comparing odds of two gaussians

Question

I am trying to compare the odds of two events happening (what are the odds of one happening first). I know that the first one occurs in an average of 10 months with a sigma of 3 months. The second occurs in an average of 84 months with a sigma of 30 months. First, is there an analytical way of solving this? Second, I attempted to solve this by simulating it (Monte-Carlo), but it is not giving me the expected results. In particular, if I ask it to just simulate just one of the events, the reported average is way before the expected average. I tried simulating both with a gaussian and a cumulative gaussian. What am I missing?

import math
import random
from scipy.stats import norm

otherStart = 2*365                              #estimate for earliest donation
otherEnd = 8*365                                #estimate for latest donation
otherMu = (otherStart + otherEnd)/2             #average
otherSig = (otherEnd - otherStart)/5            #get some semblance of a sigma

meStart = 7*31
meEnd = 13*31
meMu = (meStart + meEnd)/2
meSig = (meEnd - meStart)/5

def cdfDist(x,mu,sig):                          #cdf function
    y = norm.cdf(x,mu,sig)
    return y

#def gaussDist(x,mu,sig):
#    y = (1/(sig*math.sqrt(2*math.pi)))*math.exp(-0.5*((x-mu)/sig)*((x-mu)/sig))
#    return y

def rollDiceUntilWin():
    day = 0                                             #start at day 0
    while True:
        chancesOther = random.uniform(0, 1)                 #roll a dice
        otherOdds = cdfDist(day+2*365,otherMu,otherSig)     #find odds for that day (it has been 2 years since that counter started)
#        otherOdds = gaussDist(day+2*365,otherMu,otherSig)
        if (chancesOther < otherOdds):                          #did it win
#            print(day)
            return 1
        chancesMe = random.uniform(0,1)
        meOdds = cdfDist(day,meMu,meSig)
        if(chancesMe < meOdds):
#            print(day)
            return 2
        day = day + 1                                   #if noone won, go to next day
#        if (day > 10000):
#            day = 0

if __name__ == "__main__":
    numRolls = 100                                      #number of rolls
    meGive = 0                                          #counter of wins
    otherGive = 0
    for i in range(numRolls):
        outCome = rollDiceUntilWin()
        if outCome == 1:
            otherGive = otherGive + 1                   #if win, add 1
        if outCome == 2:
            meGive = meGive + 1

    chances = meGive / (meGive + otherGive)             #get chances
    print(chances)

Answer 1

Comment: the time to a future event cannot be gaussian because a gaussian random variable has a non-zero probability of being negative, but it could be a reasonable approximation in some cases.

Do you want the probability or the odds of the event?
If $X$ is gaussian with mean 10 and standard deviation 3 and $Y$ is gaussian with mean 84 and standard deviation 30, and they are independent, then $X-Y$ is gaussian with mean -20 and standard deviation $\sqrt{10^2+30^3}$. The probability that $X<Y$ is $$P[X<Y]=P[X-Y<0]=P\left[\frac{X-Y-(-20)}{\sqrt{10^2+30^2}}<\frac{0-(-20)}{\sqrt{10^2+30^2}}\right]$$ Now, find the probability that a standard normal random variable is less than the number $\frac{0-(-20)}{\sqrt{10^2+30^2}}$. After you have found the probability (p) of the event, the odds of the event is p/(1-p).