Rusty with probability here...
The question is: Flip a coin 11 times. If you get 8 tails or less, I will pay you \$1.
Otherwise, you pay me \$7.Step 1. Find the expected value of the proposition. Round your answer to two decimal places.
Step 2. If you played this game 615 times how much would you expect to win or lose? (Losses must be entered as negative.)
I think that the expected value of the proposition is the sum of the probability of getting 8 tails or less times times the 1 dollar value, plus probability of getting more thand 8 tails times the $7 value.
I would like a walkthrough of the mindset and reasoning, not just the answer so I can learn how to approach the problem. I am also learning how to solve with R.
There are $2^{11}$ possible outcomes, of these there are $\binom{11}{9}+\binom{11}{10}+\binom{11}{11}=67$ outcomes in which I pay you $7$ dollars.
Therefore the expected cash the other player gets is:
$\frac{2^{11}-67}{2^{11}}-7\frac{67}{2^{11}}=\frac{1512}{2^{11}}\approx0.73$.
So I don't recommend you play the game.
The expected cash after playing the game $n$ times is $0.73\times n$