Question

The chance of a white Christmas (defined as at least one inch of snow on the ground) in Spokane is approximately 60%. Run simulations for each of the following.

a) Run a simulation for 10 years and record how many times a white Christmas occurs. Describe and record your results below.
b) Repeat the above simulation 9 more times. Record your results below.
c) Compare your results with the 60% theoretical probability and comment.

Answer 1

I am guessing what is meant is that you use some random number generator to simulate the random (or pseudorandom) processes that lead to a WCinS, record the results, and then compare to the probability. The idea is probably to demonstrate to you that the probability of an event computed from a finite length of a stochastic sequence is rarely equal to the theoretical probability that would be computed from an infinite length of that process, and in fact approaches it quite slowly -- the error goes something like 1/sqrt(N), where N is the length of the process, IIRC.

So, for example, if you had a 10-sided die, you could roll it, and say if the number comes up 1 through 6 you get WCinS, and if it comes up 7 through 10 you get !WCinS. You do this 10 times in a row, write down the results, and approximate the probability of WCinS as n(WCinS)/N, where N=10 is the number of times you roll the die, and n(WCinS) is the number of times it came up WCinS. You're unlikely to get exactly 0.60. Then you roll the die 9 more times, record the results, and again compute p(WCinS) the same way. You are again unlikely to get exactly 0.60, but you are likely to get closer to that number.

It's hard to get a physical random number generator, like a coin or die, that will cleanly give you an outcome that has a 60% chance of success, because 60 is a weird number. You may have to use a random number generator on a computer. If I write a little code using the random number generator on my computer I get the following stochastic sequence for WCinS:
year 0: WCinS = 1, p(WCinS) = 1/1 = 1.000000
year 1: WCinS = 1, p(WCinS) = 2/2 = 1.000000
year 2: WCinS = 1, p(WCinS) = 3/3 = 1.000000
year 3: WCinS = 1, p(WCinS) = 4/4 = 1.000000
year 4: WCinS = 1, p(WCinS) = 5/5 = 1.000000
year 5: WCinS = 1, p(WCinS) = 6/6 = 1.000000
year 6: WCinS = 1, p(WCinS) = 7/7 = 1.000000
year 7: WCinS = 1, p(WCinS) = 8/8 = 1.000000
year 8: WCinS = 1, p(WCinS) = 9/9 = 1.000000
year 9: WCinS = 1, p(WCinS) = 10/10 = 1.000000
year 19: WCinS = 1, p(WCinS) = 17/20 = 0.850000
year 29: WCinS = 0, p(WCinS) = 22/30 = 0.733333
year 39: WCinS = 0, p(WCinS) = 27/40 = 0.675000
year 49: WCinS = 1, p(WCinS) = 33/50 = 0.660000
year 59: WCinS = 0, p(WCinS) = 39/60 = 0.650000
year 69: WCinS = 0, p(WCinS) = 44/70 = 0.628571
year 79: WCinS = 1, p(WCinS) = 51/80 = 0.637500
year 89: WCinS = 1, p(WCinS) = 56/90 = 0.622222
year 99: WCinS = 0, p(WCinS) = 64/100 = 0.640000
year 109: WCinS = 0, p(WCinS) = 69/110 = 0.627273
year 199: WCinS = 1, p(WCinS) = 122/200 = 0.610000
year 299: WCinS = 1, p(WCinS) = 184/300 = 0.613333
year 399: WCinS = 1, p(WCinS) = 245/400 = 0.612500
year 499: WCinS = 0, p(WCinS) = 310/500 = 0.620000
year 599: WCinS = 0, p(WCinS) = 373/600 = 0.621667
year 699: WCinS = 0, p(WCinS) = 430/700 = 0.614286
year 799: WCinS = 1, p(WCinS) = 491/800 = 0.613750
year 899: WCinS = 1, p(WCinS) = 545/900 = 0.605556
year 999: WCinS = 1, p(WCinS) = 608/1000 = 0.608000

As you can see, I get WCinS for the first 10 years in a row, so if I lived in Seattle I'd conclude p(WCinS) = 1 after 10 long years. I only get something approaching the true probability if I live in Seattle and observe a thousand years. As I said, this demonstrate the extremely slow approach of probabilities computed from stochastic processes to their asymptotic result.