I know beer and math are not topics that are discussed together often, but I thought it would be fun to start this offbeat series on trying to mend that gap a bit and bring in some neat concepts and ideas.
This first post will use a recent beer to launch into the topic. Evil Twin and Omnipollo collaborated on a beer they called Russian Roulette that had a twist. There were actually two different beers brewed, an IPA and a Black IPA, and idea is they were both bottled in the same bottle, with the same label. This meant that you did not know which one you had until you popped the cap. This made me think of a sort of probability problem, not dissimilar to the Martingale betting system.
In the Martingale system, a player first choses a bet that has a near 50-50 chance of success, such as red/black on a roulette table. The strategy begins by betting a small amount, and then with each successive loss doubling their bet so that an eventual win will recover all of their losses and net them a profit. However, as anyone who has ever tried this system will eventually find out, there are problems that occur. More specifically, an event that has a 50-50 chance, does not guarantee that a event is guaranteed to occur. If you need proof for yourself, see how many times you can flip a tails on a coin without a heads coming up. Now, think about having to double your bet for each of those tails. Unless you have an infinite wallet (in which case, can you take me out for some drinks?), you will eventually run yourself bankrupt.
Getting back to beer here, this idea tangentially came to mind when I was thinking about picking up some bottles of the Russian Roulette. More specifically, I was thinking “How many bottles would I have to buy to guarantee that I get to try both of these beers?” The answer to that question is of course an infinite number of bottles(and no, you cannot really tell which one it is by holding i up to the light…). However, you can use statistics to answer the similar question of “How many bottles would I have to by to give myself a 75% chance of getting to try both beers?” Here is where the math comes in. Skipping over some of the details, the buying of these beers (and the later drinking of them!) turns out to be what are called Bernoulli trials, and we can use that idea to answer that question. I will put some wikipedia links at the end to fill in the details, but the results turn out to:
To be 50% sure I will get both beers, I need to buy 2 bottles.
To be 75% sure I will get both beers, I need to buy 3 bottles.
To be 90% sure I will get both beers, I need to buy 5 bottles.
And so on. As you can see, if you want to increase you chances of getting both beers, you need to buy quite a few bottles. And at around $13 a pop, even though I wanted to get both, the quick math I did above quickly justified just buying one and dealing with the results. And spoiler alert, mine was a Black IPA.
Links for those looking for some more information:
http://en.wikipedia.org/wiki/Martingale_(betting_system)
http://en.wikipedia.org/wiki/Bernoulli_trial
