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Can You Win The Lotería?

Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. There are two types: Riddler Express for those of you who want something bite-size and 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 next week’s column. If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

Riddler Express

From Taylor Firman, the unluck of the draw:

Lotería is a traditional Mexican game of chance, akin to bingo. Each player receives a four-by-four grid of images. Instead of a comically large rotating bin of numbered balls, the caller randomly draws a card from a deck containing all 54 possible images. If a player has that image on their grid, they mark it off. The exact rules can vary, but in this version, the game ends when one of the players fills their entire card (and screams “¡Lotería!”). Each of the 54 possible images can only show up once on each card, but other than that restriction, assume that image selection and placement on each player’s grid is random.

One beautiful day, you and your friend Christina decide to face off in a friendly game of Lotería. What is the probability that either of you ends the game with an empty grid, i.e. none of your images was called? How does this probability change if there were more or fewer unique images? Larger or smaller player grids?

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Riddler Classic

From Ben Wiles, a mathematical trip across the pond:

My favorite game show is “Countdown” on Channel 4 in the UK. I particularly enjoy its Numbers Game. Here is the premise: There are 20 “small” cards, two of each numbered 1 through 10. There are also four “large” cards numbered 25, 50, 75 and 100. The player asks for six cards in total: zero, one, two, three or four “large” numbers, and the rest in “small” numbers. The hostess selects that chosen number of “large” and “small” at random from the deck. A random-number generator then selects a three-digit number, and the players have 30 seconds to use addition, subtraction, multiplication and division to combine the six numbers on their cards into a total as close to the selected three-digit number as they can.

There are four basic rules: You can only use a number as many times as it comes up in the six-number set. You can only use the mathematical operations given. At no point in your calculations can you end on something that isn’t a counting number. And you don’t have to use all of the numbers.

For example, say you ask for one large and five smalls, and you get 2, 3, 7, 8, 9 and 75. Your target is 657. One way to solve this would be to say 7×8×9 = 504, 75×2 = 150, 504+150 = 654 and 654+3 = 657. You could also say 75+7 = 82, 82×8 = 656, 3-2 = 1 and 656+1 = 657.

This riddle is twofold. One: What number of “large” cards is most likely to produce a solvable game and what number of “large” cards is least likely to be solvable? Two: What three-digit numbers are most or least likely to be solvable?

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Solution to last week’s Riddler Express

Congratulations to рџ‘Џ Adam Martin-Schwarze рџ‘Џ of Sequim, Washington, winner of last week’s Riddler Express!

Last week we met a soccer coach who was trying to assemble a team of 11 players in a very specific way. He had an infinite pool of players to choose from, each of whom wore a unique number on their jersey such that there was one player for every number. That number also happened to be the number of games it took on average for that player to score a goal. The coach wanted his team to average precisely two goals per game, and he also wanted his weakest player to be as good as possible. What number does the ideal weakest player wear? What are the numbers of the other 10 players the coach should select?

The weakest player selected for the team wears the number 24. The other 10 players wear the numbers 1, 5, 6, 8, 9, 10, 12, 15, 18 and 20.

Let’s quickly check that everything adds up correctly. If a player’s number is 5, say, then that player scores an average of 1/5 goals per game. If their number is 6, they average 1/6 goals per game, and so on. So our team as a whole averages 1/1 + 1/5 + 1/6 + 1/8 + 1/9 + 1/10 + 1/12 + 1/15 + 1/18 + 1/20 + 1/24 = exactly 2 goals per game, just like the coach wanted!

I’m not aware of a more elegant method of solving this coaching conundrum than basic guess-and-check. There are many possibilities to consider, but one thing we do know is that the fractions we’re adding up to try to get to a sum of 2 are of a specific type: 1 divided by a whole number, which are also known as Egyptian fractions. These made a prominent appearance in the so-called Rhind papyrus, which also happens to be the oldest known collection of math puzzles.

We also know that we want the worst player on the team to be as good as possible — that is, to have as big a fraction as possible — and that there are 11 players on the team. So first we might check 1/1 + 1/2 + 1/3 + … + 1/11, i.e., the best possible team, but that sum equals about 3, too big for our coach. We then might check the sums of the possible sets of fractions between 1/1 and 1/12, and then the possible sets between 1/1 and 1/13, and so on, rejigging the sums until we find something that gets us to exactly 2. It turns out that none of these will add up to exactly 2 until we get to testing out those fractions between 1/1 and 1/24, and specifically those fractions listed above.

