My Math Forum  

Go Back   My Math Forum > College Math Forum > Advanced Statistics

Advanced Statistics Advanced Probability and Statistics Math Forum


Reply
 
LinkBack Thread Tools Display Modes
February 4th, 2012, 05:21 AM   #1
Newbie
 
Joined: Feb 2012

Posts: 1
Thanks: 0

Solve my argument

I play lawn bowls, and when they run knock out competitions they invariably have to have whats called an alignment round in the first round to get the number of teams back to a number that is 2^something so that it reduces back from 32 to 16 to 8 to 4 to 2 to a winner...... in tennis when they have first round byes they are usually given to the seeded players and spread evenly through the draw. In bowls, it is a random draw so that the 2 best teams could play each other in the first round.
If there was say 48 teams in a bowls competition they would have to reduce that to 32 teams and to do this 32 teams would play a first round match and the other 16 teams would have a bye. In bowls this alignment round is invariably played in the top half of the draw, so that the top half of the draw has 32 teams in it, and the bottom half of the draw has 16 teams in it with a first round bye.

Everyone is arguing that it doesnt matter as the draw is random and the same number of teams have to play a first round match! But my argument is that the person who makes the final from the top half of the draw is the best of 32 players and the player from the bottom half is the best of 16 players. Or if they were all equal ability then the chance of making the final from the top half of the draw is 1/32 and the bottom half is 1/16.

All players have differnt abilities so therefore there is a chance that all the better players are drawn in one side or the other or even playing each other in the first round! But the probability is that there will be 2 good teams in the top half of the draw for every good team in the bottom half of the draw.

How do I explain that the first round byes should be spread evenly through the draw?


Cosi
Cosi is offline  
 
February 4th, 2012, 08:49 PM   #2
Math Team
 
agentredlum's Avatar
 
Joined: Jul 2011
From: North America, 42nd parallel

Posts: 3,372
Thanks: 233

Re: Solve my argument

It may be informative to consider how many games a player has to win to finish 1st, so let's assume the draws have already been made.

From the top draw a player must win 6 games. If the probability is 1/2 for each game we have (1/2)^6 = 1/64. So a player from the top draw has a 1 in 64 chance of winning the final.

From the bottom draw a player must win 5 games. If the probability is 1/2 for each game we have (1/2)^5 = 1/32. So a player from the bottom draw has a 1 in 32 chance of winning the final.

Apparently, your chances DOUBLE if you get the first round buy.


Let's consider what happens before the initial draws are made. You have a 2/3 chance of being in the top draw. If that happens then you have a 1/64 chance of winning the final so (2/3)(1/64) = 1/96

You have a 1/3 chance of getting a buy. If that happens then you have a 1/32 chance of winning the final so (1/3)(1/32 ) = 1/96

Hmmmm..... the numbers seem offfff... apparently everyone has equal probability in the very beginning but 1/96 seems too low.
agentredlum is offline  
Reply

  My Math Forum > College Math Forum > Advanced Statistics

Tags
argument, solve



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Is this a valid argument? edwinh Calculus 3 August 24th, 2012 12:28 AM
combinatorial argument lamhmh Applied Math 2 July 28th, 2011 02:03 AM
What is the argument of z near infinity Singularity Complex Analysis 0 October 13th, 2010 03:50 PM
problem with Argument timlo Complex Analysis 1 July 24th, 2010 07:41 AM
Find the argument! soulnoob Complex Analysis 3 January 9th, 2010 02:17 PM





Copyright © 2019 My Math Forum. All rights reserved.