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Please check my work, expected value calculations are often of the sort where you get an answer but there's no "check", so to speak.

Imagine you have a scenario with the following rules:

  • Inputs determine several possible outcomes
  • Your outcome for each trial is determined randomly
  • The different outcomes have different probabilites
  • Repeat the same inputs over and over until you win, see below
  • Each outcome has a different $score$
  • Higher scores are bad
  • You stop when you get the "correct" outcome
    • Don't count your score for the correct outcome, in other words, $score=0$

Example: for given inputs, you have (fail probabilities are the same but they don't have to be):

  • $P(success) = 6.6667\%$
  • $P(fail_1) = 23.333\%$, $Score(fail_1) = .5$
  • $P(fail_2) = 23.333\%$, $Score(fail_2) = 4$
  • $P(fail_3) = 23.333\%$, $Score(fail_3) = 6$
  • $P(fail_4) = 23.333\%$, $Score(fail_4) = 20$

Calculate: What is the expected score for a given list of failed outcomes?

My answer:

Let $FS_0$ = failed score on one trial that fails

$$ FS = \sum\limits_{trials}{P(trial)Score(trial)} $$

If we succeed on the first try, score is zero.

If we succeed on the second try, score is $(P(fail_{all})*FS$

If we succeed on the third try, score is $(P(fail_{all})*(FS + (P(fail_{all})*FS)$

This implies the infinite sum:

$$ EV = P(fail_{all}) * \Bigg( FS + P(fail_{all}) * \bigg(FS + P(fail_{all}) * \Big(FS + P(fail_{all}) ... $$ $$ = P(fail_{all}) * FS * (1 + P(fail_{all}) + P(fail_{all})^2 + P(fail_{all})^3 + ... ) $$ $$ = P(fail_{all}) * FS * \left(\frac{1}{1-P(fail_{all})}\right) $$ $$ = FS * \left(\frac{1}{P(success)} - 1\right) $$

So in the case of our example above,

$$ EV_{example} = 7.11667 * \left(\frac{1}{6.67\%} - 1\right) $$ $$ EV_{example} = 99.63333 $$

How did I do?

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up vote 1 down vote accepted

As a general hint, you should never use percentages when describing probability problems - it can leas to simple mistakes.

The Expected score of a single failed game

$$ \begin{align} S &= \frac{\sum_{i=1}^4{p_is_i}}{1-p_0}\\ \end{align} $$

In this case $S=7.625$

Games continue until there is a success with $p_0=0.0\dot6$. This is a geometric series and the expected number of failures before a success is given by


So the expected score of the game described is $7.625\times14=106.75$

I think you got the second part right, but I think that you included the success criteria when working out the expected payout for a failed game i.e. you did not divide by $(1-p_0)$. So, right method, wrong numbers!

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Good point about the percentages, I'll keep that in mind. I don't understand why I'm supposed to divide by $\left(1-p_0\right)$. Could you explain further? – durron597 Apr 16 '13 at 4:48
Because the expected number of games $\frac{1-p}{p}$ is the number of failures before the first success. You could use the alternative formulation of the geometric series which is the number of rolls including the first success, $\frac{1}{p}$. You you get the same answer because the two terms cancel in the first formulation and aren't there in the second! Either is fine, you just need to be consistent. – Dale M Apr 16 '13 at 5:08
Yes but... shouldn't I be multiplying by $1-p$, not dividing by it? Because that's what I did, that's the outer $P(fail_{all})$ in the original question. – durron597 Apr 16 '13 at 5:34
What you did in the expected number of games is correct - multiply by $(1-p)$. In order to get the expected value of a game given that you have failed, you need to eliminate the chance of success - so divide by $(1-p)$. These two terms cancel out when you multiply them - which is why you need to be clear about which of the alternate formulations of geomatric distribution you are using. See – Dale M Apr 16 '13 at 5:48

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