What is the probabilty of filling out the march madness basketball bracket completely correct?

I need the equation used to figure out the probability of filling out the basketball bracket completly correct, the equation for picking the sweet sixteen correct, the elite eight, and the final four. What equation do I use to figure out the probability of getting any of these correct?What is the probabilty of filling out the march madness basketball bracket completely correct?
for each game, you have the possibility of getting it right 1/2 of the time. The total probability would be (1/2)^x, where x is the number of games played.What is the probabilty of filling out the march madness basketball bracket completely correct?
There are two basic ways to do this type of problem 1) wit combunations or 2) with probability.



This wasn't one of the questions but assume you want the probability of picking the winner. ( assume there are 64 teams, forget the play in game) The combinations method says there is 1 winner out of 64 teams so your probability is 1 / 64

the probability method says that your picked team has to win six games in a row so the probability is (1/2)^6 = 1/64



to pick the final four you need to pick the winner of 4 separate tournaments each with 16 teams.

the probability of picking each tournamen is 1/16 so the probability of getting all 4 is (1/16)^4



similarly the probability for the top 8 is (1/8)^8



the probability for the top 16 is (1/4)^16



the probability for the top 32 that is the first round is (1/2)^32 which is what you expect = 2.3*10^(-10)





the bottom line then is that you must pick the winner of each of 63 games to get the entire bracket correct. he probability is (1/2)^63 = 1.08*10^(-19)



If you think the probability of picking a game correctly is more than 1/2 then use that probability instead of 1/2 in the above.

the probability is p1 * p2 * p3 * ... * p63 where each pi is probability of picking that particular game.What is the probabilty of filling out the march madness basketball bracket completely correct?
Assuming that each game has a 50/50 chance of either team winning (which they don't). It would just be (.5)^63 (because there are 63 games). Think about it the odds of picking two games right is 1 in 4 or 25% because there are 4 possible out comes: You could pick both games right (1), both games wrong (2), the first game right and second wrong (3), or the first game wrong and second right (4) = (.5)^2 = .25 = 25%. Every game you add you just multiply by .5 again.