help with these tiebreakers please?
96150
I voted resonably on these answers.  I did not just chose the first one.  I hope this helps you out.  GOOD LUCK!!!
993676
here are the tiebreaker procedures:\n    The ranking of each team in each group will be determined as follows:\n    a) greatest number of points obtained in all group matches;\n    b) goal difference in all group matches;\n    c) greatest number of goals scored in all group matches.\n\n    If two or more teams are equal on the basis of the above three criteria, their rankings will be determined as follows:\n    d) greatest number of points obtained in the group matches between the teams concerned;\n    e) goal difference resulting from the group matches between the teams concerned;\n    f) greater number of goals scored in all group matches between the teams concerned;\n    g) drawing of lots by the Organising Committee for the FIFA World Cup.\n\n\nSo I guess end the end they would have to draw lots....
988467
The 4 teams in each group play each other.  The two teams with the most points (win 3 pts, undecided 1 pt) advance to the next round.  If 2 teams have the same number of points, they look at their goal difference (goals made vs. goals against).  If that is also the same, they look at the result of the game played between the 2 teams.  If they played to an "undecided", then they raffle off the next round.  Sounds unfair, but what else is there - a "drink-off"??  THAT would be cool, though....
1135375
Write two fractions on the board , column A  and  B.\n\nWhich is greater A or B or are they equal.   Choose teams.Keep score.\n\nYou(as the teacher) can easily tell the answer quickly:  \nA .......... C\n--- vs---- -----\nB ...........D\n\nIf AxD is greater the correct answer is A\nifBxC is greater the correct answer is B
236336
a tie between d and the 2nd c
785056
First Tie Breaker is division head to head then conference head to head then total points scored, then total points scored in common opponent games.  There is a 1 in 4 Billion chance that it would need to go further than that.
268952
Game1-5/12\nGame2-6/11\nGame3-7/10\nGame4-8/9\n\nGame5-1/W4\nGame6-2/W3\nGame7-3/W2\nGame8-4/W1\n\nGame9-W5/W8\nGame10-W6/W7\n\nGame11-W9/W10\n\n\n\nWhere it says "W#" the "W" is the winner of that game number.\n\nBasically, give the top 4 seeds a bye in the 1st round.
1017522
If every team has the same number of points, the same goal difference and the same number of goals for and against then the following method is employed:\n\nGreatest number of points obtained in all group matches, Goal difference in all group matches, Greatest number of goals scored in all matches.\n\nIf two or more teams are equal on the basis of the above three criteria, their rankings will be determined as follows: \n\nGreatest number of points obtained in the group, matches between the teams concerned, Goal difference resulting from the group matches between the teams concerned , Greater number of goals scored in all group matches between the teams concerned, Drawing of lots by the organising committee for the Fifa World Cup.
714830
In the group table, the goal difference is only the third deciding factor. The first is the points total. The second is the head-to-head record.
947747
Pick one knight anywhere.\nIf you pick either of the knights directly adjacent to him, you are done.  This happens 2/24 of the time.\n....OOXXOO...\n\nIf you pick a knight that is separated by exactly one seat (2/24 of the time), then there are 3 out of 23 seats that are adjacent.\n...OXOXO....\nIf you pick a knight that is more than one seat away (20/24), then there are 4 out of 23 seats that are adjacent.\n...OXO...OXO...\n\nSo the probability is:\n1/12 + 1/12 * 3/23 + 10/12 * 4/23\n\nAdding it up:\n23/276 + 3/276 + 40/276, or 66/276 = 33/138\n\nThis is approximately 23.91%
168600
Well, you have 8 teams right? So team 1 could play w/ team 2, team 3 could play w/ team 4, team 5 could play w/ team 6, and team 7 could play w/ team 8. Make it double elimination. If team 1 loses and team 3 loses then they will play against each other. The loser of that team will have lost their 2nd game which makes them out and the winner will play the loser of the other side. If you need any more help i can help you. Just email me.
965631
If there are 22 teams playing then there are 11 games. Assuming that the palray is such that there WILL be a winner (ie no draws) then there are two possible results for each game .. a win or a loss (Team A Wins  - Team A loses). So there are a total of 2^11 possible combinations each round.\n\nIf a draw is possible then there are 3^11 possible combinations W - D - L on each game (Team A Wins  - Team A Draws -Team A Loses).\n\nP(all correct) = 1/2^11\n= 1/2048\n≈ 0.05%\n\nIf a draw is possible then there are 3^11 possible combinations W - D - L on each game (Team A Wins  - Team A Draws -Team A Loses).\n\nThen P(all correct) = 1/3^11\n= 1/177147\n≈ 0.0006%
988227
a team get 3 points for a win, 1 for a draw and 0 for a lose
1059925
1.total # of points in all matches\n2.goaldifference in all matches\n3.total # of goals scored in all matches\n4.result in between teams tied according to tie breakers 1-3\n5.goaldifference in between teams tied according to tie breakers 1-4\n6.goals scored in between teams tied according to tie breakers 1-5\n7.drawing of lots
1142821
There is no tiebreaker. The case is reheard. See Hudson v Michigan that was decided today. \nSandra Day O'Connor retired before a decision was made; so under court rules, her vote did not count leaving a 4-4 tie. The case was reheard, and the Court decided that police can now enter your home within seconds of announcing their presence and without knocking to gain your attention.
156986
First, I am not sure why you call this factoring. This is not factoring. \n\nThink of it this way. If each of the teams play each other once, there are G1=[N^2-N]/2 games. To see that this is true, just compute either \n\nN choose 2 = N!/[(N-2)!2!]=[N^2-N]/2 \n\nor \n\nmake a matrix with the N teams as both rows and columns. Every element of the matrix constitutes a game between two teams, one in the corresponding row and one in the corresponding column. One has to disregard the N diagonal elements, because a team cannot play itself, so there are a total of N^2-N games in the matrix, without the diagonal. Since this includes games when A plays B as well as B plays A, we have to divide by 2, to get again [N^2-N]/2.\n\nIf each of the teams play each other 2 times, there are G2=N^2-N games. To see this, just imagine the matrix of team-labelled columns and rows again, and subtract the diagonals again, so that both the games between A and B and between B and A are included, for a total of 2 games between each pair of teams. The number of elements in the matrix is N^2-N, ignoring the diagonal elements again. \n\nFor the teams to play each other 3 times, they can just play the one game schedule, and then the 2 game schedule, for a total of 3 times between each pair of teams, so:\n\nG3=G1+G2=(3/2)*[N^2-N]\n\nFor N=7, this is just 1.5*[49-7]=1.5*42=63.\n\nThere are 63 total games played.
1159012
The numbers in brackets are the half time score