M&Ms in a Jar
May 22, 2009
I recently found myself at a break-room style work gathering that featured a popular party game: guess the number of objects in a clear container. There are many variations of this game, perhaps the most common is to guess the number of M&Ms in a jar.
I'm the type of person that likes to know how games are played and whether there are any optimal strategies. I submitted my guess. I lost. I wanted to know what I could have done to improve my chance of winning.
M&Ms in a jar is such a frustrating game because it seems too simple, and yet the ability to guess accurately is incredibly difficult. With enough players, the range of guesses can be pretty extreme. When I played this game I saw people attempt to utilize a few interesting strategies. One person counted the number of candy pieces on the edge of the jar and used the equation for the volume of a cylinder to calculate the total M&Ms. But that doesn't really work if you don't know the dimensions of the jar or if you don't count the M&Ms on the edge accurately. Another person tried to weigh the the jar in one hand an a known number of M&Ms in another; but that doesn't work if you don't know the weight of the jar itself. Often times the winner of this game seems to be someone who simply got lucky.
In order to generate an edge, you have to know the expected value of playing. Let's say the winner gets a $50 Starbucks gift card. Let's also say that there are 50 players. Further, we'll assume that no player is more skilled than another, so the winner is basically chosen randomly. Because there isn't any cost to play, the expected value would look like:
E(playing) = (probability of winning)(payoff)
E(p) = (1/50)($50 Starbucks card)
E(p) = $1.00 in Starbucks or about half of a cup of coffee
So if an individual can't really become "skilled" at this game, is there anyway to improve the expected value of playing? At first I didn't think there was, but then it hit me. Players can generate an edge by tapping into the wisdom of crowds.
Now imagine that ten players secretly form a consortium in a back room. Each member writes a guess on a piece of paper and then they calculate the mean value of the guesses. The consortium submit their guess under a single name and agree to split the prize evenly if they win.
What does the expected value for these players look like? Well, it depends. Because the consortium will have a better chance of winning than individual players (essentially, it will be more "skilled" at the game and the more players in the consortium, the higher the skill level) the expected value will be higher than $1. But it also depends on how many people are playing the game, how many people agree to pool their guesses and split the prize, and whether or not any other groups have formed. If all 50 players agree to work together and split the prize, then the expected value returns to only $1 and the game isn't any fun at all. My guess is that there is a normally distributed type curve that can take these variables into account to determine an optimal strategy. If someone is willing to crunch the numbers and figure out what it is (or if someone already has), I will be impressed.
I'm the type of person that likes to know how games are played and whether there are any optimal strategies. I submitted my guess. I lost. I wanted to know what I could have done to improve my chance of winning.
M&Ms in a jar is such a frustrating game because it seems too simple, and yet the ability to guess accurately is incredibly difficult. With enough players, the range of guesses can be pretty extreme. When I played this game I saw people attempt to utilize a few interesting strategies. One person counted the number of candy pieces on the edge of the jar and used the equation for the volume of a cylinder to calculate the total M&Ms. But that doesn't really work if you don't know the dimensions of the jar or if you don't count the M&Ms on the edge accurately. Another person tried to weigh the the jar in one hand an a known number of M&Ms in another; but that doesn't work if you don't know the weight of the jar itself. Often times the winner of this game seems to be someone who simply got lucky.
In order to generate an edge, you have to know the expected value of playing. Let's say the winner gets a $50 Starbucks gift card. Let's also say that there are 50 players. Further, we'll assume that no player is more skilled than another, so the winner is basically chosen randomly. Because there isn't any cost to play, the expected value would look like:
E(playing) = (probability of winning)(payoff)
E(p) = (1/50)($50 Starbucks card)
E(p) = $1.00 in Starbucks or about half of a cup of coffee
So if an individual can't really become "skilled" at this game, is there anyway to improve the expected value of playing? At first I didn't think there was, but then it hit me. Players can generate an edge by tapping into the wisdom of crowds.
Now imagine that ten players secretly form a consortium in a back room. Each member writes a guess on a piece of paper and then they calculate the mean value of the guesses. The consortium submit their guess under a single name and agree to split the prize evenly if they win.
What does the expected value for these players look like? Well, it depends. Because the consortium will have a better chance of winning than individual players (essentially, it will be more "skilled" at the game and the more players in the consortium, the higher the skill level) the expected value will be higher than $1. But it also depends on how many people are playing the game, how many people agree to pool their guesses and split the prize, and whether or not any other groups have formed. If all 50 players agree to work together and split the prize, then the expected value returns to only $1 and the game isn't any fun at all. My guess is that there is a normally distributed type curve that can take these variables into account to determine an optimal strategy. If someone is willing to crunch the numbers and figure out what it is (or if someone already has), I will be impressed.
