Please break these 9 tiebreakers. Thanks!! Ending in 1 hour!!?
597995
I'm with you, we as country have let things go to far. The is no personal accounting, something needs to done; bring back Kangaroo courts and pubic stoneings. May God save this great country. keep up the good work
421531
hold ur breathe and count backwards 100 99 98 97 96 95 94 93 92 91 90.........................
698606
Hmm, interesting question, actually, and a practical application of math.  Hey, all you folks who ask what this stuff is good for, here is a real-life situation where you can use it!\n\nGetting to the problem.  Obviously, the contestant to go last spins again if his score is lower than the best score.  If tied, there is a spin-off where his winning chance is 50%; so, if there is a tie, he should respin if he has 45 or less (giving him at least 11 out of 20 good numbers), not spin if he has 55 or more (giving him at most 9 out of 20 good numbers - better to go to the spin-off, where he has a 50% chance), while 50 is a "push".   \n\nNow, for the contestant to go 2nd to last: First, let's assume the first player busts (of course, the second player respins if his score is lower than the first player), which essentially makes this a two-player game.  Based on what number the second player stops on, this is the winnning percentage I calculate for the last player:\n\n100 - 4.875%\n95 - 14.375%\n90 - 23.375%\n85 - 31.875%\n80 - 39.875%\n75 - 47.375%\n70 - 54.375%\n65 - 60.875%  \n60 - 66.875%\n55 - 72.375%\n50 - 77.375%\n\nSo, you may think the second player should respin whenever his odds are under 50%, but that is actually incorrect.  For example, if the second player has 70, giving him a 45.625% chance of winning, respinning would give him a 70% chance of going bust - he is better off stopping with the 45.625% chance.  To determine whether or not respinning improves the second player's chances, a second calculation is needed that considers each of the 20 scores after the second spin - some of them bust the player but the others improve his chance of winning.  Based on that, here is a new table, giving the last player's chance of winning if the second player respins with the score given:   \n\n95 - 95.24375%\n90 - 90.9625%\n85 - 87.13125%\n80 - 83.725%\n75 - 80.71875%\n70 - 78.0875%\n65 - 75.80625%  \n60 - 73.85%\n55 - 72.19375%\n50 - 70.3125%\n45 - 69.68125%\n\nSo, comparing the two tables, the critical points are 55 and 60.  If the second player has 60, he should stop - the last player's chance of winning improves if the second player spins.  At 55, however, the second player should spin, improving his odds by about 0.2%.\n\n\nMoving on to the three player game... now it gets really messy.  The player to go first needs a score that can beat the other two players.  Does he stop on 55 again, hoping the other two players both bust?  He'll need a better score than that to have a 50% chance of winning, but, as was noted in the two-player analysis, stopping with a 30% chance can be better than making a second spin.  As for how to do a complete analysis, uh, beats me.  This may require a computer to do all the math.\n\n* * * * * *\n\nThinking about the three-player game, maybe it isn't as messy as I initially thought - the only additional consideration for the first player, as opposed to the second player, is that either player will stop if the first player's score is beaten.  This can be calculated from the first table.  For this table, listed is the chance of the first player losing (i.e. either player beats him) if he stands on the score given:\n\n100 - 9.51% \n95 - 26.68%\n90 - 41.29%\n85 - 53.59%\n80 - 63.85%\n75 - 72.31%\n70 - 79.18%\n65 - 84.69%\n\nSimilar to the two-player game, I can now calculate the odds if the first player decides to respin.  Since I’m getting a little tired crunching these numbers, for this last table I’ll only list the critical ones:\n\n70 - 83.4%\n65 - 82.3%\n\nIn other words, the first contestant should respin on 65, stop on 70.\n\nIn conclusion, I think the optimum strategies are.....\n\nFirst player: spin on 65, stop on 70\n\nSecond player: spin on 55, stop on 60 (assuming his score at least matches the first player)
654474
Actually, the head-to-head fantasy baseball game does have ties.  My league has 10 categories.  If I lead in 5, we tie in one category, my record is 5-4-1.\n\nThe league standings disregard the ties, looking only at winning percentage.  My percentage above would be .555\n\nSince I found this question under "Fantasy Sports", I felt it was safe to assume this is what you meant.
1172674
Probably a lottery or a coin flip.\n\nFrom the Wiki, the tiebreakers are\n\n"If teams are even on points at the end of group play, the tied teams will be ranked as follows:\n\n-greater number of points obtained in matches between the tied teams \n-goal difference in matches between the tied teams \n-greater number of goals scored in matches between the tied teams \n-goal difference in all group matches \n-greater number of goals scored in all group matches \n-a play-off on neutral ground, with extra time and penalties if necessary (in qualifying) \n-drawing of lots (at the final event)"
491020
1-20 the quickest gets the points! and i mean 1234..... lol.?
894091
None. Two odds make an even
149355
omg your kidding me... thats a trick question. the set would already  be over if it is 6-4.
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....
1070320
In the first round, if a game ends in a tie, that is how it ends.  There is no sudden death nor is there overtime.
821077
once we chose 5 persons for one committe then other committe members get fixed.\n\nThe 5 members can be chosen in 10C5 ways = 10*9*8*7*6/120\nor 6*7*6 or 252 ways
860554
27 ways to pick the first person in the group\n26 ways to pick the second\n25 ways to pick the third\n24 ways to pick the fourth\n23 ways to pick the fifth.\n\nHowever, order is irrelevant, so we need to divide by 5!.\n\nThis is 27 choose 5\n= 27 x 26 x 25 x 24 x 23 / 120 \n= 80,730 ways
978079
Each match eliminates exactly 1 person. That is, there's a bijection between matches and contestants eliminated. Since there are n-1 contestants eliminated, the total number of matches is just:\n\nM(n) = n - 1
949670
trying to even the score.....
1065918
To get your answers, imagine what has to happen.  \n\na) First, you have to pick 1 of the 4 white balls out of the 12 balls there.\n\n2nd, you have to pick 1 of the 3 white balls left out of the 11 balls there.\n\nSo, in order to get a red and then a red, you need to get 4 out of 12 followed by 3 out of 11 or 4/12 times 3/11 or 12/132 = 1/22.\n\nb) First, how many ways can 5 people line up?  5!.  How many ways are correct (assuming they all have different names, only 1).  So, the prob is 1/5! or 1/120.
295748
Total joke.  I mean how stupid is it to root for one team to win so your team can advance, but if that team doesn't win to root for a certain amount of runs to be scored so your team can win the tie breaker.  There has got to be a better way of breaking ties, like oh I don't know, play another game.
2126
Games already over!!!!!!!