what is the highest amount one person can win in one game of jeapordy?
356782
Taking into account the newer dollar amounts ($200-1000 for 1st round, $400-$2000 for Double Jeopardy)...theoretically, the maximum amount of money can be won if the last clue picked is a Daily Double and it is the lowest dollar amount on the board.\n\nSo, for the first round, with five different dollar amounts and six categories:\n\n$200 x 5 = $1000 (this is taking into account the Daily Double)\n$400 x 6 = $2400\n$600 x 6 = $3600\n$800 x 6 = $4800\n$1000x 6 = $6000\n\nTotal before Daily Double in the 1st round: $17,800\nThen make it a 'true' Daily Double: $35,600\n\nFor Double Jeopardy the totals are:\n\n$400  x 4 = $1600 (this is taking into account both Daily Doubles)\n$800  x 6 = $4800\n$1200 x 6 = $7200\n$1600 x 6 = $9600\n$2000 x 6 = $12000\n\nTotal before Daily Doubles: $70,800\nDouble the winnings: $141,600\nThen again for the second Daily Double: $283,200\n\nAssuming they've got the Final Jeopardy answer right, they can double their score one last time to get a maximum score of:\n\n$566,400\n\nAccording to Wikipedia, the odds of all of this happening is about 3,288,600 to 1 against.
81369
189\nAnd One person had 100 points
427487
How much could you do?..there, you have the answer! u r a winner
403576
In game one between A and B, the outcomes are:  A wins, B wins, or they draw, equaling 3 outcomes.\n\nIn 14 games, there are 3^14 outcomes.
974181
go to gamewinners.com
1012413
Shoot out until there is a winner....Just keep going...
376282
The correct answer is 68%. Good luck on the game!
913338
To win 1 number, you have 36!/35! = 36 different single numbers.  Since you are picking 6 numbers in one bet, you can cover 6!/5! = 6 numbers per bet; therefore, you get full coverage of those 36 numbers in 36/6 = 6 bets.\n\nTo win 2 numbers, you have 36!/(2!34!) = 18*35 = 630 different pairs of numbers.  Since you are picking 6 numbers in one bet, you can cover 6!/(2!4!) = 15 pairs per bet; therefore, you get full coverage of those 630 pairs in 630/15 = 42 bets.\n\nTo win 3 numbers, you have 36!(3!33!) = 6*35*34 = 7140 different sets of three numbers.  Since you are picking 6 numbers in one bet, you can cover 6!/(3!3!) = 20 sets of three numbers per bet; therefore, you get full coverage of those 7140 sets in 7140/20 = 357 bets.\n\nIn general, when choosing c numbers from a set of n numbers, to win x numbers at least, you need:\n\nn!(c-x)! / (c!(n-x!))   bets.
493279
Sorry, missed it,\nI was at the extreme underwater dragon riding pudding toss championships.\nXenji Nyonsonmanzicizichimanschlecher- 12.327 def.\nThe Baron Jacque Herman Gustof Gustov Tinkerman- 11.169
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)
894091
None. Two odds make an even
1019434
enough to crush a human
746392
I don't know but I'd love to be one to win all that money! Someone that doesn't live far from me won a lucky case. I trye once in a while.Oh, thanks for your answer on one of my last questions :-).
950841
well , it is like wresting...anyone can win the game ......
941038
The  goal is to win the prize. The prize will be given to the winner....POSSESSION.
967466
5 9 17 26 37 and mega number 16\nGood luck!\nTell me if you win something! ; )
41454
To gather as much toys as possible. Who ever dies with the most toys wins.