Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,1 and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter or send me an email.
You are at The Riddler Casino, and you are betting on a horse race. The casino provides betting odds (in the American format) for each horse. For example, odds of -150 means that for every $150 you bet, you win an additional $100. Meanwhile, odds of +150 means that for every $100 you bet, you win an additional $150.
To break even, a horse with -150 odds should win 60 percent of the time, while a horse with +150 odds should win 40 percent of the time. (Yes, both +100 and -100 correspond to a 50 percent chance of victory.) Of course, most casinos rig the odds such that betting on all the horses in a race will cause you to lose money.
But not The Riddler Casino! Here, a horse with -150 odds has exactly a 60 percent chance of winning, and a horse with +150 odds has exactly a 40 percent chance.
Today, a five-horse race has caught your eye. The odds for three of the horses are +100, +300 and +400. You can’t quite make out the odds for the last two horses, but you can see that they’re both positive multiples of a hundred. What are the highest possible odds one of those last two horses can have?
From Chris Gerig comes a variant of the famed Monty Hall problem:
In a game show, there are three identical doors arranged in a row from left to right. The host of the show, “Monty,” chooses one of the doors and places a prize of $100 dollars behind it. There is no prize behind the other two doors. You are not present when Monty chooses the door and places the money behind it, so you cannot say for certain which door the prize is behind.
You are then brought to the stage and must select one of the three doors to open. If the prize money is behind it, then you win! But if you guess incorrectly, all is not lost. You can pay $80 to pick a second door. Before you make that second selection, however, Monty will give you a hint, telling you whether the prize is behind a door that’s to the left or to the right of your first choice. (Note that this hint is only helpful when you previously selected the middle door.) If the prize isn’t behind that second door, you can pay another $80 to try a third time.
Assume that both you and Monty play with optimal strategies — you to maximize your expected net earnings (prize winnings minus payments for hints), and Monty to minimize the same. How much net earnings can you expect to make on average?
Extra credit: Suppose Monty has you pay $80 up front before selecting your first door, and each subsequent selection (if you choose to make it) continues to cost $80. If the prize money remains $100, this game isn’t worth playing. How much should the prize money be to make the game worthwhile?
Solution to the last Riddler Express
Congratulations to 👏 Ryan Steel 👏 of West Chester, Pennsylvania, winner of last week’s Riddler Express.
Last week, as a citizen of Riddler Nation, you were visiting the United States. Upon landing at an American airport, you wanted to exchange your 100 Riddlerian rupees for some American currency. Fortunately, you noticed a currency exchange station where it was possible to make a profit.
At the time, the dollar was known to be more valuable than the rupee. Now this station said they would give you D dollars for each rupee, where D was a decimal less than 1 that went to the hundredths place. So D could have been 0.99, 0.50 or 0.37, but not values like 0.117 or 1/𝜋. And when exchanging dollars back into rupees, the station used an exchange rate of R, where R was equal to 1/D rounded to the nearest hundredth. (Yes, that last part was very important.)
For example, suppose D was 0.53. In this case, when you traded in 100 rupees, you received $53. When trading the $53 dollars back, the station used an exchange rate of 1/0.53, or 1.88679…, which they rounded up to 1.89. And so returning the $53 got you 100.17 rupees — a net profit!
What value of D would have earned you the greatest profit for your 100 rupees? (Remember, D was a decimal that went to the hundredths place and was less than 1.)
First off, where in this process did the profit-making actually occur? Because D was a decimal rounded to the nearest hundredth, when you exchanged 100 rupees you always received a number of dollars of equal value. It was the exchange back from dollars to rupees where profiting was possible.
Many solvers derived an algebraic expression for how much money you made (or lost). You started with 100 rupees, and after the first exchange you had 100·D dollars. The exchange rate back was 1/D rounded to the nearest hundredth, which could be expressed mathematically as round(100/D)/100, where “round” is a function that rounds to the nearest whole number. In the end, you had 100·D·round(100/D)/100 rupees, which simplified to D·round(100/D).
From there, most solvers either wrote code to find the maximum of this function or, like Lise Andreasen, used a spreadsheet to test the values of D from 0.01 to 0.99. Solver Bryce Manifold plotted the percent difference between this function and your initial 100 rupees, as shown below. There were some intriguing behaviors in this graph, including some apparent oscillations and a conic shape that widened for larger values of D. The maximum occurred when D was 0.93. In this case, your initial 100 rupees netted you 0.93·round(100/0.93), or 100.44 rupees — a whopping 0.44 percent increase.
While 0.44 percent may not sound too impressive, that little bit of leverage was all you would have needed to get your arbitrage scheme off the ground. If you had your computer make 158 such trades in rapid succession, you’d double your money. After 10,000 such trades, you’d have more money than the annual GDP of the entire world (although someone would likely have caught on to your scheme by that point).
Solution to the last Riddler Classic
Congratulations to 👏 Bill Neagle 👏 of Springfield, Missouri, winner of last week’s Riddler Classic.
Last week, there was a parking lot behind your office building with 10 spaces that were available on a first-come, first-serve basis. Those 10 spaces invariably filled up by 8 a.m., and the parking lot quickly emptied out at 5 p.m. sharp.
Every day, three of the 10 “early birds” who snagged spots before 8 a.m. left at random times between 10 a.m. and 3 p.m. and did not return that day. Knowing that some early birds left during that five-hour window, nine “stragglers” drove by the lot at random times between 10 a.m. and 3 p.m. If there was an available spot, a straggler immediately parked in the spot and didn’t leave until 5 p.m. If there was no open spot, a straggler immediately drove away from the lot and parked somewhere else, and didn’t return that day.
Suppose you were a straggler arriving at a random time between 10 a.m. and 3 p.m. What was the probability that you got a spot in the lot?
This puzzle was an exercise in combinatorics. In total, there were 12 people to keep track of, all of whom took their action — three leaving and nine arriving — at a random, independent time within the same interval. One way this scenario could have played out was with all the early birds leaving, followed by all the stragglers (S) arriving, the last of which was you (Y). We can document this as EEESSSSSSSSY — three early birds (E), followed by eight stragglers (S) and you (Y). In this case, you would not have gotten a spot in the lot. But for other orderings, like EESYESSSSSSS, you would have gotten a spot.
The total number of ways this parking scenario could have played out was equal to the number of ways to arrange three Es, eight Ss and one Y, which was 12!/(3!·8!·1!), or 1,980.
Among these equally likely cases, solvers like Bradon Zhang and Marissa Weichman divided the ones where you got a spot into four categories, based on the occurrence of the following four subsequences of letters:
After some careful consideration, you should be able to convince yourself that these four categories are mutually exclusive (i.e., an ordering can’t contain more than one of these sequences) and that the orderings with one of these subsequences are precisely the orderings in which you’d be able to park in the lot.
From there, you had to count up how many of the 1,980 sequences included each of these subsequences. For example, to find the number that included EY, out of the 12 total spots for letters, there were 11 in which the EY subsequence could have started. The remaining 10 letters included two Es and eight Ss, of which there were 10!/(2!·8!), or 45, ways to order them. And so there were 11·45, or 495, such orderings. Meanwhile, there were 72 orderings containing EESY, seven orderings containing EEESSY, and similarly another seven containing EESESY.
In total, there were 581 orderings that contained one of the four subsequences, which meant your probability of snagging a spot in the lot was 581/1,980, or about 29.3 percent.
Want more puzzles?
Well, aren’t you lucky? There’s a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. It’s called “The Riddler,” and it’s in stores now!
Want to submit a riddle?
Email Zach Wissner-Gross at firstname.lastname@example.org.