The Riddler: Competitive Coin Flipping

One of the things I do every Friday evening or Saturday morning is check the math puzzle column that FiveThirtyEight runs, called the Riddler. It ranges in difficulty from merely ridiculous to something very near impossible, as far as I'm concerned, but then it's been a while since I did any kind of detailed study of math.

(This is also why I'm not doing a weekly series about this - because about half the time I read it and immediately decide that I'm not going to spend my time slamming my forehead into whatever wall they've come up with.)

But, the express puzzle for this week actually seems interestingly doable, so here we go!

For this puzzle, I decided to plot out the different ways this game could develop. Since it's based on coin flips, there actually aren't that many of them - 4 possible results for the first flip from each team, and then 4 more options for the second, for a total of 16. While I would expect a fair number of games to require more than two flips, any additional game states will be identical to the conditions at the start or following one of those first flips.

For example, one of the options for the first flip - both teams getting tails - leaves the situation for subsequent flips no different than what we had at the start of the game. Another option - tails for the Blue team, heads for the Red team - doesn't allow either team to complete their sequence, but can start Red's sequence; if it shows up anywhere, I know the game is in that particular state for the next flip.

Plotting out all the states fairly quickly leads to the conclusion that Red has an advantage. Assuming the coins are fair, there's a 25% chance each of ending up in a state where either Red or Blue can win on the next flip and the other team can't. But on the second flip, even if Blue has the advantage, they can't hold on to it; they'll either win, hand the advantage over to Red, or reset the game. If Red got the advantage, they can hold on to it; they will either win, hold on to the advantage, or put both teams on equal footing. And if neither team got an advantage on the first flip, that's either a reset or - if both teams got heads - then there are equal chances of Blue winning, Red winning, a reset, or Red gaining the advantage.

Simply put, there are a lot more ways for Red to gain an advantage and to hold on to it. A better mathematician than me could probably come up with the exact odds, but this looks clear enough for me to conclude that Red is a safer bet. (At least, assuming I didn't screw something up.)

And now I can start working on the longer classic puzzle... yeah, don't expect me to be posting about that one any time soon.