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View Full Version : Function relating SD,EV, and probability of random event? Long


GTSamIAm
06-29-2005, 11:54 PM
I realize this does not make seem to make much sense at first. Say my friend and I decide to "gamble" with a die. If it comes 1-4, he pays me a dollar. That's the gamble, there is no advantage for my friend. So my EV would be:

4/6 * +$1 + 2/6 * $0 = +$2/3, call it about 67 cents

4 out of 6 times I win a dollar, 2 out 6 I win nothing. My question is this: Is there some way to figure out what the standard deviation on my EV by simply knowing the probability and EV? This only works on completely random events, however.

Is it possible to find the standard deviation of this without running trials over and over? Also, the intersection of terms is
confusing. How can you instantly convert probability into standard
deviation? And what does that mean? I do not know how to interpret or
explain my results.

My results showed that the standard deviation of a 0.5 probability is
0.5. This doesn't make any sense since you cannot have values lower
than 0 or greater than 1 for my example.

What I did was run 100,000 trials on a computer where I used a random
number generator. I made a number that kept a record of what happened
for each . I fed it the probability of an event, and then it generate
a random number. If the random number was less than the probability, I
put a 1 for that trial. If it was more, I put a 0 for that trial. Once
done, I computed the discovered probability (not the one already known
before the trials) and the standard deviation on the set of trials.

From there, I did this for several other probabilities to try and find a trend between the probability and the standard deviation. The standard deviation relates to the discovered probability and not the known probability. I found a decent approximation relating the two. So you could use the known probability to find what the standard deviation would converge to over time. This approximation is even over x = 0.5.

SD = a * ln(b * Prob) where a = 0.10224887 and b = 265.914801

These constants change depending on which data points I use. I used Prob = .01 and SD = .1 and Prob = .5 and SD = .5. The SD gets a little off around Prob = .25 by no more than +- 0.03 SD. This approximation is only valid for probabilities between 0 and 0.5. Between 0.5 and 1 you reflect this equation over x = 0.5.

AaronBrown
06-30-2005, 05:56 PM
If you have a gamble such that with probability p you win W and probability (1-p) you lose L, your expected value is:

p*W - (1-p)*L

your standard deviation is:

[p*(1-p)]^0.5*(W - L)

If you do the bet N independent times, you multiply your standard deviation by the square root of N.

I think in your example, W = 1 and L = 0. Then the formula reduces to the square root of p*(1-p).

GTSamIAm
07-02-2005, 04:37 PM
Wow. Thank you.