bj42 said:
Thanks matt21. The win rate is $128 per hour and the actual SD is $1808 per hour. Not sure I follow your formula. What do you get as the probability of losing $15K or more in 15 hours?
Ok, so EV(15 hrs) = 15x$128 = $1,920
SD(1hr) = $1,808
SD(15hr) = $1,808x SQRT(15) = $7,002
to work out the probability of losing $15,000 or more we use the normal distribution.
where mean,x = $1,920 and SD=$7,002
=normdist(-15000,1920,7002,TRUE) [the formula that you enter into MS Excel]
=0.78%
there is a 0.78% chance of losing $15000 or more - i.e. this should happen about once in 128 [1/0.78] trips
=normdist(10000,1920,7002,TRUE) [the formula that you enter into MS Excel]
=87.57% - this figure represents the probability of winning $10k or less, or losing - in other words there is a 1-0.8757 i.e. 12.43% chance of winning $10k or more - so this should happen once in every 8 trips.
and now more interestingly.... EV for 50 trips - averaging 17 hours/tripEV = 50 x 17 x $128 = $108,800
SD = SQRT(40x17) x $1,808 = SQRT (680 hrs) x $1,808 = $47,146
immediately this indicates that you have reached N0, i.e. your EV ($108k) is now bigger than one standard deviation ($47k). In fact your EV is now equal to more than TWO standard deviations meaning that if you did a sample of 50 trips over and over, then only once in 100 samples of 50 trips should you make a zero profit or a loss. In other words your sample size is not really that small (680 hours!).
=normdist(0,108800,47146,TRUE) = 1.05%
you have clocked up sufficiently enough hours to really be able to expect to be well in the black.