Author Topic: A mathematical conundrum  (Read 2902 times)

Mike Morgan

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A mathematical conundrum
« on: December 17, 2009, 07:39:35 PM »
True story.

Last years works Christam Raffle sold in the region of 7,500 tickets as did this years.  Absolutely no hint of a fiddle, but the same person won both raffles.

Can anyone work out the odds? For one year it has to be 7,500:1, but to win both...
Would it be as simple as 7,500 squared?

ps just for context, last years prize was a Ford Ka and this years was ?5k

Allan Heron

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Re: A mathematical conundrum
« Reply #1 on: December 17, 2009, 08:50:52 PM »
I think it is 7500 squared
It's Just My View

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Pete T

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Re: A mathematical conundrum
« Reply #2 on: December 18, 2009, 02:12:44 AM »
Ha, think logically, and with simpler numbers.
A one in two chance of winning in one year gives either a win or a loss
a one in two chance in the second year gives either a win or a loss, so we have four possabilities over two years, win win, loss loss, win loss, or loss win. A one in four chance of winning both years. So it is (technically speaking) a one in 7,499 times 7,499 = one in 56,235,001 chance of winning both years.
The chances of winning in either year remain the same, even for the winner of the first year. Toss a coin, it comes up heads 10 times, whats the odds for the 11th throw. A head or a tail, one in two.

A nice trick for xmas parties.
Reciprocal numbers
142857
Multiply by 2, 3, 4, 5, 6, and 7 and see how the sequence changes..
Virgin places don't mean a thing to people who never bring their hearts along.

Dai Port Talbot

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Re: A mathematical conundrum
« Reply #3 on: December 18, 2009, 03:46:31 AM »
The odds change over time. Before the first raffle it was 7499 squared to 1. After the first raffle it was 7499 to 1 and after the second raffle it was infinity to 1. But only in the infinite number of universes in which these outcomes ocurred. There is a larger infinity in which, for instance, that particular individual did not participate. Both infinities include universes where I found myself in a sandwich with Kate Bush and Sonja Kristina in their prime. Didn't happen, how much bad luck can you have?
« Last Edit: December 18, 2009, 03:49:14 AM by Dai Port Talbot »

Dave Bardsley

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Re: A mathematical conundrum
« Reply #4 on: December 18, 2009, 11:37:33 AM »
I think Pete's theory works, but only if he bought one ticket for each year. 
If he bought two tickets each year would the overall odds be reduced by a half or a quarter? 
Like Dai says, does it make a difference if he bought the ticket(s) at the stat of the first year or at the start of each raffle.

I failed statisics at college.  One of my better qualities.

Mike Morgan

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Re: A mathematical conundrum
« Reply #5 on: December 19, 2009, 11:40:53 AM »
For the sake of what may pass for sanity(!) I posed the question on the basis that each of the ticket holders purchased one ticket.  I know that is not the case, but it's a daft enough question without introducing further complexities into the equation. 

ps I bought another pasty for my return home last night, but this week still didn't eat it or stop off in the chippy on the way home.  You couldn't make it up!  Wild.