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I am no mathematician, my syntax is crappy and I will know LESS than you think. Can anyone please help me with the following problem?

Compute E(X) for the following random variable X:

X=Number of tosses until getting 4 (including the last toss) by tossing a fair 10-sided die.

From a closed previous post (with which I can't interact because I don't have reputation) I have got the following result:

$E[X]= 1+ \frac1{10}\times 0 + \frac9{10}\times E[X]$

The problem is I can't understand this and although it answered the question at hand, I gained no knowledge. Can anyone please derive such expression with as many steps as possible in a way your grandmother would understand? Thank you in advance!

MDSv
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  • What parts of this answer do you understand? I mean, what is a valid starting point for an answer acceptable to you? – kimchi lover Oct 03 '20 at 17:28
  • $1/10$ of the time the length is $1$, and $9/10$ of the time, it is $1 $ + (length of a new game). Using what’s called linearity of expectation, the expected (I.e. average) length E is the weighted sum $E = (1/10)(1) + (9/10)(1+E)$ Which can be solved for $E$. – Ned Oct 03 '20 at 17:41

1 Answers1

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This formula contains a bunch of things. Let's look at them one at a time. Basically, it says: we roll the die once and see how many more times we might have to roll it after that.

  • $1$. We must roll the die at least once. Depending on the result, we might need to roll it more than that, but we must roll it at least once. That's where the $1$ comes from.

  • $\frac{1}{10}$ This is the probability that the die comes up "$4$".

  • $0$. If the die roll comes up $4$, we're done and we need no extra rolls; in that case we're done.

  • $\frac{9}{10}$. This is the probability that the die does not come up "$4$".

  • $E[X]$. If the die did not come up "$4$", we will need to keep rolling it until it turns up "$4$". This value is how many more times we are expected to need to roll it after the current roll.

Robby the Belgian
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  • Thank you so much! I guess I made a mess out of what X is and kept trying to use the formula $E[X] = \sum x_i*p(x_i)$ and failed... But why not $E[X-1]$ instead of $E[X]$? Thanks again! – MDSv Oct 03 '20 at 17:44
  • When we roll the die and it does not come up $4$, we kind of have to start all over again. So we again have to wait $E[X]$, and add our current roll to it. So, in that case we have $1 + E[X]$, and that happens $\frac{9}{10}$ of the time. We could have written it as $E[X] = \frac{1}{10} (1 + 0) + \frac{9}{10} (1 + E[X])$, if that is more clear. – Robby the Belgian Oct 03 '20 at 18:51
  • Thank you! :) I think I got it now :) – MDSv Oct 03 '20 at 21:39