## Trying to find math inside everything else

### We Didn’t Playtest This At All

Yesterday was my best friend’s birthday and his wife got him the game We Didn’t Playtest This At All, which is a very silly game that was tons of fun. (We probably played it about 15 times.) The point of the game is to win or, barring that, to make everyone else lose. And that’s all the rules there are, other than Draw 1, Play 1. Everything else is in the cards.

One set of cards in the game has players all throw out 1 to 5 fingers on the count of three:

Since you don’t know what card they are playing, even and odd really don’t matter. But winning on a prime…that’s interesting.

As I was leaving, I started to wonder if there was a best number you could throw out to maximize your chances of winnings (or, alternately, stopping to person who played the card from winning). Talking about it with another math teacher who was there, I hypothesized that, because of the lower density of prime numbers as numbers get larger, you’d want to throw smaller numbers to increase your chances of getting a prime.

But, of course, I couldn’t just leave that conjecture. I had to test it! For the purposes of this, I assumed all other players besides yourself throw out a random number of fingers, essentially becoming 5-sided dice.

It’s pretty simple to compute for two players:

• If I throw out a 1, it’ll be prime if my opponent throws 1, 2, or 4.
• If I throw 2, she needs to throw 1, 3, or 5.
• If I throw 3, she needs to throw 2 or 4.
• If I throw 4, she needs to throw a 1 or 3.
• If I throw a 5, she needs to throw a 2.

This supports my hypothesis: throwing a 1 or 2 increase the odds of a prime, and a 5 radically decreases them. (Of course, then we can get all game theoretical — if I know you’re gonna throw 5, I should throw 2. But then, if you know that, you should throw 4, etc.)

What about for more than 2 players? The game box says we can have up to 10. I worked it out somewhat in my notebook on my train ride home, but then I had the power of Excel. (It actually took me longer than I would like to admit to re-figure out how to find the probabilities of, say, getting a total of 12 when 3 people throw out. I was counting up all the possibilities for a while until I realized the recursive method for calculating those probabilities. And if Wolfram-Alpha hadn’t been so hard to use in this regard, I might not have figured it out myself.)

On the left are the probabilities that you opponents’ total will be a certain number. On the right is the number of ways you can get prime if you throw out that number.

For three players, 1 is still the champ is terms of getting you a prime, but surprisingly, 5 is second place! What had been the worst number to throw out to get primes for 2 players is now the second best with 3 players. And for 4 players, 1 and 5 are actually the worst (though only slightly), with 2, 3, and 4 coming out on top. But at this point, it’s pretty balanced. 5 players is almost equally likely no matter what you throw. It’s almost as if they playtested this?

But now, the pattern emerges.

When I extended to 6 or 7 players, though, it became clear that 1 really was the true winner and 5 the worst. Once we were out of the weeds of the prime-heavy teens, the hypothesis seems more true. (It also holds for 8 players.) Of course, I haven’t proven that it will always be true for 6+ players…but I leave that as an exercise to the reader.

### How Many Subway Transfers Do I Have to Make?

I was talking with Sam Shah and had the following exchange:

Of course, after I said that…I had to find out if it was actually true. So I pulled up a map of the subway system and started analyzing.

I realized the best way to analyze the system would be to create a matrix of connections: if I can transfer directly between two lines (or the walking transfer from 59th St/Lex to 63rd/Lex, since you don’t need to pay again), then put a 1 and make the cell green. If not, put a 0 and leave the cell white. That’ll show a chart of all the places you can get to on a single transfer.

Most of the lines have direct transfers, with a few being tricky. Breakout stars are the A, which connects with all but the 6, and the F, N, and R, which only miss some or all of the shuttles. Particularly difficult train lines are the G, the J, and the 6.

So this answers the question of where you can get with only 1 transfer. But what about two transfers? For that, we can multiply this matrix by itself. This is the result:

What do these numbers mean? Well, to explain, let’s look at the G –> 6, which I have highlighted in blue. The number there is 8. This means that there are 8 ways to get from the G to the 6 with two transfers:

G –> 7 –> 6
G –> D –> 6
G –> E –> 6
G –> F –> 6
G –> L –> 6
G –> M –> 6
G –> N –> 6
G –> R –> 6

So this chart shows that you can get from any line to any other with at most two transfers*, with one exception: the Rockaway Shuttle to the 6. However! Those stops aren’t solely serviced by the S. (The only stop in the system solely serviced by an S train is Park Pl, on the Franklin Ave Shuttle.)

Because of that, I can amend my statement to the following, which I have proven true:

During rush hour, you can get from any stop on the subway to any other with a maximum of two transfers.

But then, that gets me wondering further…this chart was just made if the connections exist, but they weren’t time-sensitive. For example, the M does not run at my stop at nights or on weekends. How would that change this chart? Especially when you consider that the E, which does not normally go to my stop, DOES at night. I leave that problem open.

———————–

```* Of course, fewer transfers doesn't always mean better. If I wanted to get
from Astoria to Greenpoint, sure, I could take the N to the G, but that
requires going all the way through Manhattan, way down into Brooklyn, and
then back up. Instead, a quick hop from the N to the 7 to the G is much
more sensible, even if it is an extra transfer.

Excel File - Subway Analysis```