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Probability Question

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frog0101:
Hi,
I have been unable to successfully solve this question:
A biased coin was flipped such that: P(T)/P(H)=q, where P(T) is the probability of flipping tails, P(H) is the probability of flipping heads, and q is a positive integer. Find the probability of flipping n heads before m tails as m approaches infinity.

I attempted to find an expression in terms of q for P(T) and P(H), however to no avail. The answer given is 1.

RuiAce:
Sorry - I previously stuffed up with this question and misinterpreted it completely. Basically, we can prove that \( P(T) = \frac{p}{p+1} \) and \( P(H) = \frac{1}{p+1}\). Basically, I've managed to prove that your required probability is this expression:
\[ \left( \frac{1}{q+1}\right)^n \times \sum_{k=0}^{m-1} \binom{n+k-1}{k} \left( \frac{q}{q+1}\right)^k \]
and I've run some R code to check that it does indeed converge to 1 as the answers have stated. But I haven't found any HS friendly method explanation for this yet.

Whereas using beyond high school techniques, we can probably somehow exploit that we have a negative binomial distribution here.

(I'm just saving my code here in case I have time eventually to revisit this problem)

--- Code: ---# arbitrary choices for initial values
n <- 10
tail_prob <- 2/3
head_prob <- 1/3

sum_given_m <- rep(0,50)

for (m in 1:50) {
  for (k in 0:(m-1)) {
    sum_given_m[m] <- sum_given_m[m] + choose(n+k-1,k) * tail_prob^k
  }
  sum_given_m[m] <- sum_given_m[m] * head_prob^n
}

sum_given_m
--- End code ---

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