I'm playing in a qualifying event for a pool team. There are 31 entrants and 16 players will make the team. Each player will play every other player 2 times for a total of 60 games per player. The top 16 players with the most total wins will qualify.

My questions are:

-How many games do I need to win to guarantee that I will be in the top 16?
-How many games for a 75% chance
-How many games for a 50% chance

If someone could explain the math behind it as well that would be great.

EDIT: disregard everything below, I made a critical mistake. I'll leave it up for laughs.

To be guaranteed a top 16 spot, it can't be possible for 16 other players to all have more wins than you, which means their average record per person can't be greater than your record. There are 31*30 = 930 total games played, so there are 930 total wins to go around. Any games you win are games other people can't have won, so they're subtracted from the 930. Like if you win all 60 games then the other players combined can only win 870 games.

If your record is x, then the highest average possible for another 16 players is (930-x)/16.

You need x >= (930-x)/16, so you need at least 55 wins.

There's only 465 games played, not 930. 2 players each time.

To be in the top half of the field you need to win half your games, which is 15 minimum.

And using Heehaw's method, if you win 15 games that puts the whole field average at 15 wins. If you win 16 games, that puts the field average at 14.96666667.

Is this true?

Six teams. Single game round robin.

A. 5-0
B. 3-2
C. 3-2
D. 3-2
E. 1-4
F. 0-5

But you play each opponent twice.

This is a neat question.

I would be amazed if there is a simple formula for this. By that I mean I would be amazed if there is a simple derivation for the right answer (I guess it is possible that the answer formula could be fairly simple).

It would be "easy" to do a brute force program to tally the number of instances of all the possible number of victories for each place finisher for smallish values for the number of teams. Not sure if this is a reasonable approach or not to gain insight into large values.

Whoops.

My logic was wrong too.

To guarantee you are in the top 16 I think you have to win 52 games.

If you win 52 that leaves 878 wins for the field. The maximum number of players who could tie or beat you at 52 wins would be 878/52 = 16 (rounded down to whole player). So you are assured of at least tying for 16th place.


And if you only win 51 games, the max number who could tie or beat you is 17, so you are not assured of qualifying.

You can't just give all the wins to the top players. Someone needs to win the game between the two worst.

All we need to know is that no more than 15 can score higher than we do, we don't care about the actual number or the distribution. Distributing those wins more broadly makes the scores lower, which can't knock us out.

This doesn't help us answer the 50% or 75% question, which will take simulations to solve. But the guaranteed qualifying is easy to solve using just a spreadsheet. If we win X, the field can only win Y, and then the maximum number of players that Y can be divided amongst and be higher than us. That's what I did.

While your number of 52 certainly guarantees that we can't be knocked out, I can also show that 51 guarantees it as well. Why?

Using your logic.
There are 930 games to be won.
You win 51 leaving 879.
The bottom 10 players play 90 games between them that someone has to win, leaving 789 games to be distributed between all players (other than you).
789/51=15 players that can tie you for wins.

But that's not even the lowest number of wins. Because you absolutely need to take into consideration the possible distribution of wins. The bottom 15 players need to total at minimum 210 wins because that's how many times they play each other and someone needs to win those games.

I am lurking and have given this zero thought.

But I don't think NOG is correct regarding 51.

By the simple fact that every game has exactly one winner and one loser, when the Top 15 play against the Top 15 (210 total times), we know that collectively they will be 210-210. Similarly for when Bottom 15 play vs Bottom 15.

So, my guess is that 52 is not the correct answer

Edit/PS: Didace said the same thing, only better, in the post right above this!

Super interesting! Would it be any different, if we solved for “best of the worst” and calculate the most wins possible by the worst 15 players and that hero must beat the “best of the worst” by a single win?

How? I think brute force of all possible distributions is the only way.

Yeah, it's more complicated than I thought.

This is probably wrong, but for fun anyway.

Scenario A) The best possible wins for the worst 15 players would be to have a record of 29-31 and the 31st loss of each worst player was to a corresponding player in the upper 15, with hero right in the middle with 30-30 record. Every player above hero has 31-29 record and every player below hero has 29-31 record.

Scenario B) The other extreme would be where every player wins every game versus the players worse than her, and loses every game to every player better than her.
The very last player would have 0 wins, next to last 2 wins, next up 4 wins....
The 15th worst player would have 28 wins, and hero is right in the middle again with 30 wins.

Now, if this symmetry is altered in either scenario, then hero situation is unchanged.


Example:

Scenario A above, the 15th best player wins an extra game versus another top 15 player.
The winner becomes the best record at 32-28 and the loser drops to a tie with hero, 30-30, hero remains in a tie for 16th place.
If a bottom 15 player gets an extra win versus another bottom 15 player, then that winner ties hero with 30-30 record and that loser drops to last place, with a record of 28-32.

What happens if a lowly bottom 15 player gets an extra win versus a top 15 player?

They both tie hero at 30-30 and hero remains tied for 16th place.

Scenario A is the closest possible outcome among evenly matched players, and scenario B is the most lopsided results among unevenly skilled players.

Since the middle of each is 30 wins, and 31 wins would guarantee a spot for hero in all cases, my guess is that the magic number is 31 wins.

Not counting games with you, the obvious worst case is that there's a perfect split where 16 people 15-15 each other and 28-0 the bottom 14. Now if you also 28-0 the bottom 14 and 16-16 the other 16, there's a 17-way tie for first at 44 points. (17*44 + 14*13 = 930, so this checks out). Depending on how tiebreaks are counted, that's either sufficient or you need 45 points to guarantee being cleanly in top 16. With 43 points, you can get 17th with this distribution and swapping in an 0-2 to one of the top 16 (he gets first at 45, 15 at 44, you in 17th at 43).

