Probability: Random variable dice problem I'm not sure how to approach the below problem using random variable. Can I consider this as distinct events and calculate the average of each and then add them together? Can I use a binomial distribution? I'm very lost...
You have 1 dice. Each face has a probability of 1/6. You roll the dice and if you get a 6, you win 50\$ otherwise you roll again. This time, if you get a 6, you get 10\$ otherwise you get nothing. The initial cost to play is 10\$. What is the average if you play 5 times?
 A: I am not sure if I understand the game correctly. But, let's assume, it has 2 rounds like you say. Let's denote rolling a $6$ by the event $H$ and any other face by the event $T$.
The possibilities are $H,TH,TT$. 
(1) $P(H)=1/6$. The payoff in this case is $50-10=40$.
(2) $P(TH)=(5/6)\times(1/6)$. The payoff in this case is $10-10=0$.
(3) $P(TT)=(5/6)\times(5/6)$. The payoff in this case is $0-10=-10$
So, the expected payoff is $(1/6)40+(5/25)(0)+(25/36)(-10)$.
A: There is quite some independence going on. Let us put it up mathematically.
There are five times you play, each independent of the other . Let $X_i, i = 1 , ... , 5$ be the amount you won in each game, so the final amount you win is $Z = X_1 + ... + X_5 - 10$. Each game also consists of two independent dice rolls. So we first focus on each game $X_i$. It is enough to focus on $X_1$ because all $X_i$ are independent and identical, they have the same definition.
Now, to calculate what you get from each game, you need a sample space. This will be the set of all outcomes of two dice rolls ; $\Omega = \{(a,b) : 1\leq a,b \leq 6\}$. The probability of each element will be $\frac{1}{36}$.
Now, define the random variable $X_1 : \Omega \to \mathbb R$ as follows :

*

*We know that if neither is a six, then we win nothing, so $X_1(a,b) = 0$ if $a,b \neq 6$.


*We know that if $a=6$ then we win $50\$$ so $X_1(a,b) = 50$ if $a = 6$.


*We know that $a \neq 6$ but $b = 6$ then we win $10\$$ so $X_1(a,b) = 10$ if $a \neq 6 , b = 6$.
Thus, $X_1$ has been defined. The average we win from each game, is $E[X_1]$, which is calculated from the formula $$
E[X] = \sum_{k} kP(X=k) = 0 P(X=0) + 10P(X = 10) + 50 P(X = 50) 
$$
So we need to calculate $P(X = 10)$ and $P(X = 50)$.
I leave you to see that $\{X = 10\}$ has five elements, and $\{X = 50\}$ has $6$ elements. Therefore, the answer is $10 \times \frac{5}{36} + 50\frac{6}{36} = \frac{350}{36}$.
But this is from one game. Using linearity of expectation :
$$
E[Z] = E[X_1] + ... + E[X_5] - E[10] = 5E[X_1] - 10 = \frac{1750 - 360}{36} = \frac{1390}{36}  = 38\frac{11}{18}
$$
So you are likely to be in profit of $38\frac{11}{18}\$$ after five rounds.
(Check the calculations , tell me if your answer is off)
EDIT : if the cost to play each game is $10$ dollars, then subtract forty dollars from the above answer.
A: For one play your expectation is $\frac{50}{6}+\frac{50}{36}-10=-\frac{10}{36}$.  Since each play is distinct, the expectation after $5$ plays is $5$ times that of one play.
You use a binomial only if you want the distribution after $5$ plays.
