How many trials to guarantee a profit?

[JayTee]

Let's say that I have someone offering me $101:$100 on flipping a fair coin. How many flips would I have to get them to agree to to be 99% sure that I will profit at least $100.

I would expect to profit $0.50 on each flip, right? So I would expect to be up $100 after 200 flips, but that's just on average. What is the probability that I am up at least $100 after 200 flips. I don't know how to solve it but I would guess 50%.

[Doug Funnie II]

This type of problem is modled by the binomial distribution. It is easier to solve a problem with the construction, "Let's say that I have someone offering me $101:$100 on flipping a fair coin. After 100 flips what are my chances of winning X dollars?"

There are lots of binomial calculators online where you set the probability of "success" (heads in a coinflip so .5) and the number of flips you attempt and it tells you the chance of having X heads come up.

[dwigt]

Since N*p will be large, a normal approximation will be valid. For a normal(0,1) P(Z<-2.33)=.01, (i.e. a 1% rare event is a 2.33 sigma event) By CLT
(S-.5*N)/sqrt(N*101^2) is approximately normal(0,1). (S=sum of N bets, E[S]=.5*N, var(S)=N*100.5^2).

Solving gives N=2.1905e+005. The EV starts out small but increases linearly where the standard deviation starts out big but increases only like square-root -- so it takes a long time for the small linear to dominate the square root.


When N=200, by symmetry the probability is 50% (or directly plug in)

[bigpooch]

When N=200, the probability is slightly more than 50%.
Since the probability of exactly half heads and tails is
C(200,100)/(2^200) and so the probability of AT LEAST half
are heads is exactly 1/2+C(200,100)/(2^201) since you have
to divide by 2. Admittedly, this will be quite small (use
Stirling's series to get a good approximation if you want to
compute this).

[dwigt]

Oops..thanks. When N=200 it is not 50%. P(exactly 100 wins) = C(200,100)/(2^200) = .0563, so it does make some difference. If you use the simple normal approximation and
P(S>99.5) you are only off by .0001