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Dropping the free double

Posted By: Maik Stiebler
Date: Tuesday, 4 December 2018, at 9:09 a.m.

In Response To: Dropping the free double (Bob Koca)

At 1a-(2n)a with n->inf, the takepoint on the free drop should be such that the very worst two-roll opening sequence is only a borderline drop. Assume that it is 31P 62S (I didn't bother to check that) and that at that point the leader, if they take, loses 36.5% singles, 20% gammons and 0.8% backgammons.

If we zoom in to the post-Crawford ME table at 1away - far away we will see something like this:
2n+1 away: x
2n+2 away: a*x
2n+3 away: b*x
2n+4 away: a*b*x
2n+5 away: b*b*x
2n+6 away: a*b*b*x
2n+7 away: b*b*b*x
2n+8 away: a*b*b*b*x
2n+9 away: b*b*b*b*x ,
where n is a very large number, x is an irrelevant placeholder, a represents the asymptotical value of the free drop (and incidentally, 1-a is the takepoint we are ultimately looking for), b the asymptotical value of the trailing opponent being two more points away (assume that b=0.6).

Then from the borderline free drop at 2n+8 away, we get b*b*b*x=0.365*a*b*b*x+0.200*a*b*x+0.008*a*x, thus a=0.833 and the takepoint at 2n+9 away is (1-a)=16.7%

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