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Skeptoid
28th January 2003, 09:04 PM
I originally posted this question in a Banter Superbowl thread that quickly got buried on page 2. Perhaps it'll have a better chance of being seen by someone who can do probabilities here.

I was in a standard 100 square pool with the following variation: While our positions in the grid remained constant we were issued 4 sets of scores, one set for each quarter. What is the probability that I would receive an identical set of scores for two quarters? My intuition tells me that the probability is rather small and I'm suspicious of hanky-panky. Are my suspicions warranted?

Walter Wayne
28th January 2003, 10:48 PM
Chance of not having any duplicates given a hundred squares.

94.1% (1.00*.99*.98*.97)

Odds of of having at least one duplicate
5.9% or 1 in 17 (1 - 0.941)

What is the probability that I would receive an identical set of scores for two quarters?

5.8% (4 choose 2) * 100 * 99 * 98 / (100<sup>4</sup>)

In the second case I am disregarding the odds of getting 3 similar or 4 similar.

Skeptoid
28th January 2003, 11:57 PM
Thanks, Walt. Suppose, for the sake of discussion and to be more general, that I were in n distinct 100 square pools (the original problem being n = 4). Further suppose that n is even. What is the probability that half of my scores would be identical? Would it be the same (5.8%)?

29th January 2003, 08:11 AM
I'm suspicious of hanky-panky.

Always consider misfeasance before malfeasance. The world has more ignorance than evil.

Walter Wayne
29th January 2003, 02:41 PM
Hi Skeptoid

I won't have a chance to answer this before heading to the Amazing meeting. If it isn't answered when I get back, I will write down a general solution for any given n.

However the answer for n = 6 and 3 identical is

100*(6 choose 3)*99*98*97/100<sup>6</sup>

This equal 0.19 %

Didn't have a chance to check my math, but I will get back to you.

Walt

edited to add: if you are using this because you think someone fixed something, wait until someone else has checked this or I have a chance to check my math.

Skeptoid
29th January 2003, 05:38 PM
Thanks again, Walt.

In the Superbowl thread, Diezel said he's seen 2 of 4 the same before so, if your math is correct, then my suspicions of a fix are not really warranted. I was thinking that the probability was probably less than 1%. Not being a regular at the establishment where the pool was taken up, my eyebrow raised when I got a pair of really lousy scores (2 - 5). I'm not normally a suspicious person but when large amounts money are involved ...

A good lesson in critical thinking though. I was skeptical of my own intuition and sought answers for or against before I acted.

If you want to post the more general solution it would be interesting but I can see that as n gets large, the probability approaches 0.

Walter Wayne
6th February 2003, 07:23 PM
Looking at the above equations that I posted, I believe they are wrong. I believe they slightly over estimate the probability as they count some combinations twice.

To simplify the equation I will make sure I over estimate and just calculate an upper bound on the probability.

p(1/2 are the same) > 100*(N choose N/2)*99<sup>N/2</sup>/100<sup>N</sup>

Walt

Number Six
7th February 2003, 08:25 PM
Are you talking about buying one square in n distinct pools, each of which gives you a different set of numbers for each of the four quarters?

Or are you talking about buying n squares in one pool that gives you a different set of numbers for each of the four quarters?

Or are you talking about something else?

Figuring out exactly what the question is seems simple but a lot of the time it is harder than it seems.

xouper
11th February 2004, 11:11 AM
bump