A really simple problem, but needs a trick or two to solve quickly.
Question: You take it in turns tossing a coin with a friend – you having the first toss – and keep going until one of you tosses heads and wins. What are the chances you’re the winner?
Answer: The chances of me winning are the sum of my chances at each toss which is: 1/2 + 1/8 + 1/32… The chances of my friend winning are 1/4 + 1/16 + 1/64…
It may be immediately obvious to you that I’ll have twice my friend’s chances of winning. So that would mean I have a 2/3 chance, and my friend a 1/3 chance.
Let’s also think about this in terms of geometric progressions. First, some basics:
when r<1:
\sum_{n=0}^\infty cr^n = \frac{c}{1-r}
and:
\sum_{n=1}^\infty cr^{n-1} = \frac{c}{1-r}
So if you can get your progression into one of the two forms above, you can just apply the formula on the right hand side.
Now start looking at this from the point of view of the guy starting second. His first toss has a probability of \small(1/2)^2, his second toss \small(1/2)^4, his third \small(1/2)^6. So we can see that on his nth toss, his probability will be \small(1/2)^{2n} with n starting at 1.
We see from the above formulas that when we have n starting at 1 we need to get our progression into the form: {\small}cr^{n-1}. So we need to expand as such: (\frac{1}{2})^{2n} = (\frac{1}{4})^n = \frac{1}{4} * (\frac{1}{4})^{n-1}. Now we have c=1/4 and r=1/4 to plug in: \frac{c}{1-r} = \frac{\frac{1}{4}} {1 - \frac{1}{4}} = \frac{1}{3}.
So of course, since someone has to win, the probability for the starter is 2/3. But let’s work this out. The starter’s probability at toss n is \small(1/2)^{2n+1} with n starting at 0. Since n starts at 0, we need to get this into the form: {\small}cr^{n}. Expanding as above: (\frac{1}{2})^{2n+1} = \frac{1}{2} * (\frac{1}{4})^{n}. Now we have c=1/2 and r=1/4 to plug in: \frac{c}{1-r} = \frac{\frac{1}{2}} {1 - \frac{1}{4}} = \frac{2}{3}.