Quote:
Originally Posted by whosnext
I do not know what your formula is purported to show.
If this is a formula for the expected number of lead changes with a fair coin (50% heads) as discussed in this thread, it is clearly wrong.
Examples of demonstrably correct percents are given above which your formula does not agree with.
It does agree with what you posted, up to 3, and then runs into problems with double precision rounding.
Quote:
Originally Posted by whosnext
Since this is easy to code up the brute force scenarios:
For 4 rounds, the distribution of the number of lead changes is:
16 have 0
220 have 1
20 have 2
0 have 3
0 have 4
Which yields an EV of 260/256 = 1.015625
R function that I have gives EV(4) = 1.0
My formula:
SUM from i = 1 to 4
i = 1
(1/2)^1 =
1/2
i =2
(1/2)^2 =
1/4
i = 3
(1/2)^3 + (2 choose 2)*(1/4)^2*(1/2)^1 =
5/32
i = 4
(1/2)^4 + (3 choose 2)*(1/4)^2*(1/2)^2 =
7/64
Total = 1/2 + 1/4 + 5/32 + 7/64 = 65/64 ( = 260/256)