I offer this Toss Up scoresheet, in which I'm proud to say I figure as "T", the winner of a three-way game against my two sons: http://www.flickr.com/photos/thane/1263170453/ I play a conservative game, never rolling 3 or fewer dice, and when in the lead, never fewer than 4. After I have the lead, the boys feel compelled to "catch up" after their rash bankruptcies and are then led into repeated subsequent bankruptcies. In most games I plod inexorably to victory. * * 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? On 8/27/07, Schroeppel, Richard <rschroe@sandia.gov> wrote:

The original paper wants x to be a vector, A to be a matrix, and b,b' to be vectors. Two obvious methods of solving it are (a) assume Left or Right for each component of Max, solve the resulting linear system, and select out the combinations which match the Left/Right choices; (b) assume a starting guess for the vector x, and iterate, hoping for convergence.

I actually did something similar once-upon-a-time, solving 2-checker Backgammon. I assigned upper and lower bounds for the probability of winning, to each Backgammon position. These were initialized to 1.0 and 0.0, and then a routine calculated updated probabilities based on the Backgammon rules. I had to leave out the doubling cube (memory constraint). The loop converged very quickly, with upper & lower probabilities being equal or adjacent. But then I faced the problem of making sense of the data, which I never really solved.

Rich

-----Original Message----- From: math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com [mailto:math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com] On Behalf Of Eugene Salamin Sent: Monday, August 27, 2007 1:57 PM To: math-fun Subject: Re: [math-fun] Toss Up Komi

As Neller and Presser say, "There is no known general method for solving equations of the form x = max(Ax+b, A'x+b')."

Assuming A, A', b, b' are fixed given numbers, the graph of y = max(Ax+b, A'x+b') is continuous, piecewise linear, and consists of two segments. Its intersection with y = x consists of 0, 1, 2 or points or a continuum, and these are the solutions.

Gene

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-- Thane Plambeck tplambeck@gmail.com http://www.plambeck.org/ehome.htm

Oh! Looking at Thane's score sheet, I realize I completely misunderstood the basic rule of the game! I was under the impression that if you rolled zero-green-nonzero-red, then you forfeited your *accumulated score for this turn* and passed the dice to the next player. But rereading Thane's original message, it seems like much more is in jeopardy: one bad roll bankrupts your entire score for the game! So choosing to stop doesn't "protect" the points you've accumulated; all it does is resets to ten the number of dice you get to roll (when it's next your turn). My misinterpretation matches the usual rules for pig and the paradigm of other similar jeopardy dice games like the excellent but long out of print Can't Stop ( http://www.boardgamegeek.com/game/41 ), and the description of the rules at the link I posted earlier. I'd like to see the official rules of Toss Up, just to be sure of the facts this time, but I can't seem to find them on-line.

From the mathematical point of view, it actually makes a big difference. The game is hard to solve because it is potentially loopy: to decide whether to stop or roll again, you need to compare P(win|stop) with P(win|roll) = P(going bankrupt)*P(win|state after bankruptcy) + other terms. In the small-penalty version that I was thinking about, state-after-bankruptcy is not so far from current state; you can do an induction on the total number of "banked" points of both players and only need to solve a handful of piecewise-linear equations each time.

But in Thane's big-penalty version, every single decision depends on the probability associated with a bankrupt state in which your total score has reverted to zero, so there's no stratification to help tame the dependencies. In this case solving formally really is hopeless, and the iterative methods we were talking about earlier probably are your only hope. --Michael Kleber On 8/29/07, Thane Plambeck <tplambeck@gmail.com> wrote:

I offer this Toss Up scoresheet, in which I'm proud to say I figure as "T", the winner of a three-way game against my two sons:

http://www.flickr.com/photos/thane/1263170453/

I play a conservative game, never rolling 3 or fewer dice, and when in the lead, never fewer than 4. After I have the lead, the boys feel compelled to "catch up" after their rash bankruptcies and are then led into repeated subsequent bankruptcies.

In most games I plod inexorably to victory. * * 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?

On 8/27/07, Schroeppel, Richard <rschroe@sandia.gov> wrote:

...

The original paper wants x to be a vector, A to be a matrix, and b,b' to be vectors. Two obvious methods of solving it are (a) assume Left or Right for each component of Max, solve the resulting linear system, and select out the combinations which match the Left/Right choices; (b) assume a starting guess for the vector x, and iterate, hoping for convergence.

I actually did something similar once-upon-a-time, solving 2-checker Backgammon. I assigned upper and lower bounds for the probability of winning, to each Backgammon position. These were initialized to 1.0 and 0.0, and then a routine calculated updated probabilities based on the Backgammon rules. I had to leave out the doubling cube (memory constraint). The loop converged very quickly, with upper & lower probabilities being equal or adjacent. But then I faced the problem of making sense of the data, which I never really solved.

Rich

-----Original Message----- From: math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com [mailto:math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com] On Behalf Of Eugene Salamin Sent: Monday, August 27, 2007 1:57 PM To: math-fun Subject: Re: [math-fun] Toss Up Komi

As Neller and Presser say, "There is no known general method for solving equations of the form x = max(Ax+b, A'x+b')."

Assuming A, A', b, b' are fixed given numbers, the graph of y = max(Ax+b, A'x+b') is continuous, piecewise linear, and consists of two segments. Its intersection with y = x consists of 0, 1, 2 or points or a continuum, and these are the solutions.

Gene

________________________________________________________________________ ____________ Got a little couch potato? Check out fun summer activities for kids. http://search.yahoo.com/search?fr=oni_on_mail&p=summer+activities+for+ki ds&cs=bz

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-- Thane Plambeck tplambeck@gmail.com http://www.plambeck.org/ehome.htm

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-- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.

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.

Joshua Zucker wrote, about Toss Up at 99-to-99:

This argument (that going first is good) doesn't make sense.

The second player won't adopt the same strategy as the first: the second player adopts the strategy of stopping when they get at least one point past the first player (or maybe stopping on a tie, if the first player was ambitious enough Knowing how many points the first player earned is a huge advantage. If the first player busts, you have an almost automatic win by just rolling once.

If the first player busts, then the game isn't yet about to end! It becomes the second player's turn, and the score is still 99-to-99, and we're back in the original game state with the players' identities swapped. (Perhaps I wasn't clear that I'm analyzing the original rules, in which busting only loses you the tentative points you've accumulated so far this turn. Points you've "banked" by voluntarily passing the dice are yours forever.) If the first player doesn't bust, then of course player 2 will adopt the strategy "beat player 1 or bust trying." But p2's probability of busting trying is just the same as was p1's of busting before reaching that point value in the first place. --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.

Joshua Zucker wrote:

That hardly seems fair, for player 1 to get an extra turn ...

Quite right, and I guess Thane's original komi question asks just how unfair this really is. It all depends on how quickly the play tends to forget who went first. The fact that player 1 only has a 53%-47% advantage with optimal play from 99-99 makes it much more plausible, to me, that the game is close to fair from the beginning. Actually, I think the ending condition Joshua brought up, where player 2 gets to go again after player 1 crosses 100 but not the other way around, is probably *less* fair, for just the reason he mentioned: player 2 has a much easier time winning, since he could safely stop at 100, while player 1 would need to go much higher out of fear. So "fair ups" isn't so fair after all. (Sort of the equivalent of baseball's home field advantage...) --Michael -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.