Time for a semi-annual talk on statistics. First there was the "Next" Markov chain problem, now we have the Secret Santa problem.
The question is : in a "secret santa" in an office, what are the chances that two people draw each other? In a pool of our readers, what are the chances that Derek draws Leland and Leland draws Derek?
So - let's say N is the number of people in the pool
Chances that Derek picks Leland: 1/(N-1)
Chances that Leland picks Derek: 1/(N-1)
So chances of a specific 2-pair is: 1/(N-1)^2
But how many possible pairs are there? Isn't this (N)*(N-1)/2? (Or, in statistics, N-pick-2)
So we mulitply the chance of a specific 2-pair by all the possible pairs?
(N)*(N-1)/2*(1/(N-1)^2)) = N/(2*(N-1))
Which works if N =2; it equals 1, which it should.
N=3; probability is 3/4. Hmmmmm....seems high.
So if N = 10, the chance of a pair is 10/18? That seems high, too. Over 50%?
Any help out there? Don - can you set up another spreadsheet to look at this?
Class is over. This reminds me of my sophomore year - when I actually was thinking about become a math major - and I took a "Discrete Math" course. The professor tried to do a proof, and after 45 minutes of work on the board, he came to this conclusion:
1 = 0 .
He quickly said: "Class dismissed". Right then I decided against being a math major.
7 comments:
I think you're overcomplicating things. The chance of Derek drawing Leland is 1/(N-1). The chance of Leland drawing Derek is also 1/(N-1). The chance of both of those events happening together is exactly the product of the chances of each event happening individually. Which means the chance of both happening is 1/(N-1)^2.
So for N=2 we get 1, still correct. For N=3 we get 1/4. For N=10 we get 1/81. These numbers seem much more realistic.
I think what threw you off was adding in the total number of possible 2-pairs that could occur. The total number of possible 2-pairs is completely irrelevant to the chances of any individual 2-pair occurring.
that's my boy!!!
Oops, looks like I answered the wrong question. I answered the first part of the question which Eric had already detailed. The chance of two specific people getting each other is 1/(N-1)^2.
Eric, you are 100% correct in your final answer.
The chance of Derek getting Leland and vice versa is 1/(N-1)^2. The chance of Derek getting Leland and vice versa OR Derek getting Eric and vice versa is the sum of the chances for each individual event:
2*(1/(N-1)^2) = 2/(N-1)^2
The chance of any one of three possible pairs would be:
3/(N-1)^2
Expand that to the total possible number of pairs and you get N/(2*(N-1)). Exactly as Eric (or Captain Stats as I like to call him) predicted.
We can verify this by taking an example using 4 people:
Chance of 1<->2 = 1/9
Chance of 1<->3 = 1/9
Chance of 1<->4 = 1/9
Chance of 2<->3 = 1/9
Chance of 2<->4 = 1/9
Chance of 3<->4 = 1/9
So the chance of ANY ONE of those happening is:
1/9 + 1/9 + 1/9 + 1/9 + 1/9 + 1/9 = 6/9 = 2/3
N/(2*(N-1)) = 4/2*(4-1) = 4/6 = 2/3
Well, that's if the question is "What are the chances that Derek and Leland end up with each other?"
But would a pair of Eric-Jon count? I thought the question was to see the chances for any pair, not just for one specific pair.
I think you are absolutely correct for the first question, though.
Oops - strike that last comment. I posted before Brad's next posting was up!
Busy website today...
Having seen Derek and Leland together in college and disregarding for a moment that Leland is married and they live on opposite sides of the country, I'm inclined to say that the chances of Derek and Leland "ending up together" is pretty high.
Of course, I'm basing this solely on the fact that Derek is gay.
hfwThe is the nerdiest and gayest thing ever.
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