s office building with 10 spaces that are available on a first-come, first-served basis. Those 10 spaces invariably fill up by 8 am, and the parking lot empties quickly by 5 pm sharp.

Every day, three of the 10 “early risers” who got spots before 8 am leave at random times between 10 am and 3 pm and don’t return that day. Knowing that some early risers leave during that five-hour window, nine “stragglers” pass through the lot at random times between 10 a.m. and 3 p.m. If a spot is available, a straggler immediately parks in the spot and doesn’t leave. until 5 pm If there is no space available, a straggler immediately drives away from the lot and parks elsewhere, and does not return that day.

Suppose you are a latecomer who arrives at a random time between 10 am and 3 pm, what is the probability that you will get a spot in the lot?

send your answer

## Solution to the last Riddler Express

Congratulations to 👏 Billy Mullaney 👏 from Minneapolis, winner of last week’s Riddler Express.

Before the 2013-14 season, the NBA changed the format of the NBA Finals, a best-of-seven series. Previously, the Finals used a “2-3-2” format: Games 1, 2, 6, and 7 were played in the higher-seeded team’s home arena, while Games 3, 4, and 5 were played in the lower seeded arena. – seeded team. With the change, the finals moved to the “2-2-1-1-1” format: games 1, 2, 5, and 7 were at the home of the highest-ranked team, while games 3, 4, and 6 were at home. . the home of the lower-seeded team.

Last week, you were playing for the highest ranked team going into the Finals. Although his team had a better record during the regular season, the two teams were evenly matched: at a neutral site, both teams had an equal odds of winning a game. But, of course, no Finals game is played at a neutral site. Both teams had a 60 percent chance to win each game at home and a 40 percent chance to win each game away from home.

Which format, 2-3-2 or 2-2-1-1-1, gave your team a better chance of winning the final? (Or were they the same?)

Some solvers, like Emily Kelly, worked on the details, but that turned out not to be necessary with careful thought from the start. We tend to break down best-of-seven series into different cases depending on how many games are played, which could be four, five, six, or seven, each with different probabilities of occurring. But another way to imagine all this would be for the two teams to play *the seven games*, even if one of the teams has already won four of them. Playing those potentially extra games would have no effect on either team’s probability of winning the series, since the *first* The team that wins four out of seven must be the *only* team to win four of seven.

So if you imagine playing all seven games no matter what, four of which are home and three away, then the order of which games are home and which are away will have no effect on your chances of winning. Therefore, both formats (2-3-2 and 2-2-1-1-1) gave you **the same opportunity** of win

In fact, the series could have been 1-1-1-1-1-1-1 (alternating home and away each game), 4-3 (the first four home games) or 3-4 (the last four home games). None of this mattered: 7 pick 4 (or 35) arrangements gave you an equal chance of winning the series.

To earn additional credit, you had to determine the chance of winning. This time, it was necessary to work on the details as Emily did. Again, it may have been simpler to imagine that she played all seven games instead of stopping as soon as a team won four of them. In the end, her probability of winning was 8,313 / 15,625, or **about 53.2 percent**.

Interestingly, when the home team had a 50+*X* percentage probability of winning, then their chances of winning the series varied non-linearly with *X*. For small values of *X*, the probability of winning a series was around 50 percent. We already said that you had a 53.2 percent chance of winning the series when the home team won 60 percent of the time. When the home team won 70 percent of the time, you won the series about 57 percent of the time. When the home team won 80 percent of the time, you won the series about 62.6 percent of the time. When the home team won 90 percent of the time, you won the series about 73.6 percent of the time. But when the home team won 100 percent of the time, home field advantage meant you were guaranteed to win the series.

## Solution to the last Riddler Classic

Congratulations to 👏Matthew Pitcock 👏 from Chicago, winner of last week’s Riddler Classic.

Last week, you started with just the number 1 written on a piece of paper in a hat. You were going to draw from the hat 100 times, and each time you drew, you had a choice: whether the number on the piece of paper you drew was *what*then you could receive *what* dollars or add *what* higher numbers to the hat.

For example, if the hat contained slips with the numbers 1 through 6 and you drew a 4, you could have received $4 or no money, but you added four more slips with the numbers 7, 8, 9, and 10 in the hat. In any case, the slip of paper with the number 4 would have been returned to the hat.

If you played this game perfectly, that is, to maximize the total amount of money you would receive after 100 rounds, how much money would you have expected to receive on average?

What made this puzzle interesting was the apparent tension between your options: at each stage, did it make more sense to cash in what you could, or was it better to reinvest and put larger amounts in the hat, which could increase your profits later on? ? Most solvers addressed this tension by working backwards from when there were only a few rounds left.

Suppose you are drawing for the hundredth (and last) time and there are *north* numbers on the hat. Since this was his last chance to get some money out of this game, he always took the money instead of putting more numbers in the hat. He could expect to receive the average of the numbers from 1 to *north*either (*north*+1)/2 dollars.

Now suppose you are drawing for the 99th time. Again, suppose there were *north* numbers on the hat (perhaps a different value from the *north* in the previous paragraph) and that you drew the number *what*. If you put the money in your pocket, you would get *what* dollars this round and then an average of (*north*+1)/2 dollars in the final round, for a total average of *what*+(*north*+1)/2. but if you put *what* more numbers in the hat, you would get zero dollars this round and an average of (*north*+*what*+1)/2 dollars in the final round. From *what*+(*north*+1)/2 was always greater than (*north*+*what*+1)/2, you better take the money here. In the end, you could expect to receive the average *what*+(*north*+1)/2 for all values of *what* from 1 to *north*what was it *north*+1 dollars.

Next, suppose you are drawing for the 98th time. Once again, suppose there were *north* numbers in the hat and you drew the number *what*. if you took *what* dollars, your expected total after 100 draws would have been *what*+*north*+1. If instead you added *what* more numbers to the hat, your expected total after 100 draws would be *again* has been *what*+*north*+1, exactly the same result! So in draw 98, it didn’t matter if you took the money or added more numbers to the hat. Either way, his expected final total was (3/2)·(*north*+1) dollars.

And now suppose you are drawing for the 97th time, with *north* numbers in the hat and a drawn number *what*. If he took the money, the total expected from him was *what*+(3/2)(*north*+1). if you added *what* more numbers to the hat, his expected total was (3/2)·(*north*+*what*+1). This time, you got more money on average by adding numbers to the hat, resulting in an expected total of (3/2)^{2}·(*north*+1) dollars.

Putting more numbers in the hat was also the best option when you drew the 96th time, the 95th time, the 94th time and so on, with each previous draw adding another factor of 3/2 to your expected total.

To summarize, you could maximize your expected profits by *always* put more numbers in the hat (regardless of what value *what* you drew) for the first 97 drawings. For 98, it didn’t matter if you took the money or added more numbers to the hat. And for the last two draws, you took the money. In the end, his total expected earnings were (3/2)^{98}·(1+1), or **about 361.4 quadrillion (with Q!) dollars**. For reference, this figure exceeded the world’s annual GDP by a few thousand times.

Solver Laurent Lessard further explored the *distribution* your winnings could take when you played to maximize the average. The distribution had a log-normal appearance, so the arithmetic mean appeared to the right of the maximum. While he made $361.4 quadrillion on average, he generally made less than that (and sometimes *a lot* further).

## Do you want more puzzles?

Well, ain’t you lucky? There’s an entire book full of the best brain teasers in this column and some never-before-seen head scratchers. It’s called “The Riddler”, and it’s already in stores!

## Do you want to submit a riddle?

Email Zach Wissner-Gross at [email protected].