Please help!!
Hey! I think it's easier here to take cases where he DOESN'T win in the first three throws. This would only occur if the coins read HTH, or THT. The probability of HTH is (0.5)(0.5)(0.5), as is the probability of THT. Therefore, 2(0.5)(0.5)(0.5)=0.25. The probability of WINNING is 1-P(not winning), therefore the answer to i) is 1-0.25=0.75.
Fuck Joe. Why make such a complicated game? Now, we don't want him to win on the second go (even), or on any even goes. The probability of NOT winning on an even go must be something like
P(Not winning)=HTHTHTHT...+THTHTHTHTHTHT=2(0.5)^n for n throws, where n is even. Him WINNING on an odd turn is the same as NOT WINNING on an even term (sort of? In terms of infinity, where winning is assured? I'm not actually 100% about that, but let's go with it).
Therefore, we need to sum up 2(0.5)+2(0.5)^2+...+2(0.5)^n for n=infinity. Use your sum to infinity formula to get a solution, then your answer should be 1-ans!
Let me know if this works out,
Jake