1. ## expected value

you bet $5 in a game and there is a 251/ 495 probability that you wil lose that$5 and 244/495 probabilty taht you will win that you will win $5. given that your expected value is -7.07 cents, how much would you lose in the long run? this is my working: 1st round: win -7.07 cents 2nd rounnd: win -7.07 cents nth round: win -7.07 cents thus the answer should be a GP of 7.07 /( 1- 7.07) ... but my answer in my notes states that it is 1.4 cents. may i know where i have gone wrong? thanks! 2. If you have an expected loss of 7.07 cents on each play, the total is NOT a geometric progression. It is just -7.07-7.07-7.07...= -7.07n cents where n is the number of times you played. 3. then why would the amt lose in the long run be 1.4cents? how do i get that value? 4. Originally Posted by alexandrabel90 you bet$5 in a game and there is a 251/ 495 probability that you wil lose that $5 and 244/495 probabilty taht you will win that you will win$5. given that your expected value is -7.07 cents, how much would you lose in the long run?

this is my working:

1st round: win -7.07 cents
2nd rounnd: win -7.07 cents

nth round: win -7.07 cents

thus the answer should be a GP of 7.07 /( 1- 7.07) ...

but my answer in my notes states that it is 1.4 cents.

may i know where i have gone wrong? thanks!
I think the book is giving an intermediate answer.
The problem seems unrealistic, though
(in terms of either losing the entire 5 dollars or not).

Round1 E(loss) = $\displaystyle \frac{7}{495}5\ dollars$

Round2 E(loss) = $\displaystyle \frac{7}{495}\left(1-\frac{7}{495}\right)5=\left(\frac{7}{495}-\left(\frac{7}{495}\right)^2\right)5\ dollars$

Round3 E(loss) = $\displaystyle \frac{7}{495}\left(1-\frac{7}{495}+\left(\frac{7}{495}\right)^2\right)5 =\left(\frac{7}{495}-\left(\frac{7}{495}\right)^2+\left(\frac{7}{495}\r ight)^3\right)5\ dollars$

E(loss) long term = $\displaystyle \left(\frac{7}{495}-\left(\frac{7}{495}\right)^2+\left(\frac{7}{495}\r ight)^3-\left(\frac{7}{495}\right)^4+...\right)5\ dollars$

The two geometric sequences are

$\displaystyle \frac{7}{495}+\left(\frac{7}{495}\right)^3+\left(\ frac{7}{495}\right)^5+...$

$\displaystyle -\left(\frac{7}{495}\right)^2-\left(\frac{7}{495}\right)^4-\left(\frac{7}{495}\right)^6-...$

The sum to infinity of the first is $\displaystyle \frac{\frac{7}{495}}{1-\left(\frac{7}{495}\right)^2}=\frac{0.0141414}{0.9 998}$

The sum to infinity of the second is $\displaystyle -\frac{\left(\frac{7}{495}\right)^2}{1-\left(\frac{7}{495}\right)^2}=\frac{0.00019998}{0. 9998}$

Hence the expected loss is 5 dollars multiplied by $\displaystyle \frac{0.01414-0.00019998}{0.9998}=(0.01394)5\ dollars\ =\ 1.4c\ per\ dollar$

5. may i know why you how you get the equation for round 2 and hence forth?

6. i think the last working doesnt give 1.4cents. 0.01395 (5) = 0.0697 dollars = 6.97 cents

7. Originally Posted by alexandrabel90
may i know why you how you get the equation for round 2 and hence forth?
The initial E(loss) is the "round 1" loss from $5. This value shrinks from round to round, while the probability of loss remains constant.$\displaystyle \frac{244}{495}5-\frac{251}{495}5=-\frac{7}{495}5=-0.0707\ dollars$Then in round 2 (being purely mathematical, considering that 7.07 cents had been lost), there is 5 dollars minus 7.07 cents to bet with, to put it that way! The expected loss is not 7.07 cents now, it's$\displaystyle \frac{7}{495}\$ of the "balance".

Continuing on in this way develops the pair of geometric series.

8. Originally Posted by alexandrabel90
i think the last working doesnt give 1.4cents. 0.01395 (5) = 0.0697 dollars = 6.97 cents
That's what I meant by 1.4c per dollar.
I think the book answer is pointing to this rather than a final answer of 1.4c.