Request: Probability Calculation

Soldato
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6 Jan 2013
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Rollergirl
Doing a sweep at work, choose 6 numbers for Lotto. Your numbers get recorded/highlighted on a spreadsheet each draw and the first person to get all their numbers highlighted will scoop the prize, which is everyone's entry fee.

£20 to enter
53 people have participated
Jackpot is £1060

2 participants, who are on opposite shifts, have chosen the same 6 numbers.

Edit to add: this isn't theoretical, it actually happened this week. :)
 
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Are you looking for the chances of someone picking a set of numbers that have already been picked?
Yes, essentially what's the chances of someone picking the same numbers as someone else.

Logic tells me that it's got to be the same odds of someone winning the lottery, because person "A" is representative of any random pick and person "B" is representative of the Lotto machine's pick. The fact that there's only 53 people in this pool is irrelevant.
 
I absolutely hate probability. I was a maths mechanics guy. Reading up on it a bit seems to suggest:

Calculation goes:
1. Total Number of Possible Sets: 45mil

2. Probability of No Two People Picking the Same Set:
The first person can pick any of the 45,057,474 sets, the second person must pick a different set, so they have 45,057,473 choices, and so on, up to the 53rd person who has 45,057,422 choices

P(all different)=45057473/45057474 × 45057472/45057474 etc

99.99694%

3. Probability of At Least Two People Picking the Same Set:
The complement of the probability that all pick different sets:
So, P(at least two same) = 1 - P(all different)

0.00306% (or 1 in 30,600).
I can't accept this in my head! I'm not saying it's incorrect, I'm just having trouble processing it.

The odds of me choosing a specific set of 6 numbers is 45 million to 1. The odds of the lottery machine choosing a specific set is the same, 45 million to 1. The odds of me choosing the same set as the machine draw is 45 million to 1, so logically player 1 and player 2 in our little syndicate must surely operate on the same odds basis?
 
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