<p>Think of all the possible outcomes. For example, if your first ball is red you loose 1 rupee (this has probability 1/2), if you instead get one green ball and then one red ball you end up with  0 rupees (this has probability 1/4), if instead you get at first 2 green balls and then a red one, you end up with 1 rupee (this has probability 1/8), etc. </p>

<p>The expected value is </p>

<p><span class="math-container">$$E(V)=(-1)\frac12+(0)\frac14+(1)\frac18+(2)\frac1{16}\dots=\sum_{n=1}^\infty(n-2)\frac{1}{2^n}=0$$</span></p>

<p>The last equality comes from two well-known results: <span class="math-container">$\sum_{n=1}^\infty\frac{n}{2^n}=2$</span> and <span class="math-container">$\sum_{n=1}^\infty\frac{1}{2^n}=1$</span></p>

CC BY-SA 4.0
2019-06-19T06:46:55.140Z
2019-06-19T06:46:55.140Z
Regio
<p>An urn contains equal number of green and red balls. Suppose you are playing the following game. You draw one ball at random from the urn and note its colour. The ball is then placed back in the urn, and the selection process is repeated. Each time a green ball is picked you get 1 Rupee. The first time you pick a red ball, you pay 1 Rupee and the game ends. Your expected income from this game is..</p>

<p>The answer given is 0 but shouldn't it be positive as the as the income rises if you consecutively draw the green ball? Please help me clarify this doubt.</p>

<expected-utility>
Expected Income Question
Yash Gupta
# Expected Income Question

(Asked by: username_0 on 2019-06-19)

An urn contains equal number of green and red balls. Suppose you are playing the following game. You draw one ball at random from the urn and note its colour. The ball is then placed back in the urn, and the selection process is repeated. Each time a green ball is picked you get 1 Rupee. The first time you pick a red ball, you pay 1 Rupee and the game ends. Your expected income from this game is..

The answer given is 0 but shouldn't it be positive as the as the income rises if you consecutively draw the green ball? Please help me clarify this doubt.

**Tags:** expected-utility

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## 1 answers

### Answer by username_1 on 2019-06-19. Score: 2 | Accepted answer

Think of all the possible outcomes. For example, if your first ball is red you loose 1 rupee (this has probability 1/2), if you instead get one green ball and then one red ball you end up with 0 rupees (this has probability 1/4), if instead you get at first 2 green balls and then a red one, you end up with 1 rupee (this has probability 1/8), etc.

The expected value is

$$E(V)=(-1)\frac12+(0)\frac14+(1)\frac18+(2)\frac1{16}\dots=\sum\_{n=1}^\infty(n-2)\frac{1}{2^n}=0$$

The last equality comes from two well-known results: $\sum\_{n=1}^\infty\frac{n}{2^n}=2$ and $\sum\_{n=1}^\infty\frac{1}{2^n}=1$


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