Probability question

[slickterp]

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.

[jay_shark]

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 .

[ncray]

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.

[jay_shark]

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 .

[jay_shark]

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

[rufus]

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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).

[slickterp]

Thanks guys! Very, very helpful. Basically, telling me that the odds are about 1% or less is what I am looking for.

[Copernicus]

[ 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.

[ncray]

[ 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.