# Whats the probability you lose money?

Suppose you are playing a game that costs 8 to play. You flip 10 coins and, for every head, you win $2. Whats the probability you lose money ? $$\begin{array}{c|cccccccccc} x& 1 & 2 & 3 & 4 & 5 & 6 & 7 &8 &9&10 \\ \hline p(x) & 5/512 & 45/1024 & 15/128 & 63/256 & 105/512 & \end{array}$$ Since the loosing money means getting less than 4 heads I just did an addition of the $$P_X(1)+P_X(2)+P_X(3)=0.1708$$ With pmf of X being$P_X(x)=\binom{10}{x}(0.5)^x(0.5)^{10-x}\$

And g(X)=Y=-8+2X the equation that define the net losses or winnings.

can this be a viable solution the way I presented it ? If yes do you know any good shortcut for this exercise ?

$$\frac{1}{2^{10}}\sum_{i=0}^3 \binom{10}{i}= \frac{1+10+45+120}{1024}\approx 0.1719$$