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April 17th, 2017, 03:21 PM  #1 
Newbie Joined: Apr 2017 From: Washington State Posts: 2 Thanks: 0  A problem I want to understand
Hello, I have been thinking of a problem, and I want to know how to express and solve it. Say I have two guys, A and B, who are in a number drawing contest. They pick numbers out of hats, and whoever picks the highest number wins. But they don't use the same hat. One of them picks from a hat with one range of numbers, and the other one picks from a hat with another range of numbers. So the game is rigged. The way I am thinking of it, we have two ranges of numbers. The lower bound of each range is 0, and the upper bounds of each range we will call A1 and B1. Also, for simplicity, say the sum A1 + B1 = 100. What I am interested in is, if the number drawing action is repeated millions of times, what will be the percentage of wins for A, for B, and the number of ties? The law of large numbers says the results should approach their theoretical probability. But what is that probability for each outcome, and how is it calculated? I would expect that if we set A1 = 50 and B1 = 50, we would have about the same number of wins for A and B. Likewise, if we set A1 = 90 and B1 = 10, we would have about 90 % wins for A and 10 % for B. But what about the number of ties? They would certainly be dependent on the overlap of the ranges, but how, specifically? Can anyone give me any insight into this? Thanks. 
April 17th, 2017, 07:34 PM  #2 
Newbie Joined: Apr 2017 From: Washington State Posts: 2 Thanks: 0 
More information: So, I know it has to do with the overlap between sets. In the case of A1 = 40 and B1 = 60, I think B would win about 60 percent of the time, and A would win 40  overlap/2 percent of the time. Overlap/2 percent would be ties. So if the overlap is 20, then A would win half of those 20 and B would win half of those 20, leaving the stats at A: 30% wins B: 60% wins ties: 10% This makes intuitive sense to me, but I do not know how to express it properly in terms of probability laws. Or maybe it is wrong altogether. 
April 17th, 2017, 11:54 PM  #3 
Senior Member Joined: Sep 2015 From: USA Posts: 1,757 Thanks: 900 
what you've got here is random variables $A,~B$ $A \sim U[0,A_{max}],~B \sim U[0,B_{max}]$ where $U[0,N]$ is a discrete uniform distribution on $0,1,\dots, N$ $A, ~B$ are independent. let's assume w/o loss of generality that $A_{max} \leq B_{max}$ the object you want the distribution of is a new random variable $Z = \max(A,B)$ $\begin{align*} F_Z(z) =&P[Z < z] \\ \\ &= P[\max(A,B) < z] \\ \\ &=P[(A<z)\wedge (B<z)]\\ \\ &=P[A<z]P[B<z] \\ \\ &=\begin{cases} 0 &z < 0 \\ \dfrac{(z+1)^2}{(A_{max}+1)(B_{max}+1)} &0 \leq z <A_{max} \\ \dfrac{z+1}{B_{max}+1} &A_{max} \leq z \leq B_{max} \\ 1 &B_{max} < z \end{cases} \\ \\ f_Z(z) &= \dfrac{dF}{dz} =\begin{cases} 0 &z < 0 \\ \dfrac{2(z+1)}{(A_{max}+1)(B_{max}+1)} &0 \leq z <A_{max} \\ \dfrac{1}{B_{max}+1} &A_{max} \leq z \leq B_{max} \\ 0 &B_{max} < z \end{cases} \\ \\ \end{align*}$ 
April 18th, 2017, 03:37 PM  #4  
Senior Member Joined: Oct 2013 From: New York, USA Posts: 582 Thanks: 81  Quote:
A1 is higher 780 times There are 40 ties B1 is higher 780 times Multiply those probabilities by 2/3 and adding the 1/3 where B1 always wins makes: P(A1 wins) = 39/120 = 0.325 P(tie) = 2/120 = 0.1(6 repeating) = 1/6 P(B1 wins) = 79/120 = 0.658(3 repeating) I don't know how to make the repeating decimal bar, so I put it in parenthesis. My method may not be as easy to write a generic form for, but my explanation requires less mathematical knowledge than the derivative used by romsek.  
April 19th, 2017, 02:34 AM  #5 
Senior Member Joined: Sep 2015 From: USA Posts: 1,757 Thanks: 900  

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