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Imagine a game, where you have to pass n successive levels.

$\frac{1}{2}$ of the players get past the first level.

$\frac{1}{3}$ of the players get past the second level, given that they've passed the first level.

$\frac{1}{4}$ of the players get past the third level, given that they've passed the second level.

What is the expected value of level at which a random player stops?

What I thought was: The Probability of passing the second level is $\frac{1}{3}$(probability of winning the second game) * $\frac{1}{2}$(probability of passing the first level) = $\frac{1}{6}$ .

The Probability of passing the third level is $\frac{1}{6}$* $\frac{1}{4}$= $\frac{1}{24}$.

So in general the probability of winning the n-th game is $\frac{1}{(n+1)!}$.

Knowing that the expected value of a random Variable is given by:

$$ E[X] =\sum_{i = 0}^n {iP(X=i)}= \sum_{i = 0}^n {\frac{i}{(i+1)!}} = 1 $$

Which apparently is incorrect.

Here's my professor answer (At the beginning of his supposition he says "Let X be the level")

Can someone explain how he arrives to the conclusion that the answer is indeed $e-1$, I mean why he doesn't use the normal Expected Value formula ?

Thanks to everyone who will take some of their time even to read this far.

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    $\begingroup$ $e=\sum_{i=0}^{\infty}\frac 1{i!}$. Your expression is missing the $i=0$ term. $\endgroup$
    – lulu
    Commented Sep 11, 2020 at 13:51
  • $\begingroup$ @lulu I noticed that I made more than a mistake in writing thank you. $\endgroup$
    – Jose Manuel
    Commented Sep 11, 2020 at 14:04
  • $\begingroup$ I don't understand the edits. You have $E=\sum_{i=0}^n\frac i{(i+1)!}$ which is clearly not $1$. $\endgroup$
    – lulu
    Commented Sep 11, 2020 at 14:14
  • $\begingroup$ In general, I can't follow what you wrote. The probability that you stop on level $i$ is not $\frac 1{(i+1)!}$. That isn't even true for $i=2$...the probability that you stop on level $2$ is $\frac 12\times \frac 23=\frac 13$. Your professor is using the fact that $E[X]=\sum_iP(X>i)$ which definitely simplifies the calculation. $\endgroup$
    – lulu
    Commented Sep 11, 2020 at 14:17
  • $\begingroup$ @lulu try putting on wolfram $\sum_{i=0} ^n \frac{i}{(i+1!)}$ $\endgroup$
    – Jose Manuel
    Commented Sep 11, 2020 at 14:29

2 Answers 2

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For a rv defined on non-negative integers, the mean can be calculated also in the following way

$$\mathbb{E}[X]=\sum_x P[X>x]$$

$x=0,1,2,...$

thus the solution is

$$\sum_{i=0}^{\infty}\frac{1}{(1+i)!}=\sum_{k=1}^{\infty}\frac{1^k}{(k)!}=\sum_{k=0}^{\infty}\frac{1^k}{(k)!}-1=e-1$$

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  • $\begingroup$ Thank you very much! But can you explain me where this formula comes from ? $\endgroup$
    – Jose Manuel
    Commented Sep 11, 2020 at 14:21
  • $\begingroup$ @JoseManuel : look here. math.stackexchange.com/questions/3820765/… $\endgroup$
    – tommik
    Commented Sep 11, 2020 at 14:24
  • $\begingroup$ Thank you very much, I'm actually stuying engineering and this kind of things are not taught with this depth. $\endgroup$
    – Jose Manuel
    Commented Sep 11, 2020 at 14:27
  • $\begingroup$ @JoseManuel : in the next answer I show you an alternative proof of the same property but extended in the continuous case $\endgroup$
    – tommik
    Commented Sep 11, 2020 at 14:36
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alternative proof for Continuous rv's

Suppose you have a continuous non negative rv.

$$\mathbb{E}[X]=\int_{0}^{\infty} x f(x)dx$$

Let's prove that

$$\mathbb{E}[X]=\int_{0}^{\infty} [1-F(x)]dx$$

that is the same as saying $\sum P[X>x]$ but in the continuous case

To prove it let's solve the integral by parts

$$\int_{0}^{\infty} [1-F(x)]dx=\underbrace{\Bigg[ x(1-F(x))\Bigg]_{0}^{\infty}}_{=0}+\int_{0}^{\infty}xf(x)dx$$

(to prove that the first addend is zero you can easy solve the limit with de l'Hôpital)

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