If you haven’t seen already, I have moved over to Skyd. I am going to keep this space open for little odds and ends and things that don’t quite fit or are a bit less formal than I intend the Skyd writing to be.
Here is the nerdy challenge: I am looking for a function where the input is the chance of a possession ending in a score (p). Obviously, 1 – p is the chance that a possession ends in a turnover. Then output is the probability of each of the six possible endings to a game of mini. The three ‘wins’ being: 3~1, 3~0, 3~-1 and the three ‘losses’: 1~-2, 0~-2 and -1~-2. Please ignore the Golden Goal variation for right now.
I brute forced it for p=0.5. This had the nice property that 1-p=0.5 as well, so the outcome tree was essentially symmetric and the probabilities easier to manage because I just used negative powers of 2. I quit after 5 generations and then estimated the remaining 20% of the probability based on the observed trends. Here is what I got:
3~1: 4.4%
3~0: 23.6%
3~-1: 18.2%
Winning: 46.2%
1~-2: 9.2%
0~-2: 18.6%
-1~-2: 25.6%
Losing: 53.4%
This was a bit surprising. I expected the chance of losing to be much higher. My intuition told me that it was two steps to lose and three steps to win, so it should divide out on a two-thirds, one-third ratio. But it turns out that you always have a chance to win, but you don’t always have a chance to lose.
Help? Ideas? There has to be an easier way.
