#1
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Probability question
I am admittedly bad at probabilities, stats, etc.
Here is the question: In a 12 team fantasy football league, where the draft order is determined at random each season, what are the chances that 5 of 12 teams are picking in exactly the same slots as last season? What would be the expected number of places up or down in the draft order for a team to move given random drawings (can one even determine this)? Ballpark is fine, but even w/ my limited skill, it seems odd just looking at it. |
#2
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Re: Probability question
Take a string of 12 numbers :
1_2_3_4_5_6_7_8_9_10_11_12 There are 12! different ways to arrange these numbers . Assume that the first 5 are in the same order as last year . Again there is only 1 way that can happen . Now we can arrange the 7 numbers 6 through 12 which gives us an additional 7! combos . 12c5*7!/12! = 0.00833 or just under 1 % . b) E(x)=1/12*(1+2+3+4+...12) = 6.5 where x takes on the values of your ranking order in the draft . |
#3
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Re: Probability question
12c5 * 7!/12! double counts some combos, but the answer is probably pretty close. I think you probably need some sort of inclusion exclusion.
Suppose the 5 chosen were 1, 2, 3, 4, 5, and 6-10 ended up being in the same order as well. This is double counted when you choose 6-10, and 1-5 just happen to be in the same order. |
#4
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Re: Probability question
I agree ncray .
I think it's a good enough answer . If op wants an exact answer it's not too difficult to work out . I may do it later on if I have time . |
#5
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Re: Probability question
The probability 1,2,3,4,5 are in the correct order is :
1/12*1/11*1/10*1/9*1/8= 1/12p5. We want the number of derangements of the numbers 6,7,...12 . This is just 1/2!-1/3!+1/4!-1/5!+1/6!-1/7!=36.7857% So 1/12p5 *12c5*0.367857 = 0.003065 |
#6
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Re: Probability question
[ QUOTE ]
What would be the expected number of places up or down in the draft order for a team to move given random drawings (can one even determine this)? [/ QUOTE ] In a random drawing, each possible permutation of teams has equal probability, so each team's net expected position is at 6.5. The expected change is thus (current position - 6.5). |
#7
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Re: Probability question
Thanks guys! Very, very helpful. Basically, telling me that the odds are about 1% or less is what I am looking for.
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#8
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Re: Probability question
[ QUOTE ]
[ QUOTE ] What would be the expected number of places up or down in the draft order for a team to move given random drawings (can one even determine this)? [/ QUOTE ] In a random drawing, each possible permutation of teams has equal probability, so each team's net expected position is at 6.5. The expected change is thus (current position - 6.5). [/ QUOTE ] How can that be? Doesnt that result in all teams moving up on average? Eg last years 1st team cannot move up to -5.5. |
#9
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Re: Probability question
[ QUOTE ]
[ QUOTE ] [ QUOTE ] What would be the expected number of places up or down in the draft order for a team to move given random drawings (can one even determine this)? [/ QUOTE ] In a random drawing, each possible permutation of teams has equal probability, so each team's net expected position is at 6.5. The expected change is thus (current position - 6.5). [/ QUOTE ] How can that be? Doesnt that result in all teams moving up on average? Eg last years 1st team cannot move up to -5.5. [/ QUOTE ] Think of it negatives as moving down by that # of spots. So last year's 1st team moves down 5.5 spots to 6.5 on average. Last year's 12th team moves up 5.5 spots to 6.5. |
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