$P$(lose 1 dollar)=0.7 $P$(lose 2 dollar)=0.2 $P$(get 10 dollar)=0.1 X:=the profit/lose.

$P$(lose 1 dollar)=0.7

$P$(lose 2 dollar)=0.2

$P$(get 10 dollar)=0.1

X:=the profit/lose.

$1.$ Find $E[X],\sigma^2$ , is it worth betting ?

$2.$ The player bet 100 times, find the probability he'll lose.

My solution:

$1. E[X]=-1\cdot0.7-2\cdot0.2+10\cdot0.1=-0.1 , E[X^2]=1^2\cdot0.7+2^2\cdot0.2+10^2\cdot0.1=11.5$

$\sigma^2=E[X^2]-E[X]^2=11.5+0.1^2=11.51$

I am really not sure if it is worth it , if $E[X] > 0$ is it worth it ?

$2.$ I need to find $E[X]^{100}$?

Thank you !

Your calculations are NOT correct

$$\sigma^2=E[X^2]-E^2[X]=11.5-0.1^2=11.49$$

as already noted, if he plays 100 times, you can approximate

$$Y=\Sigma_i X_i\dot{\sim}N(-10;1149)$$

thus

$$P(Y<0)=\Phi\left(\frac{10}{\sqrt{1149}} \right)\approx61.60\%$$

better, using the continuity correction factor

$$P(Y<0)=\Phi\left(\frac{9.5}{\sqrt{1149}} \right)\approx61.04\%$$

Concluding, if you have to take a decision with these information no, it is not worth betting...

1. Your calculations seems correct. But you cannot answer the question with the information at hand. You need to know how the individual trades off gains and losses. This is usually (in economics) done by specifying a utility function that maps the monetary outcome to a numeric representation of the wellbeing of the decision maker. The decision maker is then often assumed to maximize the expected utility.

2. I guess you are expected to use the central limit theorem in this case. $% X_{100\text{ }}$will be approximately normally distributed with mean $100E[X] $ and $\sigma _{100}^{2}=\sigma ^{2}* 100$.