Biased coin flip-

Assume a coin is biased toward HEADS X percent of the time.
What is the percent chance and standard deviation of a run of T trials
ending up with more HEADS. In addition, what is the percent chance for
each combination?

Example = A coin is known to come up heads 55% of the time. For a set
of 12 coin flips, what is the likelihood that more HEADS than tails
will come up after 12 flips. What are the odds for 7 heads, 5 tails. 8
heads, 4 tails. 8 tails, 4 heads. Etc.

Please answer the example and provide the formula to enable
calculation for any percent X, trials T, and odds for each
combination.

---

Clarification of Question by davidj-ga on 16 Sep 2004 07:52 PDT

Thanks, but most importantly, I still need a clear approximation of
how to calcuate the percent likely to have more "heads" for a set of T
trials with any given coin bias.

Firstly let's make X a fraction rather than a percent so instead of
55% we say 0.55.

The probability of getting N heads in T trials is T!/(N!(T-N)!) X^N
(1-X)^(T-N) where ! is factorial (i.e. M!=Mx(M-1)x(M-2)...1).

The probability of getting more heads than tails is the probability of
getting all heads+prob. getting all but one head+...+prob. getting
just more heads so you just sum the terms above for each N where
N>T-N. I don't believe there is a simple formula for computing this,
you just have to do the sum.

It doesn't make sense to ask what the 'standard deviation' of getting
more heads. Maybe you want the standard deviation of the number of
heads.

The mean number of heads is X*T
The standard deviation number of heads is sqrt(X*(1-X)*T)

Let's do the T=12 coin example with X=0.55

P(12 heads)=0.55^12=0.000766218
P(11 heads)=0.55^11*0.45*12!/11!=0.00752287
P(10 heads)=0.55^10*0.45^2*12!/(10!*2!)=0.0338529
P(9 heads)=0.55^9*0.45^3*12!/(9!*3!)=0.09232608
P(8 heads)=0.55^8*0.45^4*12!/(9!*4!)=0.16996393
P(7 heads)=0.55^7*0.45^5*12!(7!*5!)=0.222498

The sum of all these terms is 0.52693

Formula is given by Binomial Distribution.
For $2, I am afraid, you will have to type into a search engine yourself.