How does your "best possible wins" scenario for the bottom half of the field fit into a round robin that ends like this?

If I am following this correctly, in the general case of N teams (N being odd), number of wins to guarantee finishing in top (N+1)/2 places is (3N-5)/2.

Math check for N=31: (3*31-5)/2 = (93-5)/2 = 88/2 = 44.

Neato.

I just remembered something I meant to post earlier.

Some of you may be familiar with the concept of a team's "Magic Number" to clinch a playoff berth or to clinch the pennant, etc. There is a simple formula to calculate a Magic Number based only upon the contending teams' current won-loss records (and how many games each team has left to play, of course).

There is a parallel formula for a team's Elimination Number. The combination of wins and losses which would result in a team being mathematically eliminated from winning the pennant.

It turns out that the Elimination Number is not always the "tightest" number. I remember reading a baseball statistics article many years ago about a season in which a team had actually been eliminated from winning the pennant a day or so before its Elimination Number went to zero!

The reason was that teams above the team in question had to play each other in subsequent games and both teams could not lose in those matchups (one of the teams had to win).

Anyway, this thread reminded me of that interesting tidbit from the sporting world.

I ran a brute force program for N=5 teams, each playing every other team twice.

Here is a table showing in what place a team with a given number of wins finishes where ties are given the highest rank (e.g., if two teams are tied for 2nd and 3rd place, they each are considered to have finished in 2nd).

Wins	1st	2nd	3rd	4th	5th
0	0	0	0	0	20,480
1	0	0	0	2,560	161,280
2	0	0	1,200	152,480	419,760
3	0	3,040	304,800	727,840	111,200
4	38,680	747,080	623,400	24,440	0
5	695,520	439,360	12,000	0	0
6	549,120	24,320	0	0	0
7	163,840	0	0	0	0
8	20,480	0	0	0	0

From the table we see that 5 wins are required to guarantee a Top 3 finish.

Tonight I ran a simulation of the N=31 case. I ran 1,000,000 seasons. I assumed that all 31 teams are evenly matched and that every game is a 50/50 coin flip.

Here is a table showing for each possible number of team wins (0-60), how many team-seasons with that number of wins finished in the Top 16, how many total team-seasons had that number of wins, and the percentage.

Season Wins	Seasons in Top 16	Total Seasons	Percentage
0	0	0	NA
1	0	0	NA
2	0	0	NA
3	0	0	NA
4	0	0	NA
5	0	0	NA
6	0	0	NA
7	0	0	NA
8	0	0	NA
9	0	0	NA
10	0	3	0.0
11	0	8	0.0
12	0	39	0.0
13	0	142	0.0
14	0	436	0.0
15	0	1,446	0.0
16	0	3,972	0.0
17	0	10,485	0.0
18	0	24,790	0.0
19	0	55,096	0.0
20	0	112,649	0.0
21	0	215,012	0.0
22	0	379,352	0.0
23	0	630,141	0.0
24	0	969,453	0.0
25	0	1,398,184	0.0
26	0	1,877,680	0.0
27	16	2,364,937	0.0
28	12,559	2,787,012	0.5
29	741,520	3,078,943	24.1
30	2,852,740	3,181,515	89.7
31	3,075,207	3,077,451	99.9
32	2,786,552	2,786,553	100.0
33	2,366,145	2,366,145	100.0
34	1,880,792	1,880,792	100.0
35	1,394,519	1,394,519	100.0
36	969,793	969,793	100.0
37	628,476	628,476	100.0
38	380,651	380,651	100.0
39	214,351	214,351	100.0
40	113,046	113,046	100.0
41	55,344	55,344	100.0
42	25,101	25,101	100.0
43	10,433	10,433	100.0
44	3,927	3,927	100.0
45	1,490	1,490	100.0
46	462	462	100.0
47	127	127	100.0
48	34	34	100.0
49	8	8	100.0
50	2	2	100.0
51	0	0	NA
52	0	0	NA
53	0	0	NA
54	0	0	NA
55	0	0	NA
56	0	0	NA
57	0	0	NA
58	0	0	NA
59	0	0	NA
60	0	0	NA

You will see that the Pcts are very knife-edgy. They go directly from 0% to 100% with little in between. I am not sure if that is an artifact of "only" running 1 million seasons. Since this simulation took several hours to complete, I am not keen on running significantly larger simulations.

Wow nice work thanks for the time it must have taken. I suppose it seems quite logical that you should make the top 16 if you win half your games. Quite interesting to see that big grouping in the middle and how much difference 1 or 2 games will make.

Again thanks for the effort and everyone else who contributed. I started off just wanting to know how many games I had to win but ended up quite enthralled with the math behind it. Didn't come close to working out a reliable method though. Brute force seems by far the best way.

I thought that too in my original post, but it appears from the simulation that you need half plus 3 (the 100% is rounded at the 32 level), at least in the case of an odd number of players. Might be half plus 2 for even number. Or might be a coincidence at this range, but I doubt it. There should be a way to prove this theory.

Are you commenting on the minimum number of wins to guarantee being in the top half?

The simulation should not be considered dispositive evidence in that question.

TomCowley posted above that that answer is 44 in the case of 31 teams and I translated his reasoning into a formula (3N-5)/2.