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Re: Prob of flipping coin n times, at no time with #h > #t?
Posted:
Feb 7, 2013 1:52 PM
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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.
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