Date: Feb 7, 2013 1:52 PM Author: RGVickson@shaw.ca Subject: Re: Prob of flipping coin n times, at no time with #h > #t? On Wednesday, February 6, 2013 11:12:36 PM UTC-8, JohnF wrote:

> Ray Vickson wrote:

>

> > On Wednesday, February 6, 2013 9:12:51 AM UTC-8, Ray Vickson wrote:

>

> >> On Wednesday, February 6, 2013 5:42:18 AM UTC-8, JohnF wrote:

>

> >>

>

> >> > What's P_n, the prob of flipping a coin n times,

>

> >> > and at no time ever having more heads than tails?

>

> >> > There are 2^n possible h-t-... sequences of n flips,

>

> >> > comprising a binomial tree (or pascal's triangle),

>

> >> > with 50-50 prob of going left/right at each node.

>

> >> > So, equivalently, how many of those 2^n paths never

>

> >> > cross the "center line" (#h = #t okay after even number

>

> >> > of flips)?

>

> >> > Actual problem's a bit more complicated. For m<=n,

>

> >> > what's P_n,m, the prob that #h - #t <= m at all times?

>

> >> > That is, P_n above is P_n,0 here. Equivalently, how

>

> >> > many of those binomial tree paths never get >m past

>

> >> > the "center line"?

>

> >> > --

>

> >> > John Forkosh ( mailto: j@f.com where j=john and f=forkosh )

>

> >>

>

> >> Feller, "Introduction to Probability Theory and its Applications,

>

> >> Vol I (Wiley, 1968), Chapter III, page 89, deals with this

>

> >> (and many related) problems. Chapter II deals with the simple

>

> >> random walk S_k = X_1 + X_2 + ... + X_k, where the X_i are iid

>

> >> and X_i = +-1 with prob. 1/2 each.

>

> >>

>

> >> On page 89 Feller states and proves Theorem 1: "The probability

>

> >> that the maximum of a path of length n equals r >= 0 coincides with

>

> >> the positive member of the pair p(n,r) and p(n,r+1).

>

> >>

>

> >> Earlier in Chapter he gave the formula p(n,k)= Pr{S_n = k} =

>

> >> C(n,(n+k)/2)/2^n, where C(u,v) denotes the binomial coefficient

>

> >> "u choose v".

>

>

>

> Thanks, Ray. That sounds like my same problem (but what's "iid" mean

>

> in the context "X_i are iid" in your first paragraph above?).

>

> I take it my answer should be "the positive member of" p(n,0) and p(n,1),

>

> where p(n,k) = C(n,(n+k)/2)/2^n. And I'm guessing that means k=0 if

>

> n even, and k=1 if n odd.

>

> Actually, I think I already had a solution for the odd cases

>

> worked out, but much more complicated-looking than yours:

>

> P_{2n+1} = 0.5 - sum(i=1...n) { 1/(2i) * 1/(2^(2i)) * C(2i,i+1) }

>

> Note that 2n+1 is always an odd (your k=1, I think) case.

>

> Numerically, mine gives (haven't programmed yours yet, to check agreement)

>

> n | P_n

>

> ----+--------

>

> 0 | .5

>

> 1 | .375

>

> 2 | .3125

>

> . | ...

>

> 4999| .00798

>

> 9999| .00564

>

> 19999| .00399

>

> slowly going to zero, i.e., you eventually get an extra head

>

> (your r>0, I think), but it may take a lot longer than you'd

>

> naively guess (because prob goes to zero very slowly).

>

> What I still can't get, in closed form, is equating the two

>

> expressions, i.e., for your n-->2n+1 and then k=1,

>

> yours: p(2n+1,k=1) = C(2n+1,n+1)/2^(2n+1)

>

> mine: P_{2n+1} = 0.5 - sum(i=1...n) { 1/(2i) * 1/(2^(2i)) * C(2i,i+1) }

>

> Are those two really equal? (I'll get around to checking numerically)

>

>

>

> >> The answer to your "<= m" question is the sum of those probabilities

>

> >> for r from 0 to m, plus the probability that the max is < 0.

>

> >> The latter can be obtained from the expression on page 77, which is

>

> >> P{S_1 > 0, S_2 > 0, ... S_2n > 0} = (1/2)* u(2n),

>

> >> and where u(2j) = C(2j,j)/2^(2j) = P{S_2j = 0}. Note that having

>

> >> all S_i < 0 has the same probability as having all S_i > 0.

>

>

>

> Thanks again. My cumbersome derivation wasn't general enough to cover

>

> this situation.

>

> --

>

> John Forkosh ( mailto: j@f.com where j=john and f=forkosh )

"iid" means independent and identically distributed. Thus, Feller's S_k is you H - T count at the end of toss k. Yes, the positive member of the pair p(n,0) and p(n,1) uses (n,0) if n is even and (n,1) if n is odd (because you need (n+k)/2 = integer).

For n even, P{all S_i <= 0} = P{max S_1 = 0} + P{all S_i < 0} = p(n,0) + (1/2)u(n). If n is odd, the formula for P{all S_i < 0} is, of course, more complicated. First, X_1 = -1 must happen (so that S_1 < 0); then the remaining (n-1) tosses must not have their H-T count rise above 0, so that means the mutually exclusive events {max new_S = 0} or {all new_S < 0} must occur (where now we have (n-1) tosses and are starting over, counting from toss 2---that is, new_S_1 = X_2, new_S_2 = X_2 + X_3, etc. We can compute P{all new_S_i < 0} because n-1 is even. So, we finally get

P{all S_i < 0} = (1/2)[p(n-1,0) + (1/2)u(n-1)] for odd n.

I have not checked to see if this matches what you got.