Suppose you play the following game: You flip $6$ coins. If you get at least $5$ heads, you win $£100$, otherwise you lose $£100$.
(a) What should you expect your net outcome to be after a hundred games?
(b) Suppose we change the rules so that you get to repeat the process $n$ times, and you now win $£200$ if you get at least $4$ heads in at least one of the n rounds, but lose $£300$ if you don’t. How many rounds (how big an $n$) must you demand at least in order not to lose money (on average)?
For (a) I worked out $\dfrac{\dbinom{6}{5}+\dbinom{6}{6}}{64}$ to be the probability of at least $5$ heads and then worked out the probability of losing by simply doing $1$ minus this number. After this I did ($\dfrac{7}{64}\times100\times100+(\dfrac{57}{64}\times100\times-100)= - £7812.5 \to - £7900$ which I hope is the correct answer.
For (b) so far I worked out the probability of at least $4$ heads as $\dfrac{\dbinom{6}{4}+\dbinom{6}{5}+\dbinom{6}{6}}{64}$ then the probability of losing is just $1$ minus this. I think the next step is to find out the probability of losing in each of $n$ rounds, which I worked out as $\left(1-\dfrac{11}{32}\right)^n$ and then the probability of winning in at least one of $n$ rounds as $1-\left(1-\dfrac{11}{32}\right)^n$.
But after this I hit a brick wall, any advice on how to continue would be great. Also any correction on the work I've already done would be greatly appreciated.
P(win a round) = $\frac{\binom{6}{4}+\binom{6}{5}+\binom{6}{6}}{64}= \frac{22}{64} = \frac{11}{32}$
So P(lose a round) = $\frac{21}{32}$
P(lose n rounds) = $({21\over32})^n$
P(win at least 1 round in n) = $1 -({21\over32})^n$
$E[x] = 200\cdot(1 - ({21\over32})^n) - 300\cdot({21\over32})^n$
You can experiment with a few values, and get the answer. Minimum n will not turn out to be very large, and as it increases, you will gain more and more.
See what wolframalpha says !
PS
We want $200\cdot(1 - ({21\over32})^n) - 300\cdot({21\over32})^n > 0$
$200 > 500\cdot({21\over32})^n$
$0.4 > ({21\over32})^n$
Since right hand side will decrease as n increases, equate the two sides, take logs to solve for n and take $\lceil n\rceil$, i.e. if any rounding is needed, round up
n = 2.175..
$\lceil n\rceil$ = 3