Just wanted some input for the following expected value questions:

Suppose a person offers to play a game with you. In this game, when you draw a card from a standard 52-card deck, if the card is a face card you win $3, and if the card is anything else you lose 1 dollar. If you agree to play the game, what is your expected gain or loss?

I computed the following:

Total number of face cards in a deck = 12.

So the expected value would be:

$\frac{1}{52}\left[\left(3\cdot 12\right)+\left(-1\cdot 40\right)\right]$

...which comes out to -1/13, or an expected loss of around 7.7 cents, right?

The other question I have is a lot harder, and I'm not sure how to solve it:

A fair coin is tossed until either a head comes up or four tails are obtained. What is the expected number of tosses?

I'm used to expected value questions where the number of times you participate in the activity is explicitly givens, so I'm not sure how to begin this one.

  • 2
    $\begingroup$ First one looks good. For the second, just notice that the game can't take more than four tosses to decide. Just work out all four cases by hand. $\endgroup$ – lulu Dec 10 '16 at 23:24

First one is right. For the second one note the following: the probability that the game ends after one toss is the same as the probabilty for head, i.e. $\frac 12$. So far, so goood.

If the game ends after two tosses, it has to be that the first toss was tail (because otherwise the game would have ended after the first toss) and the second toss was head (because otherwise the game would still continue). So the probability that the game ends after two tosses is the probability for T H, which is $\frac 12 \cdot \frac 12= \frac14$.

Similarly, for three tosses we have to have T T H , so the probability is $\Bigl(\frac12\Bigr)^3=\frac 18$.

Finally, if the game ends after exactly four tosses, then the first three tosses had to result in tail (because otherwise the game would have stoped before). And the forth toss can either be heads or tails: if it is heads then the came stops, and if it is tails then I have four tails in a row. Hence the probability is $\Bigl(\frac12\Bigr)^3 \cdot \Bigl(\frac12 +\frac12 \Bigr)=\frac 18$.

The exptected value is then given by: $$ E[X]= 1 \cdot \frac 12 + 2 \cdot \frac14 + 3\cdot \frac18 + 4\cdot \frac18= 1.875 $$

  • $\begingroup$ Ah I see! Thank you for this. So in this case, the "value" we are computing is the number of tosses, right? $\endgroup$ – Chris T Dec 11 '16 at 18:57
  • $\begingroup$ @ChrisT right :) $\endgroup$ – Cettt Dec 12 '16 at 11:47

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