Solver David DeSmet shared a handy computer program he wrote to churn through all these possibilities, and Martin Piotte shared his thorough accounting of the possible teams.

Solution to last week’s Riddler Classic

Congratulations to рџ‘Џ Curtis Bennett рџ‘Џ of Long Beach, California, winner of last week’s Riddler Classic!

Last week, three astronauts were at the edge of their Mars lander, staring down at the surface of the red planet. Each wanted to be the first human to step foot on the planet, but they wanted to pick who it would be using a fair and efficient method. They could, for example, use a fair coin, assign each astronaut an outcome — heads-heads, heads-tails and tails-heads — and flip the coin twice. If the result was tails-tails, they could simply restart the process. However, that method could take a long time and there was exploration to be done.

Another approach, however, would be to use an “unfair coin” — one for which the probabilities of heads and tails are not equal. Is it possible to make a fair choice among three astronauts with a fixed number of flips of an unfair coin? You were able to set the coin’s probability of heads to any number you like between 0 and 1. You could flip the coin as many times as you like, as long as that was some known, fixed number. And, you could assign any combination of possible outcomes to each of the three astronauts.

Indeed it was possible, though who knew coin flips could get so complicated. This puzzle’s submitter, Dean Ballard, walks us through his solution:

What makes this problem interesting is that at first glance it appears to require searching an overwhelmingly large space of possibilities. The trick to solving this more easily lies in a simplifying assumption. Instead of dealing with three different probability functions for the three different astronauts, we can assign two of them sets of head-tail combinations that will give them the equal chances of winning, independent of the weighting of our coin. This way, we can only worry about two things at once rather than three. We will need at least four coin flips to make this work.

Let \(p(H)\) be the probability that our specially designed coin lands heads. With four flips we have 16 possible outcomes: HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH and TTTT. Note that subsets of these 16, such as {HHHT, HHTH, HTHH, THHH}, all have the same probability — in this specific case, \(p(H)^3 (1 – p(H))\). So let’s call these “3H1T” — three heads, one tail.

Expressing the 16 outcomes this way gives us: one 4H, four 3H1T, six 2H2T, four 1H3T and one 4T, or in another piece of shorthand, [1, 4, 6, 4, 1]. Let’s say we assign Astronaut A just the HHHH and TTTT outcomes, and evenly divide the other 14 between Astronauts B and C. This gives us A = [1, 0, 0, 0, 1], B = [0, 2, 3, 2, 0], and C = [0, 2, 3, 2, 0]. Each of these defines a probability function for each astronaut. (Note that the sum A + B + C = [1, 4, 6, 4, 1], so all outcomes have been assigned.) The probability for Astronaut A equals 1 when \(p(H) = 0\) or \(p(H) = 1\), so it is concave up. The function for B and C equals 0 when \(p(H) = 0\) or \(p(H) = 1\), so it is concave down. Since both functions are continuous, as long as A’s value is less than B’s and C’s value when \(p(H) = 0.5\) (which is true in this case), there must be a solution between 0 and 0.5, and another between 0.5 and 1.

Here is what those solutions look like graphically. On the x-axis is \(p(H)\) and on the y-axis is the probability that an astronaut wins the contest.

Let’s quickly check the solution at point A, where the \(p(H)\), the probability of our coin landing heads, equals about 0.24213. As we mentioned above, Astronaut A gets to take the first step if the coin lands with four heads or four tails. This happens with probability \(0.24213^4 + (1-0.24213)^4\) = 0.3333, or a third, which is exactly what we want. Since the other two astronauts have been assigned probabilistically equivalent outcomes, we know they must have equal chances, which must also be a third. So we’ve successfully devised a fair method that will give us a result in a known and fixed number of unfair coin flips!

For extra credit, you faced the same question but with five astronauts. I’ll spare you all the gory details, but suffice it to say that solver Zach Wissner-Gross devised one method that used eight flips of a coin that came up heads about 81.7 percent of the time. The colors correspond to the lucky astronaut who will get to make history based on the flips shown on the axes of the diagram. So, for example, if the coin came up eight straight heads — relatively likely given the weighting of the coin — then Astronaut Yellow gets to take those first steps.

Want more riddles?

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?

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  1. Important small print: For you to be eligible, I need to receive your correct answer before 11:59 p.m. Eastern time on Sunday. Have a great weekend!

Oliver Roeder was a senior writer for FiveThirtyEight. He holds a Ph.D. in economics from the University of Texas at Austin, where he studied game theory and political competition.