In this position, if we call the average number of movies remaining E, half the time we get a square leaving the same situation with 1 more move, and half the time we get a triangle that we can place but would end the game. Thus E = ½(E+1) + ½(1), or E=2. |

In this position, half the time we get a square which ends the game immediately, and half the time we get a triangle that we can place upside-down to leave the previous example. Thus E = ½(0) + ½(1+2), or E=3/2. |

In this position, half the time we get a square which leads to our first example, and half the time we get a triangle that leads to our second example. Thus E = ½(2+1) + ½(3/2+1), or E=11/4. |

Can you find how long the game will last on average from the positions below, which are enough to completely solve the width 2 cases? How about positions of width 3? Warning: some of them are tricky because it is unclear what the best placement of a piece is. And there are infinite chains you will have to deal with for seemingly simple positions.

end | ||||

↑ | ||||

3/2=1.5 | ← | 11/4=2.75 | ← | 37/8=4.625 |

↓ | ↙ | ↑ | ↕ | |

2 | 19/8=2.375 | ← | 9/2=4.5 | |

↺ | ↘ | ↓ | ||

0 |

Here are my solutions for expected lifetimes for width 3 positions. Evert Stenlund corrected some of my calculations.

2 | ← | 223/36=6.194+ | ← | 295/36=8.194+ (ES) | ← | 463/48=9.645+ (ES) | |||||

↙ | ↙ | ↙ | ↖ | ||||||||

← | 13/2=6.5 | ← | 151/18=8.388+ | ← | 295/36=8.194+ | ← | 655/72=9.097+ | ← | 10949/1008=10.862+ (ES) | ← | 709/63=11.253+ (ES) |

↓ | ↓ | ↓ | ↓ | ↓ | ↗ | ||||||

← | 51/8=6.375 | ← | 149/18=8.277+ | 6 | ← | 8 | ← | 1339/126=10.626+ (ES) | |||

↓ | ↓ | ↺ | ↘ | ↺ | |||||||

these lead to a collection of towers where squares are put on the left and triangles are put on the right | 2 | ← | 4 |

If you can extend any of these results, please e-mail me. Click here to go back to Math Magic. Last updated 11/23/18.