This week’s problem of calculating the probability of the last pin being knocked down in a rhombus of bowling pins is difficult or impossible to calculate exactly without simulation for large N. I started by using a simplified probability model which turns out to have a lot of the same features as the correct simulated calculation and turned out to be what looks like an an upper bound.

We start by calculating a 2D grid with valid entries where the N^2 pins are located and each entry represents the probability of a pin falling down. The combined probability P of a pin being knocked down by either of its one or two predecessors, conditional on either of them having fallen is:

P(pin falling) = 1 – [ (1 – p*P(pin 1 above falling)) * (1 – p*P(pin 2 above falling)) ]

If pin 2 does not exist, as in the edge cases, P(pin 2 above falling) = 0 and the calculation can be identical for all the valid pin entries in a square grid, with entries without pins set to 0. **This calculation is not exactly correct as the probabilities of the nodes in the 2D multiply connected grid are not really independent. **

Propagating these probabilities for n=5, with the first pin going down with certainty, we see some interesting behavior already:

There are a few interesting things about the above conditional probability calculation:

- At probabilities around .5 and below it appears that the last pin will show a rapidly decreasing probability of being knocked down.
- It is possible for a pin to have a greater probability of being knocked down than its immediate two predecessors.
- There is a condition where the probability of a pin falling equals the two pin probabilities above it and the grid will tend to “lock-up” at a constant probability in the middle. This condition can be calculated by setting all P’s in the first equation above equal, and is:

P = (2p-1)/ p^2

By calculating much larger grids and observing the results it appears that for large N and p > .5 this happens to all grids in the center region as the edge probabilities become irrelevant, and this probability ends up being the probability of the last pin falling down. For p <=.5 the probability of the last pin falling down appears to go to zero for large N.

Below are the last pin falling probabilities estimated with the stability condition and a grid calculation at N= 500. Again, remember these are not exact results but probably more like **upper bounds**:

p(pin falls) | (2p-1)/p^2 | P(last pin falls): Grid N = 500 |

0.500 | 0.000 | 0.003 |

0.600 | 0.556 | 0.556 |

0.700 | 0.816 | 0.816 |

0.800 | 0.938 | 0.938 |

0.900 | 0.988 | 0.988 |

1.000 | 1.000 | 1.000 |

Focusing in on the critical p=.50 at larger N:

p(pin falls) | (1-2p)/p^2 | P(last pin falls): Grid N= 5000 | P(last pin falls): Grid N = 10000 |

0.495 | n/a | 0.000000 | |

0.496 | n/a | 0.000000 | |

0.497 | n/a | 0.000000 | |

0.498 | n/a | 0.000000 | |

0.499 | n/a | 0.000000 | |

0.500 | 0.000000 | 0.002526 | 0.000134 |

0.501 | 0.007968 | 0.007551 | 0.007968 |

0.502 | 0.015873 | 0.015157 | 0.015872 |

0.503 | 0.023715 | 0.023362 | 0.023715 |

0.504 | 0.031494 | 0.031370 | 0.031494 |

And we note that crucially the p = .500 values are decreasing with increasing N, while values above p =.500 are still stabilizing and actually increasing at this stage. So this imperfect upper bound calculation shows a sharp transition point at .p= .500.

To get exact values we will need to run a simulation. Note we will need 2 random numbers per pin to account for the separate probabilities of striking the two pins underneath.

The simulation behaves a bit differently than our upper-bound calculation and it appears to show a transition point appears at around .600 or so (this sim is N=50 at 1000 reps):

Zooming in:

And here are results with N=100 and 5,000 reps:

p(pin falls) | P(last pin falls): Simulation Grid, N= 100, 5,000 Reps |

.500 | 0.000 |

.600 | 0.004 |

.700 | 0.573 |

.800 | 0.870 |

.900 | 0.975 |

1.000 | 1.000 |

It is difficult to know how stable these results will be with large N and e.g. how sharp the stability cut-off is. The assumption is it gets sharper as N gets larger judging by our simplified probability model results.

Finally, I ran one simulation for p = .700 at N=200 and 10,000 reps and got a result of ** P= .579 ,** so I guess that is the end of it for now and we decide that Fosse will win this Battle of the Bowlers 57.9% of the time with the remaining 42.1% being draws. Probably.

Code for the probability calculation:

Somewhat inefficient code for the simulation: