My Math Forum  

Go Back   My Math Forum > High School Math Forum > Probability and Statistics

Probability and Statistics Basic Probability and Statistics Math Forum


Reply
 
LinkBack Thread Tools Display Modes
April 17th, 2017, 02:21 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.
Kallaste is offline  
 
April 17th, 2017, 06:34 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.
Kallaste is offline  
April 17th, 2017, 10:54 PM   #3
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: CA

Posts: 1,238
Thanks: 637

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*}$
romsek is offline  
April 18th, 2017, 02:37 PM   #4
Senior Member
 
Joined: Oct 2013
From: New York, USA

Posts: 522
Thanks: 74

Quote:
Originally Posted by Kallaste View Post
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%
B1 has a 1/3 chance at winning by picking a number greater than 40. Given that both numbers are 40 or lower, there are 40^2 = 1,600 possibilities.

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.
EvanJ is offline  
April 19th, 2017, 01:34 AM   #5
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: CA

Posts: 1,238
Thanks: 637

Quote:
Originally Posted by EvanJ View Post
(3 repeating)
of course
romsek is offline  
Reply

  My Math Forum > High School Math Forum > Probability and Statistics

Tags
problem, understand



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Ok, I don't understand what this problem is asking me to do. Angelwngs26 Elementary Math 4 April 10th, 2016 08:38 PM
Help me understand this problem blackhype Calculus 3 March 24th, 2015 04:35 PM
I don't understand this problem drunkelf22 Elementary Math 10 February 1st, 2015 10:22 AM
Help me understand this problem. SMARTYPANTS Algebra 11 June 18th, 2014 10:15 PM
Couldn't understand this problem.. pnf123 Calculus 0 April 4th, 2014 10:02 PM





Copyright © 2017 My Math Forum. All rights reserved.