Steady Eddy
Legend
Any hockey fans out there? Math nerds? Both? Great!
Consider a team who's average score against an opponent is 3-3. That's a total of 6 goals per 60 minutes, or .1 goal per minute. The probability that a goal will be scored in "t" minutes is given by (1 - e^-.1t)
Since our team scores half of the goals, the probability that our team will score the next goal is .5(1 - e^-.1t)
If you're one goal behind and there's plenty of time, the probability of tieing the score (maybe only a temporary tie) is close to .5, but as time starts running out, your probability of doing this gets smaller.
Suppose you know that with your goalie out of the game, and an extra attacker replacing him you'd lose 15 - 45 on average. That's a total of 60 goals in 60 minutes, or 1 goal per minute. The probability that a goal will be scored in "t" minutes is (1 - e^-t). But since now our team is only scoring on fourth of the goals, the chance that we score the next goal is .25(1 - e^-t). If a lot of time is left, our chance of tieing the game is only .25, but if time is short, this expression will give higher values than the other.
To find the optimal time to yank the goalie, let y = .25(1 - e^-t) - .5(1 - e^-.1t), find y', and set it equal to zero. You'll get t = ln(5)/.9, or about 1.7 minutes.
BTW, this doesn't depend on the parameters I used. They are only for illustration. They can be replaced with variables and you can still derive the optimal time in the manner I demonstrated.
BTW II, I've looked on the net and haven't seen a formula. I've seen empirical data, instead. But that data doesn't contradict this formula.
Consider a team who's average score against an opponent is 3-3. That's a total of 6 goals per 60 minutes, or .1 goal per minute. The probability that a goal will be scored in "t" minutes is given by (1 - e^-.1t)
Since our team scores half of the goals, the probability that our team will score the next goal is .5(1 - e^-.1t)
If you're one goal behind and there's plenty of time, the probability of tieing the score (maybe only a temporary tie) is close to .5, but as time starts running out, your probability of doing this gets smaller.
Suppose you know that with your goalie out of the game, and an extra attacker replacing him you'd lose 15 - 45 on average. That's a total of 60 goals in 60 minutes, or 1 goal per minute. The probability that a goal will be scored in "t" minutes is (1 - e^-t). But since now our team is only scoring on fourth of the goals, the chance that we score the next goal is .25(1 - e^-t). If a lot of time is left, our chance of tieing the game is only .25, but if time is short, this expression will give higher values than the other.
To find the optimal time to yank the goalie, let y = .25(1 - e^-t) - .5(1 - e^-.1t), find y', and set it equal to zero. You'll get t = ln(5)/.9, or about 1.7 minutes.
BTW, this doesn't depend on the parameters I used. They are only for illustration. They can be replaced with variables and you can still derive the optimal time in the manner I demonstrated.
BTW II, I've looked on the net and haven't seen a formula. I've seen empirical data, instead. But that data doesn't contradict this formula.