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Test OLM Th 12/17/09

Posted By: Jason Lee
Date: Friday, 18 December 2009, at 6:21 a.m.

In Response To: Test OLM Th 12/17/09 (Jason Lee)

My very first inclination was, dude, this is easy, just dump the checkers, you cannot afford to give up 17 fatal shots by running out of there now.

Say we stay and dump checkers in our board with 6/2 6/1. Then next turn, assuming White doesn’t roll doubles and move up, we’re faced with getting squeezed off into a single shot with 61, 62, 63, 64, 65, 54, 53, 52, 43, and squeezed into a double shot with 51, 42, 41. Eighteen numbers leave a direct six shot, for 17 hitters apiece. Two numbers (51), played 18/17 18/13 leaves 22 returns. Two numbers (42) played 18/16 18/14 leaves 23 returns. Two numbers (41) played 18/17 18/14 leaves 23 returns. Therefore, when we return to this similar position next time, we face getting hit 17*18 + 2*22 + 2*23 + 2*23 = 442 times out of 1296.

If we leave now, we get hit 17/36 = 612/1296 of the time, so we’d get hit 170/1296 more by leaving now, which is about 13%.

This is not exactly right, because sometimes White DOES improve by rolling a good set.

I think I can do all that over the board pretty close.

What I don’t know is how much dumping the two checkers costs us in the race. It has to cost a lot, because in the calculation I did above, while 24 numbers leave shots, some of the non-shot leaving numbers force us to dilly around inside AGAIN (32, 31, 21).

Then there’s the question of the cube. By running now, when missed, we’ll be able to send the cube. When we dump and then get away with things later, we’ve cost enough in the race that we may not be able to cube.

I guess if I were smarter, this would be easy. OK, knowing full well this could be a ridiculous whopper, I’m going to stick with my original play, only because I don’t know enough about the numbers to prove to myself that running is correct. I suspect I’ve computed everything I need, but don’t seem to be able to assemble it into one coherent picture.

6/2 6/1

JLee

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