What is the probability that if you pick 22 correct college winning teams (all outcomes must be correct)?
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%
1108310
25, 26, 33, 40, 41, 47 bonus ball will be10
219821
This cannot be ascertained without knowing how many are being selected in all...\n\n\nhttp://www.futureaccountant.com/probability/\nhttp://www.schoolingkids.com/
438115
Okay, first you have to calculate how many the combinations of the first three teams are. At this point, arrangement is irrelevant. So, it'll be 15C3. Sorry, I don't exactly remember the full formula. Then, you'll get 455. Now, the arrangement is becoming relevant, since, the three teams could be number one, two, or three. So, at this point, permutation takes place. You are to multiply the previous answer with 3P3(or 3!, same things, aren't they?!). So,the answer will be 2,730! I know, it could be similar to another or other answer(s), but in math, the way of solving a question is more important than the answer itself! (That's what i think!)
257322
It's very simple. Every taster has to choose one out of 5 cups, so on every round the probability of a single taster choosing A is 1/5.\nSince the event is repeated 4 times independently, the overal probability will be 1/5 of 1/5 of 1/5 of 1/5.\nThat is, (1/5)^4 = 1/625.
1120894
For years that they played more than 30 games: \n\n1994 Team was 49-9 - Win % = .845\n1966 National Champion team was 27-6-1 -  Win % = .808\n1991 Team was 52-13 - Win % = .800
131088
try this site\n\nhttp://www.futureaccountant.com/probability/
224785
NO!!  ON ESPN THEY SAID THAT OUT OF THEIR 3 MILLION BRACKET ENTRIES NO ONE GOT IT ALL RIGHT.  CRAZY RIGHT.  ONLY 3 PEOPLE GOT LIKE 15
967346
This probability problem can be used using Binomial Distribution.  \n\nGiven a number of 'x' success, the probability in a binomial experiment is:\n\nP(x) = [n choose x]*p^x*q^(n-x)\n\nwhere:\n\nn choose x is a combination problem: n!/[(n-x)!x!]\np = probability of success\nq = probability of failure\nn = total selection\nx = successes\n\nIn this problem, we are looking for at least one case has a success (which is one of the highest 9 numbers).  In essence, we are solving for:\n\nP(x=1) + P(x=2) + P(x=3) + P(x=4) + P(X=5).  \n\nHowever it would be easier to find the probability of the failures and subtract from the total probability because we only have to calculate 1 probability.  This would be:\n\n1 - P(x=0) is what we are looking for.  Now all we need is to calculate the probabilty.\n\nFor P(x=0), we have the following:\n\nn = 5\nx = 0\np = (9/26) = probability of success\nq = (17/26) = probability of failure\n\nPlugging into P(x), we get:\n\nP(x=0) = (5!/5!)(9/26)^0(17/26)^5\n\nSolving, you get P(x=0) = 0.12\n\nTherefore, \n\n1 - P(x=0) is ------>\n\n1- 0.12 = 0.88\n\nBelow is a website devoted to Binomial Distribution exercises\n---------\n\nHope this helps
152796
try this site...\n\nhttp://www.futureaccountant.com/probability/
245215
There are five tosses with two choices each.\nEach toss is a 50% chance of coming up heads regardless of the results before it or after.\nThe probability of getting a certain number of heads would be calculated as such: number of heads/number of choices.\nWe are looking for two heads and there are ten choices (heads and tails of each coin)\nThe probability of it would be 2/10 or 20%\n\nIt is correct in that each toss is 50% chance, but in this case we are speaking of of the chances of a group of coins having a certain number turn up heads.  We are also speaking of the probability of this before any actions are taken.  The odds will change with each coin toss till the last one which will be 50%
192950
The odds have to do with how much money you need to bet to win 100 on a favorite, or how much you win on a $100 bet on an underdog. A plus is for the underdog (team expected to lose) and a minus is for the favorite (team expected to win).\n\nSo, if a team was -150, you would have to bet $150 to win $100 (would get $250 back). If a team was +150, if you bet $100 and they won, you would win $150 (get back $250).
1043446
There are 32 teams who\nthink they can win!(~~)
185383
There are a total of 63 games to be played (every team will lose exactly once, except the entire winner of the tournament).\n\nIn each game, there are exactly two possible outcomes (given the results of the games leading up to it).\n\nTherefore the number of completely-filled brackets is 2^63, that is 2 to the 63rd power, that is 2*2*2*...*2 (63 times).\n\nThis is an *enormous* number.  To give you an idea, it is 19 decimal digits long.\n\nIn general, if there are n teams, then there are 2^(n-1) ways to fill out the bracket completely.  This formula matches your examply (although, with 2 teams there are actually 2 possibilities).
600596
There are 6 possible outcomes for the roll. There is only one outcome when the number is 4. Therfore the chances of getting a 4 are 1/6.
258297
Each game you have a 50/50 chance of being right, so that's 1/2^1(^is the sign for an exponent), there are 32 first round games, so for picking the perfect first round the chance is 1/2^32, for the second there are 16 games, you multiply the odds of picking these right (1/2^16) and continute for the number of games each round, so your formula will be, 1/(2^32*2^16*2^8*2^4*2^2*2^1) which will equal 1/9,223,372,036,854,775,808.  There is 9,223,372,036,854,775,808 different combinations of winners, and only one is correct for the perfect bracket.
262364
There's 60 picks in the draft.  About 40 of them will be college players, maybe 20 of them are in the tournament, with the others coming from teams that didn't make it.  The rest will be foriegn players.  A few may try out as unrestricted free agents as well.