# Probability Question (1 Viewer)

#### frog0101

##### New Member
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 flippings 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.

#### InteGrand

##### Well-Known Member
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 flippings 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.
$\bg_white \noindent To get an expression for P(T) and P(H) in terms of q, note that P(T) = 1-P(H). Substitute this into \frac{P(T)}{P(H)} = q and solve for P(H). However, we don't actually need to know exactly what P(H) is in terms of q to do this question. It is enough to know that P(H) is strictly between 0 and 1, which is guaranteed by the fact that \frac{P(T)}{P(H)} is equal to a positive number.$

$\bg_white \noindent Now, if we denote p = P(H)\in (0,1), the probability of having \textbf{exactly} n heads before m tails is$

$\bg_white P(\text{exactly }n\text{ heads before }m\text{ tails}) = \binom{n+m-1}{n}p^{n}(1-p)^{m},$

$\bg_white \noindent which you can show combinatorially as an exercise (and would need to prove in the HSC if using it).$

$\bg_white \noindeint For any fixed n, since p\in (0,1), it turns out that \textbf{\underline{this expression tends to 0 as m\to \infty}}. If the problem meant find the probability of obtaining \emph{at least} n heads before m tails, the expression for this is$

\bg_white \begin{align*} P(\text{at least }n\text{ heads before }m\text{ tails}) &= 1-P(\text{less than }n\text{ heads before }m\text{ tails})\\ &= 1 - \sum_{k=0}^{n-1} P(\text{exactly }k\text{ heads before }m\text{ tails}).\end{align*}

$\bg_white \noindent Assuming you've managed to prove the statement in bold and underlined above (which you should do as an exercise!), we know that as m\to \infty, we have$

$\bg_white P(\text{exactly }k\text{ heads before }m\text{ tails})\to 0$

$\bg_white \noindent for any k, whence (using the fact that the limit of a finite sum is the sum of the limits)$

$\bg_white \lim\limits_{m\to\infty}P(\text{at least }n\text{ heads before }m\text{ tails}) =1 - \sum_{k=0}^{n-1}0 = 1,$

$\bg_white \noindent so the answer is 1.$

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