People should Speak Up if they want me to Shut Up about Toss Up. Thane wrote:
On the "original" rules pointed out by Michael Kleber: suppose two players both have 99 points. I think both players would want to "pass", since it's got to be better to let the other guy cross 100 first, and see where he lands up before rolling your own dice? Or maybe not?
That appealed to me at first, but upon sober reflection I don't think it's true. The game doesn't end just because I cross the 100 line during my turn; it only ends if I *stop rolling* when over. So if it's 99 to 99 and my turn, I think I'd want to roll aggressively: if I bust, it's the other guy's turn, but if my aggressive chance-taking pays off, I'll stop with a big score and it's very likely I'll win. Of course, if I'm too aggressive, the other guy will likely get his chance to do the same, and I've lost the advantage of getting the first bite at the apple. So I should moderate my aggression. Here, let's play a simpler game, which I'll call "Tail". The goal of Tail is to "earn" the higher number in [0,1]. The two players take turns naming arbitrary numbers p in [0,1], and then picking a point x uniformly in [0,1]; if you pick an x which is larger than your chosen p, then you have earned p. Once you earn a number, the other player gets just one chance to earn a higher one, and then the game ends. As long as neither player has earned a number, play passes back and forth. So the higher a p you pick, the less likely that you will earn it, but if you do then the more likely it will result in your winning. I think this does a decent job modelling 99-99 Toss Up, where p stands in for "how likely is it that someone playing the strategy 'N points or bust' will bust?" Not quite, because in Toss Up you can aim for N points and end up over, but it's pretty close. So what p should you pick in Tail? Well, there is some optimal p, and we can assume both players will use it, which simplifies things. I go first, earn p with probability 1-p = q, and if I do, my opponent tries to earn p (well, p+epsilon) and fails with probability p. So I win with probability q p and lose with probability p p on my turn. If I fail, then my opponent does the same. At the end of our two turns, my probability of having won is qp + pqq, and my probability of having lost is qq + pqp; if neither of these two has happened (probability pp), we're right back where we started. Overall, then, I win with probability (qp+pqq)/(1-pp), which has a local maximum at... (<scribble>)... p = sqrt(3) - 1 =~ .732. The probability of winning is approx 0.535898. For Toss Up, it's easy to inductively calculate the probability that the strategy 'N or bust' leads to a bust. The probability that '39 or bust' fails is .711, while '40 or bust' goes sour with probability .763. Plugging both into (qp+pqq)/(1-pp), it looks like '39 or bust' is better. So if you're in a game of Toss Up tied 99-99, I think you should roll until you gain 39 or more points. Against an optimal player (who does the same thing), whoever goes first will win with probability 0.535639. Of course, if your opponent is too tame, you're even more likely to win! --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.