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September 3rd, 2009, 06:42 AM  #1 
Newbie Joined: Sep 2009 Posts: 2 Thanks: 0  Pick the correct number from the correct group
Forgive me if I've chosen the wrong forum, but I have absolutely no background in mathematics to even realize where my question fits. I'll bravely ask anyway. You have 10 groups of numbers: Group 01: { 91, 72, 59, 57, 54, 48, 35, 32, 22, 21 } Group 02: { 91, 76, 73, 51, 40, 39, 35, 32, 30, 23 } Group 03: { 113, 103, 49, 46, 39, 36, 35, 33, 31, 28 } Group 04: { 97, 73, 71, 59, 59, 59, 57, 51, 40, 40 } Group 05: { 59, 59, 42, 39, 36, 34, 22, 21, 21, 20 } Group 06: { 73, 42, 41, 36, 36, 31, 30, 28, 27, 21 } Group 07: { 86, 71, 65, 59, 57, 57, 52, 45, 42, 36 } Group 08: { 78, 63, 48, 46, 46, 37, 35, 29, 26, 25 } Group 09: { 82, 82, 59, 53, 51, 43, 37, 32, 28, 27 } Group 10: { 74, 73, 72, 40, 32, 32, 28, 23, 22, 21 } The task is that by random order, 10 people will begin picking a number, until all the numbers have been picked. Each person can only pick one number from each group. After all the numbers have been selected, the person with the highest sum of his or her numbers is the winner. Without thinking, most people will start picking the highest number available from a group that they have not yet selected from. My question is, is there some kind of formula I can come up with to get an advantage? And how does this formula change as numbers are selected ? How much more valuable is that 73 in Group 6 compared to that 73 in Group 4? Does anyone have any ideas that may help give me an advantage in a game such as this? Thanks in advance. Last edited by skipjack; June 19th, 2015 at 07:00 PM. 
September 3rd, 2009, 09:35 AM  #2  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Pick the correct number from the correct group Quote:
1. Each person puts their name on a slip of paper. The papers are removed at random; people line up in the order in which their names are drawn. The first, second, ..., tenth people pick, then start over from the first. 2. Each person puts their name on a slip of paper. The papers are removed at random; people line up in the order in which their names are drawn. The first, second, ..., tenth people pick. Now the slips of paper are put in the hat again and the order is determined again. 3. Each person puts their name on ten slips of paper. People pick as their names are drawn. 4. Each person puts their name on a slip of paper. The papers are removed at random; people line up in the order in which their names are drawn. The first, second, ..., tenth people pick. Next the tenth, ninth, ..., first people pick, then start over from the first. 5. Each person puts their name on a slip of paper. The papers are removed at random; people line up in the order in which their names are drawn. The first, second, ..., tenth people pick. Next the tenth, ninth, ..., first people pick. Then the process is repeated five times with the names in the hat. . . . Regardless, what matters is not how high a given number is but how different it is from the second, third, ... highest number remaining from that group. So given a choice between 97, 92, 90, 90 and 50, 30, 25, 20 I would certainly choose the 50 over the 97. Last edited by skipjack; June 19th, 2015 at 07:02 PM.  
September 3rd, 2009, 11:20 AM  #3 
Newbie Joined: Sep 2009 Posts: 2 Thanks: 0  Re: Pick the correct number from the correct group
Thanks for the reply, Actually each person would be placed in random order initially. They would pick in this order for round one, then reverse in round two and so on through the 10 rounds. if the sets were A: { 90, 85, 80, 80 } and B: {50, 30, 20, 20 } obviously that 50 in 'B' would be highly valuable. I believe with the sample groups in the original post, the differences are quite a bit more subtle, and I was just trying to come up with some sort of formula that could equate the differences in numbers across the sets, and indicate which one is the most valuable. Then, I would need the formula to adapt as numbers are selected so when it was my turn to pick I would always know the most valuable one remaining. 

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