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Topic: Prob of flipping coin n times, at no time with #h > #t?
Replies: 10   Last Post: Feb 14, 2013 2:25 AM

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JohnF

Posts: 172
Registered: 5/27/08
Re: Prob of flipping coin n times, at no time with #h > #t?
Posted: Feb 8, 2013 2:36 AM
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Ray Vickson <RGVickson@shaw.ca> wrote:
> 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"?

>> >>
>> >> 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.

>
> "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.


Thanks a lot for additional explanation, Ray.
Your p(2n+1,k=1) = C(2n+1,n+1)/2^(2n+1), taken from above, indeed matches
my P_{2n+1} = 0.5 - sum(i=1...n) { 1/(2i) * 1/(2^(2i)) * C(2i,i+1) }
for n=0,1,2, as per numerical table above. I'll have to do a little
more work to check the large n cases, but would guess agreement.
And I fooled around a little algebraically, but so far failed to prove
them equal.
--
John Forkosh ( mailto: j@f.com where j=john and f=forkosh )



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