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Suppose that in a slot machine there are $n+1$ possible outcomes $A_1,\dots,A_{n+1}$ for a single play. A single play costs $\$1$. If outcome $A_i$ occurs, you win $a_i$, for $i=1,\dots,n$. If outcome $A_{n+1}$ occurs, you win nothing.

(c): The slot machine owner wishes to pay $da_i$ dollars when outcome $A_i$ occurs, where $a_i = 1/p_i$ and $d$ is a number between $0$ and $1$. The owner also wishes his or her expected profit to be $\$.05$ per play. (The player’s expected profit is $-.05$ per play.) Find $d$ as a function of $n$ and $p_{n+1}$ . What is the value of $d$ if $n = 10$ and $p_{n+1} = .7$?

Great. So all the rules suddenly change for part (c), if I read it correctly. It is easy to derive the formula $$\sum^n_{i=1} \frac{d}{p_i}=0.95$$

My brain probably simply made a dumb short-circuit can't see how I can introduce $p_{n+1}$ into the expression...

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1 Answer 1

up vote 1 down vote accepted

You are missing 2 things

a) that

$$\sum_{i=1}^{10}p_i=1-p_{n+1}$$

b) your formula is wrong, the expected payout is

$$\sum^n_{i=1} \frac{dp_i}{p_i}=nd$$

Which is the sum of the products of the payout and the probability

So

$$d=\frac{0.95}{n(1-p_{n+1})}$$

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Could you explain how you got $$dp_i/p_i$$? –  user54609 Apr 13 '13 at 0:50
    
payoff is $\frac{d}{p_i}$, the probability is ${p_i}$, expected value is the sum of the product of these –  Dale M Apr 13 '13 at 3:27
    
Could you please minor-edit your posts so I can undo the downvote? :) –  user54609 Apr 13 '13 at 11:20

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