|
BGonline.org Forums
Followup queston
Posted By: Maik Stiebler In Response To: Followup queston (Bob Koca)
Date: Friday, 21 December 2018, at 8:44 a.m.
Let the match equity at M away, N away be x (or in a shorter notation q(M,N)=x). Let q(M-1,N-1)=x+s (think of s as match shortening value) and q(M-1,N)=x+p (think of p as point value). Generally, q(M,N)=(q(M-1,N)+q(M,N-1))/2. Then it follows that q(M,N-1)=x-p [1], q(M-2,N)=x+2p-s [2], q(M,N-2)=x-2p-s [3].
Thus, for the first player the risk of taking at (M,N) is q(M,N-1)-q(M,N-2)=p+s, while risk+gain is q(M-2,N)-q(M,N-2)=4p. TP=risk/(risk+gain)=(p+s)/(4p). Now, if M=N, s is obviously 0, so TP=25%. Intuitively, match shortening should benefit the leader [4], thus s > 0 and TP > 25% if M < N, while s < 0 and TP < 25% if M > N.
---
[1]:
q(M,N)=(q(M-1,N)+q(M,N-1))/2
x=(x+p+q(M,N-1))/2
q(M,N-1)=2*x-x-p
[2]:
q(M-1,N)=(q(M-2,N)+q(M-1,N-1))/2
x+p=(q(M-2,N)+x+s)/2
q(M-2,N)=2*(x+p)-x-s
[3]:
q(M,N-1)=(q(M,N-2)+q(M-1,N-1))/2
x-p=(q(M,N-2)+x+s)/2
q(M,N-2)=2*(x-p)-x-s
[4]:
I think that for a proof it suffices to show that for the trailer at N away, M away it is more likely to reach 2away, 1 away than 1 away, 2 away later in the match.
|
BGonline.org Forums is maintained by Stick with WebBBS 5.12.