# How do I go about calculating the expected profit/winning per play of this game

I designed this game for my Data class but I'm having alot of trouble calculating the expected profit per play of this game. We have taken up examples involving lottery tickets in which you have to calculate the profit and winning per ticket bought but I can't seem to apply the same concepts to my game due to the various stages and rules.

This is how the games goes

You pay $20 to start and there are three rounds Round 1: There is a 1/2 chance you win and move on, and a 1/2 chance you lose Round 2: There is a 2/6 chance you move on to the next round, and 1/6 chance you win$10 AND move on. (3/6 chance you lose and get nothing)

Round 3: There is 1/12 chance you win $100, and if you lose you get nothing Here is my attempt at it After multiplying every correlating x and y value i get E(X) = -18.5, meaning the profit the ones running the game make is$18.5

Help is greatly appreciated!!!

## 1 Answer

You pay $\$20$and start playing. Let your profit be$x$which is initially$-20$. • With a probability of$\frac12$, you lose:$x = -20$. • Otherwise, with a probability of$\frac12$, you proceed with unchanged$x$. • With a probability of$\frac12$, you lose:$x = -20$. • With a probability of$\frac16$, you proceed with a new$x = -10$. • With a probability of$\frac{11}{12}$, you lose:$x = -10$. • Otherwise, with a probability of$\frac{1}{12}$, you have$x = 90$. • Otherwise, with a probability of$\frac13$, you proceed with unchanged$x$. • With a probability of$\frac{11}{12}$, you lose:$x = -20$. • Otherwise, with a probability of$\frac{1}{12}$, you have$x = 80. Summing along this tree of probabilities, \begin{align*}E(x) &= \frac12 (-20) + \frac12 \bigg(\frac12 (-20) + \frac16\bigg(\frac{11}{12}(-10) + \frac{1}{12}(90) \bigg) + \frac13\bigg(\frac{11}{12}(-20) + \frac{1}{12}(80) \bigg)\bigg) \\ &= -\frac{205}{12} = -17\,{}^1/_{12}\end{align*} Thus you will expect to lose\$17\,{}^1/_{12}$ in this game, which is the profit of the people running the game. Your calculation probably went wrong somewhere, the best way to avoid this is to make a tree like I did (perhaps more graphically a `tree').

• Thank you!, I also got the same answer using the prize and probability chart. – 2247382 Nov 22 '16 at 3:23
• I was wondering, in the bracket why do you put 1/2 (-20) in again, wasn't the probability of losing the 20 dollars already dealt with at the beginning of the solution? – 2247382 Nov 22 '16 at 22:16
• @chaosdig5 At the beginning, we deal with the probability of losing in the first round. We must consider also the probability that we lose in the second round after advancing from the first, which is counted separately. – shardulc Nov 22 '16 at 23:40
• Oh right, the 3/6 becomes 1/2. Thanks so much man, you're a life saver – 2247382 Nov 23 '16 at 0:09
• @chaosdig5 You're welcome! You can upvote the answer if you like it by clicking the up-arrow and accept it by clicking the green tick-mark. – shardulc Nov 23 '16 at 1:09