Math gurus...possibility of identical fantasy teams?

[brandoman]

Fun exercise? I'm curious what the likelihood of 2 people drafting the exact same team? 12 team league, Yahoo! players, 23 players drafted?

[Padsin05]

lets just say it rounds down to 0

[bigken117]

lets just say it rounds up to 0

FYP, I think they're lower than that.

Just go with 1/12 each round, or 8% in the first round, and keep multiplying 1/12 22 more times, which in excel is 1.50949E-25. So yeah...

[Padsin05]

lets just say it rounds up to 0

FYP, I think they're lower than that.

Just go with 1/12 each round, or 8% in the first round, and keep multiplying 1/12 22 more times, which in excel is 1.50949E-25. So yeah...

acutally mathematically it isnt 1/12 each round. Your scenario assumes every player being drafted in the same rounds of every draft.

since there are 30 teams x25 per roster, the true odds are 1/750 per player per pick

so the math is 1/750*1/738*1/726 etc, repeating downward 23 times (subtracting 12 players each time that other teams picked)

mathmatically speaking the odds of me and another person taking the same player with the first pick overall is 1/750 or .133%

so basically the "True odds" of teams having the same team is 0

[Padsin05]

btw if you want to factor in rankings, the odds still round down to 0

[bigken117]

actually I know it isn't 1/12 every round and actually it's not 1/750 for the first pick overall either. I was rounding numbers to make the math easier. Our answers remain consistent though.

[The_Show]

What is this math you speak of?

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I hate statistics (although I never actually took a class in it).
With 12 teams and 23 players each, that's 276 players.

276 choose 23 gives 2,084,633,151,901,356,196,159,615,812,153,600 different combinations.

So the probability of having one of those combinations occurring twice is ~9.594 × 10^-34

Maybe...? Not sure if that makes sense. It all goes down the crapper anyways when you consider drafts aren't (or shouldn't be ) random.

[Skin Blues]

lets just say it rounds up to 0

FYP, I think they're lower than that.

Just go with 1/12 each round, or 8% in the first round, and keep multiplying 1/12 22 more times, which in excel is 1.50949E-25. So yeah...

acutally mathematically it isnt 1/12 each round. Your scenario assumes every player being drafted in the same rounds of every draft.

since there are 30 teams x25 per roster, the true odds are 1/750 per player per pick

so the math is 1/750*1/738*1/726 etc, repeating downward 23 times (subtracting 12 players each time that other teams picked)

mathmatically speaking the odds of me and another person taking the same player with the first pick overall is 1/750 or .133%

so basically the "True odds" of teams having the same team is 0

It's not this simple. There are also position limitations, likelihood of being drafted in certain rounds (ie: extremely likely the teams are identical if they both pick first, second, or third overall, so it's definitely not 1/750). Also they don't need to pick the same players in the same round on order to end up with the same team.

[brandoman]

FYP, I think they're lower than that.

Just go with 1/12 each round, or 8% in the first round, and keep multiplying 1/12 22 more times, which in excel is 1.50949E-25. So yeah...

acutally mathematically it isnt 1/12 each round. Your scenario assumes every player being drafted in the same rounds of every draft.

since there are 30 teams x25 per roster, the true odds are 1/750 per player per pick

so the math is 1/750*1/738*1/726 etc, repeating downward 23 times (subtracting 12 players each time that other teams picked)

mathmatically speaking the odds of me and another person taking the same player with the first pick overall is 1/750 or .133%

so basically the "True odds" of teams having the same team is 0

It's not this simple. There are also position limitations, likelihood of being drafted in certain rounds (ie: extremely likely the teams are identical if they both pick first, second, or third overall, so it's definitely not 1/750). Also they don't need to pick the same players in the same round on order to end up with the same team.

You're right, it isn't that simple...at all. If you calculated each round on 1/750 basis...well that isn't correct. There would have to be an algorithm that accounted for average draft position.