Back in the 90's I was playing PBM games. In the zine "Novelty" by David Tittle, there was a game called "Beat the Boss".

RULES

Players are employees trying to arrive as late as possible to the office without being caught. You enter a time between 9:00 and 9:59. So does The Boss. If you arrive before the Boss you win the number of minutes after 9:00. Otherwise you lose that same amount.

The Boss wins double the minutes, minus 60 (that's his lunch hour busy filling timesheets...).

Originally, the game is played in 10 or 15 turns ("days") and whoever comes in earliest becomes the Boss for next day. (Ties drawn randomly.)

But for the time being, let's concentrate on a simple version : one day, one Boss and one player.

QUESTION

What is the optimal strategy for each?

(Note: there is no Nash equilibrium in pure strategies; you must resort to mixed strategies. Translation: the optimal strategy is to select your time at random, and the question is "according to what probability distribution?")

The number of minutes after 9:00 that the employee arrived?

The number after 9:00 that the boss arrived?

Something else?

If it's the middle one, then it seems to me that the answer is, "The boss arrives at 9:00, since the boss score is unaffected by the player, so the player also arrives at 9:00 as any other time results in losing points."

If it's not, then what would the player's arrival time have to do with how much time the boss has to fill out timesheets?

Twice the number of minutes that the Boss arrived after 9:00Quote:ThatDonGuyQuestion: what do you mean by, "The Boss wins double the minutes, minus 60"?

The number of minutes after 9:00 that the employee arrived?

The number after 9:00 that the boss arrived?

Something else?

The interaction is in the zero-sum nature of the game. The object of the game is not to maximise your own score, but to end up with MORE POINTS THAN your opponent.Quote:ThatDonGuyIf it's the middle one, then it seems to me that the answer is, "The boss arrives at 9:00, since the boss score is unaffected by the player, so the player also arrives at 9:00 as any other time results in losing points."

If the Employee plays 9:58 and the Boss plays 9:59, they both get 58 points. Game value = 0. There are smarter things to do: e.g. the Boss arriving on 9:58 makes him 56 pts and -58 for the Employee. Net gain = 114.

Forget it, it's just window-dressing to justify the minus 60. Make it 30 or 45, it shouldn't change the optimal play, as long as it is not dependent on the Employee's time.Quote:ThatDonGuyIf it's not, then what would the player's arrival time have to do with how much time the boss has to fill out timesheets?

Best thing Employee can do is tie by coming in 9.58, everything else is a loss

So if we are talking one day, Boss should do just that, hoping Employee does not know optimal strategy. Worst case is a tie.

Now, Boss might assume that Employee anticipates the tie strategy and will come in 9.58 accordingly.

So Boss might come in 9.58 instead to turn that tie option into a win.

This opens an Employee win option: 9.57

And thus the metagame goes on and on...

These are my thoughts so far....

Let P be the number of minutes after 9:00 that the employee, or "player", arrives, and B the number of minutes after 9:00 that the boss arrives

If B < 20, then P wins if P < 60 - 2B

If 20 < B < 30, then P wins if P < B

If B > 30, then P wins if 2B - 60 < P < B (only 2B - 60 is an equilibrium point)

On an XY graph where one axis is P and the other is B, the total result area is 60 x 60 = 3600, and the area where P wins is 1500.

Choose a random real number X in [0, 1500]

If X < 800, then the P value where the winning area = X is P = 30 - (900 - X)

^{1/2}

If 800 <= X <= 1500 OOPS - that should be 1050, then the P value where the winning area = X is P = (1200 - 2X)

^{1/2}

If X > 1050, then the P value where the winning area = X is P = 60 - (3000 - 2P)

^{1/2}

However, my game theory is not as good as I would like it to be, so who knows how accurate that really is.

You both proceeded with a value function {+1,-1}, i.e. win or lose. This is totally in line with the game as presented.

I was busy on a value function ScoreBoss - ScoreWorker. This is because I anticipated going further with the repeated game, where scores are added.

Clearly the answer differs from one to the other.

Don:

I do not clearly see what your X represents. What should the player do?

Quote:kubikulannInteresting.

You both proceeded with a value function {+1,-1}, i.e. win or lose. This is totally in line with the game as presented.

I was busy on a value function ScoreBoss - ScoreWorker. This is because I anticipated going further with the repeated game, where scores are added.

That's because you said, "The object of the game is not to maximise your own score, but to end up with MORE POINTS THAN your opponent."

Quote:kubikulannDon:

I do not clearly see what your X represents. What should the player do?

Select a random real number X in the range 0 <= X <= 1500.

The number of minutes to arrive after 9:00 is P; apply the appropriate equation for P depending on whether X < 800, X is between 800 and 1050, or X > 1050.

For example, if the player chooses X = 260, then arrive P = 30 - (900 - 260)

^{1/2}= 10 minutes after 9:00.

Nice.

A next step would be to compute the value of the game, as to determine the best amount of boss penalty such as to make the game equilibrated. I remember players thought 60 was too much.

Quote:kubikulannI see.

Nice.

A next step would be to compute the value of the game, as to determine the best amount of boss penalty such as to make the game equilibrated. I remember players thought 60 was too much.

Those players didn't get the game. It must be 60, otherwise Boss coming in 9.59 is an automatic win. Maybe it should be more to eliminate the sure tie option for Boss.

Quote:CanyoneroBoss comes in 9.59 - 58 points

Best thing Employee can do is tie by coming in 9.58, everything else is a loss

You're making the assumption that the employee can't arrive at 9:58:30, scoring 58.5 points.

Of course, in that case, the boss arrives at 10:00, scoring the maximum 60 points - and going by a strict reading of the rules, "tie goes to the boss", so the player scores either less than 60 (by arriving before 10:00) or -60 (by arriving at 10:00).

Boss coming at 9:59 cannot make a Nash equilibrium. Employee would come at 9:58, to which the Boss' best response is NOT 9:59.

You're right.Quote:CanyoneroThose players didn't get the game. It must be 60, otherwise Boss coming in 9.59 is an automatic win. Maybe it should be more to eliminate the sure tie option for Boss.

According to my calculations, the value of the game (before penalty) is around +80 for the Boss.