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Recall my weighted random walk:
“ | Starting at initial value j, take a weighted random walk as follows. At each step add +6.375 with 80% probability, -25 with 20% probability. This time the game does not end. | ” |
Now here are generalized question:
1.
a. For a positive integer k, what is the probability p that I will touch or go below j-k at least once in the next n steps of my walk?
b. What is the expected current value after n steps, starting at j?
c. What is the standard deviation of this value? Does it roughly follow a bell curve (is normally distributed) or if not, how is it distributed?
2.
a. For a positive integer k, what number of steps m imply a probability p of 0.5 that I will touch or go below j-k at least once during the m steps, starting at value j? And for any probability p?
b. If m is however many steps result in touching or going below j-k, what is the average value and standard deviation of m (Sorry if this is in fact the same question as question part (a), that part is about probability 0.5, this one is the average value of m before it in fact touches. I guess this might be the same?)
c. What is the standard deviation of the two values in question part (b)? Do they roughly follow a bell curve (are approximately normally distributed) or if not, how are they distributed?
3. For a given positive integer k, if I would like to take q steps with a p probability that I will never touch or go below j-k, what should j be?
If it is much easier to answer the above questions if j = k (meaning that I touch or go below zero by touching or going below j-k) then you can go ahead and do that.
I have very little background in statistics and I'm trying to find the above answers in monte carlo form, but it's just taking too long. I'm hoping someone who knows statistics can give me formulas. Thanks!
--80.99.254.208 (talk) 06:31, 7 March 2012 (UTC)
I can't give you the formulas, not sure there are simple formulas, but instead of using the monte carlo method, you can calculate all propabilities starting with one throw, then two, then three etc. writing a program for this shouldn't be too hard. Let's say the chance of winning p times out of N is (p,N). you know (1,1)=0.8 and (0.1)=0.2 ; the you calculate (0,2) (1,2) and (2.2) and so on. In general, (p,N)= 0.2 * (p,N-1) + 0.8 * (p-1,N-1). Start of with 0$ to keep it simple, you only need to know how much money is lost or won for every case (p,N) namely 6.375*(N-p) - 25 * p; calculate up till a reasonable value N and use that array to get the numbers you want. You'll have to do some extra work to get the chance of going below a value at least once, because you'll have to exclude values (p,N) that reached that limit so you don't use them for calculating (p+1,N+1), winning after you went bust doesn't count. The average value after N will be N * 0.02 (? I think); the standdev will be almost the same as for the case 6.25 and -25 (with mean value 0). The difference with a fair game (=average=0) is so small (diference between winning 50 times or 51 times) that you'll have to go to high number of throws before you'll notice much difference. Since a player in a fair game having limited wealth will always go broke against a bank with unlimited capital, it's no surprise that your simulation went bust so many times.. 84.197.178.75 (talk) 14:08, 7 March 2012 (UTC)