Odds of two people picking the same team in a NCAA fantasy pool

I'm in a pool that determines the winner by selecting any 10 players from the NCAA Tournament and aggregating points based on points+rebounds+assists.

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

I am not a math major and maybe I am falling for a gamblers fallacy, but to me the obvious answer is that you need a minimum number of 2 entries to have 2 entries exactly match.

Let 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

Quote: ThatDonGuy

Let 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: sc15

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.

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: dwheatley

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.

Well yeah, the original question is very vague.

I would say the answer's 2 then.

I don't think a random number generator will be able to tell you the answer, because the picks are not random. Most people would pick certain players over other players because their numbers indicate they should be picked

My 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.

Quote: Lando

My 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.

As per dwheatley's post, yes it seems Birthday Paradox-ish (I forgot to quote and just hit reply, so his original isn't included in this edit)

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.