You can pick any player from any team, including a whole team, and one must take into consideration the fact that if a player's team is defeated, that man can no longer score. That is, you want the best players from the teams you think will last the longest so you can score more points.

The question is this:

What's the minimum number of entries required such that 10/10 selections are EXACTLY the same for two different entries?

Let's say the number of different players selected tops out at 40 players. If you can plug 35 or 45 as the max distinct players used over all the entries, that would be cool too.

Thanks in advance

Lando

Assuming there are 40 players from which you can choose 10, how many entries do there have to be before the probability that at least two have the same 10 players is 100%?

If this is the case, then there are (40)C(10) = 847,660,528 unique ways to do this, so, by the pigeonhole principle, there need to be 847,660,529 entries.

Here is a table for each possible number of players that can be chosen from 25 to 50:

Choices | Number needed for duplicate |
---|---|

25 | 3268761 |

26 | 5311736 |

27 | 8436286 |

28 | 13123111 |

29 | 20030011 |

30 | 30045016 |

31 | 44352166 |

32 | 64512241 |

33 | 92561041 |

34 | 131128141 |

35 | 183579397 |

36 | 254186857 |

37 | 348330137 |

38 | 472733757 |

39 | 635745397 |

40 | 847660529 |

41 | 1121099409 |

42 | 1471442974 |

43 | 1917334784 |

44 | 2481256779 |

45 | 3190187287 |

46 | 4076350422 |

47 | 5178066752 |

48 | 6540715897 |

49 | 8217822537 |

50 | 10272278171 |

Quote:ThatDonGuyLet me see if I have the question correct:

Assuming there are 40 players from which you can choose 10, how many entries do there have to be before the probability that at least two have the same 10 players is 100%?

If this is the case, then there are (40)C(10) = 847,660,528 unique ways to do this, so, by the pigeonhole principle, there need to be 847,660,529 entries.

Here is a table for each possible number of players that can be chosen from 25 to 50:

Choices Number needed for duplicate 25 3268761 26 5311736 27 8436286 28 13123111 29 20030011 30 30045016 31 44352166 32 64512241 33 92561041 34 131128141 35 183579397 36 254186857 37 348330137 38 472733757 39 635745397 40 847660529 41 1121099409 42 1471442974 43 1917334784 44 2481256779 45 3190187287 46 4076350422 47 5178066752 48 6540715897 49 8217822537 50 10272278171

Incorrect, this is assuming the picks are random.

The probability of 2 entries being identical are fairly high. If 2 people use the same methodology for their picks (eg, using some ranking list and just going with the top 10), they'll end up with the same picks.

Quote:sc15Incorrect, this is assuming the picks are random.

The probability of 2 entries being identical are fairly high. If 2 people use the same methodology for their picks (eg, using some ranking list and just going with the top 10), they'll end up with the same picks.

That doesn't matter for this question. Either the answer is 2 or DonGuy's exhaustive approach is correct, depending on what the OP meant.

Now if we wanted to know how many entries were needed before there was say a 50% chance of a copy (like birthday paradox), then you would be right, we'd need the individual player probabilities.

Quote:dwheatleyThat doesn't matter for this question. Either the answer is 2 or DonGuy's exhaustive approach is correct, depending on what the OP meant.

Now if we wanted to know how many entries were needed before there was say a 50% chance of a copy (like birthday paradox), then you would be right, we'd need the individual player probabilities.

Well yeah, the original question is very vague.

I would say the answer's 2 then.

Clearly, the minimum one needs is 2, but the odds are exceedingly against 2 and only 2 entries being exactly the same when there are about 35-45 "handicapped" selections possible (that's just my guess based on my own expertise and what people MIGHT do).

So the question is, re-stated:

What's the minimum number of entries that would produce at least one identical match of 2 entries with a 95% probability? I'm presuming, again, that our data set is about 40 possible distinct players that anyone would select, but you could use 50 if you like. That is a contingency but should be easily accounted for. Obviously, as you stated, 2 is possible but I would think the odds would be .01% or, 10000 entries needed (1/.0001).

Is that more clear?

Thanks.

Quote:LandoMy apologies guys, but thanks for the responses so far.

Clearly, the minimum one needs is 2, but the odds are exceedingly against 2 and only 2 entries being exactly the same when there are about 35-45 "handicapped" selections possible (that's just my guess based on my own expertise and what people MIGHT do).

So the question is, re-stated:

What's the minimum number of entries that would produce at least one identical match of 2 entries with a 95% probability? I'm presuming, again, that our data set is about 40 possible distinct players that anyone would select, but you could use 50 if you like. That is a contingency but should be easily accounted for. Obviously, as you stated, 2 is possible but I would think the odds would be .01% or, 10000 entries needed (1/.0001).

Is that more clear?

Thanks.

Again, even with the choices narrowed down this is NOT a lottery with randomly generated numbers.

If 2 people use the same methodology, or same ranking list to make their picks, the odds of 2 entries matching is exceedingly high.

Yes, I can see that this is getting more at the point.

If you need this data, let's say:

The first 3 everyone will take (Okafor, Kaminsky, Towns/Cauley-Stein), so 1.00

The next (4-10) will be 0.9

11-20 0.75

20-30 0.55

30-40 0.33

As percentage plays or "ratings"/probabilities

I hope that helps, it seems fairly complex.