# Game in which you filp $6$ coins, get $£100$ for at least $5$ heads and lose $£100$ otherwise

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.

• It's OK to have an "expected outcome" that's not possible to attain, so I'd eliminate the step that asserts that -7812.5 = -7900. (You should never write such a thing anyway; it's not true!) – John Brevik Aug 12 '15 at 14:43
• Thanks John I can see now that it's best to leave that step out. Tru blue anil, thanks so much for the help, eh sorry but I'm still pretty much stuck after this point, any help at all would be great. – Ralph Aug 12 '15 at 15:50

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

• Ok thanks so much for the help,sorry for basic questions but maths isn't my first subject, so from here do I just try different values for n? Should I be using logs or something? Thanks again. – Ralph Aug 13 '15 at 11:52
• I thought you'd be able to manage from here. you take logs. I'll add a ps. – true blue anil Aug 13 '15 at 14:47