Question 1012911
<font face="Times New Roman" size="+2">


If you roll a fair die once, the probability of getting a 1 or a 2 is 1/3.  The probability of 4 successes in 4 trials where the probability of success on one trial is:  


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_4\left(4,\frac{1}{3}\right)\ =\ {{4}\choose{4}}\left(\frac{1}{3}\right)^4\left(1\,-\,\frac{1}{3}\right)^{4\,-\,4}\ =\ \left(\frac{1}{3}\right)^4\ =\ \frac{1}{81}]


So over the long run you win 1 time every 81 times you play, and you lose 80 times every 81 times you play.  When you win, you pay $10 and get $50 back for a net gain of $40.  When you lose, you pay $10 and get nothing back for a net gain of -$10.


Hence your per game net is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{81}\,\times\,$50\ +\ \frac{80}{81}\,\times\,(-$10)\ \approx\ -$9.3827]


And if you played 810 games, you should expect to lose about $7600 overall.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

</font>