How the Mexican Game Lotería Is Providing Comfort During a Pandemic
The traditional game of chance is uniquely suited for this uncertain time.
- La Lotería, or Mexican bingo, is one of the games people are turning to while practicing social distancing.
- Artists are reimagining traditional cards to better represent modern day.
- Lotería can offer both a distraction and a sense of comfort.
In this time of quarantine and social distancing amid the coronavirus pandemic, puzzles and board games have become a way to pass the time with loved ones (virtually and not), while offering a creative outlet. There is, however, one game in particular that feels uniquely suited for this uncertain time: La Lotería.
A traditional game of chance, lotería—the Spanish word for lottery—is often referred to as Mexican bingo, where illustrated cards depicting the Mexican aesthetic replace bingo balls. Latinx and Hispanic communities have been playing this game for hundreds of years, but in the past decade, it has become increasingly visible in the United States, according to Google Trends.
At present, artists like Rafael Gonzales, Jr., and Millennial Lotería creator Mike Alfaro are reimagining lotería cards to capture our “new normal,” including versions that represent hand sanitizer, working from home, and other coping mechanisms. Elsewhere, Latinx creators, brands, and even former presidential candidate Julián Castro, have created their own cards or merchandise inspired by the game. And just this past December, Google invited Mexican and Mexican-American artists to reimagine and reinvent the cards for an interactive Google Doodle to celebrate the 106th anniversary of its copyright in Mexico.
Brooklyn-based author and illustrator Cecilia Ruiz, 37, was one of the guest artists invited to work on the Google Doodle—and to her, part of the game’s persistent appeal is purely nostalgic. The cards themselves are retro and charming, she tells OprahMag.com, and the game reminds her of growing up in Mexico City. “It was one of the few games that we could all play,” she says. “My grandparents, my parents, my cousins. it’s amazing in that way.”
The pandemic has certainly triggered a similar sense of nostalgia, as more people physically isolate themselves in an effort to flatten the curve. It’s not surprising, then, that lotería is among the games people have turned to. In fact, nostalgia can be a powerful coping mechanism, studies posit.
And yet, nostalgia is just one reason why Ruiz thinks there’s been an increase in visibility. She notes that today, there is also a greater Latinx and Hispanic population in the U.S., many of whom grew up playing the game. (Census data estimates that the U.S. Hispanic population reached a record 59.9 million in 2018.) The game is still sold at mercados in the U.S. and Mexico, and there is also this little thing called the internet.
You could say a combination of all three is how Alfaro ended up making Millennial Lotería. In 2017, the 31-year-old creative director found his old lotería games while visiting his family in Guatemala. He told OprahMag.com that he felt nostalgic, yes, but he also thought some of the traditional card concepts were outdated. This was around the time the #MeToo movement was starting to gain traction, and a card like “La Dama” (“The Lady” or “The Lady in Waiting”) felt “so reductive for the time that we were going through.” Since then he’s reimagined that card to be “La Feminist,” plus several others to illustrate concepts and issues that millennials can better relate to.
“Tinder dates? That’s a card,” Alfaro said. “Technology is a big one, like hashtags. I think a lot about issues that affect me as an immigrant, too; coming to America and how hard it was to navigate the system.”
One such card was “La Border Wall.” He didn’t want people to think that he was supporting it, so he drew a ladder to show that he was overcoming it. It’s important to him to be as authentic and honest as possible, he said, especially when targeting millennials, who he believes are good at detecting bull.
So far, it’s been a huge success: Millennial Lotería has sold over 60,000 copies and is currently a number one best-seller on Amazon. An Instagram filter that randomly selects cards in Alfaro’s game has received 1.3 million impressions so far. And in addition to the pandemic-related cards he’s been posting on Instagram (as well as hosting live lotería games), he has plans to release a new version, the Shiny AF edition. It will introduce some cards while phasing out others, and each one will look more holographic and glitter-y. “It’s like if Lotería and Lady Gaga had a baby,” he says.
You can see the enthusiasm for Alfaro’s game in his Instagram comments, especially among young people who are frequently tagging their friends, asking for prints and even more variations of the cards. But under a recent post for “El TikTok,” there was a comment that asked him why he was ruining the game. Alfaro said he’s gotten this kind of pushback from “boomers” before, but, as was the case with the border wall, he meets these situations with humor. “If it makes them upset, it’s going to make Latino millennials more excited. It’s not just my abuelo’s game. It’s mine.”
It’s not just that lotería has become more visible in the U.S., but it’s also become more accessible—across platforms and generations. It’s a game, but the fact that the cards are in Spanish also makes it a learning tool. The cards are typically presented with a short verse or riddle while playing, so it promotes philosophical thinking and perspective, too. As artists like Ruiz and Alfaro continue to reimagine the decks that they grew up with, it will make that intellectual social commentary more relevant and impactful. Lotería checks a lot of boxes, and, as Yvette Benavides wrote in her Creative Nonfiction: Issue #72 last year, it’s life-giving.
Here is a little more about the history of the game, and how you can play yourself.
Who invented lotería?
As Amherst college professor Ilan Stavans explains in his 2003 paper, “¡Lotería! or, The Ritual of Chance,” the game has a complex history. It originated in Italy during the 15th century—the Italian word is “lotto”—before it made its way to “New Spain,” the name for modern Mexico at the time, in 1769. King Charles III of Spain established “la lotería nacional,” which started out as a hobby for the elite before traveling “ferias” or fairs were introduced for the masses to come and play.
In 1887, French entrepreneur Don Clemente Jacques published the “Don Clemente Gallo” version of the game with ten boards and 80 cards, including “un naipe” or a joker, according to Stavans. These games would be included in care packages for soldiers at the time, but it wasn’t until they returned home and played the game with their families that it really become popular.
Modern decks now include Spanish names for each illustration, plus fewer cards, but Jacques’ version still remains one of the most recognizable to date.
What do lotería cards mean?
There is a randomness to the cards, but traditionally, each has been a window into Mexican history and culture: “El Bandolón” (“The Mandolin”), “El Nopal” (“Prickly Pear Cactus”), and “La Muerte” (“The Death”), the latter of which is among Ruiz’s favorites. For the Google Doodle, she reimagined some other classic cards, like “El Sol” (“The Sun”), “La Luna (“The Moon”), and “El Pajaro” (The Bird). She was inspired by the traditional illustration, but she did take some liberties when drawing (including a new card for “El Guacamole”), especially for the sun and moon. “The original look more serious and kind of scary, so the ones I did were happier,” she said. “More joyful.”
More than 100 years after the game came to Mexico, la lotería is reaching new audiences in the U.S. and providing both distraction and comfort in trying times.
Can You Win The Lotería?
Edited by Oliver Roeder
Illustration by Guillaume Kurkdjian
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,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!
“> 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.
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?
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?
Solution to last week’s Riddler Express
Congratulations to 👏 Adam Martin-Schwarze 👏 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 👏 Curtis Bennett 👏 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 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
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 or , so it is concave up. The function for B and C equals 0 when or , 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 (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 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 , 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 = 0.3333, 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!
Welcome to The Riddler.