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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: 164
Registered: 5/27/08
Re: Prob of flipping coin n times, at no time with #h > #t?
Posted: Feb 7, 2013 2:12 AM
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Ray Vickson <RGVickson@shaw.ca> 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 )



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