Autistic Teen Picks First Two NCAA Rounds Perfectly
nbcchicago.com
nbcchicago.com
"CBSSports.com can not confirm Alex's entry -- the company doesn't track entries to their Bracket Manager program."
In other words: think of the tournament's seeding as a bracket filled out by the NCAA itself. There's only a 1 in 13 million chance that it will be right through the first 2 rounds.
What this kid did is amazing. Its more than the inevitable result of enough random guesses.
The main problem with your logic is that most people make the same choices. For example, only 1% of ESPN brackets had Cornell surviving the first 2 rounds this year, while 73% thought Kansas would make the championship.
To my knowledge, no one on ESPN.com has ever predicted all 65 games in the first 2 rounds correctly.
You are right, my analysis is terrible. Boom, inferiority admitted.
That said if he is that good reliably, there is lots of money to be made on sports betting.
For starters, in the lottery:
1) No selection (number) is better than another. i.e., the winners are random
2) Each round has no effect on the next
3) The conditions of the experiment will not change
The NCAA tournament is the opposite:
1) Some selections (teams) are clearly better than others. Furthermore, the tournament is structured to give the best teams the best chances of winning.
i.e., selection is not random.
2) Each round limits the possible outcomes of the next. To have a chance of picking Round 2 correctly, you must have already aced Round 1.
3) Each game can be affected by an unlimited number of variables. For instance, if John Wall gets hurt it would alter the chances of a Kentucky national championship. More subtly, so could a delayed flight.
Even picking the correct lottery numbers twice in a row is not a fair comparison, because the results of the first experiment have no bearing on the second.
I'd love to play the lottery if I knew certain balls were weighted more heavily than others.
If you think non-random events are difficult to predict, give me your longitude and latitude and I'll predict with in five minutes what time the sun will rise tomorrow. Want to bet $1000 against?
The real reason the it is more difficult is because the odds randomly are 1 in 4 billion for the first round, where as a lottery like the 6/49 is 1 in 13,983,816. However, the NCAA is not completely random, it would be difficult to discern what the real odds are but the fact that many players are within a couple games should give you some clue to the decrease in odds given the extra information.
If 100 million brackets are filled out and the realistic odds are 1 in 1.5 billion (given the extra information) then someone should fill out a perfect bracket every 15 years. If you look at the prizes being offered for a correct bracket, it should give you some indication as to the greatly reduced odds that one actually faces. If the odds are really 2^48 for a correct 2 round bracket then it would stand to reason that you could offer a 1 trillion dollar prize for a correct bracket with a ticket price of $1 and still have a house margin of 99.64%
100 - ((1 / ((2^48) / 1 trillion)) * 100) = 99.644
This simply isn't the case. (My math may be way off, I was never a stats major)
1 / (2^126)
I'll have to leave it to Wolfram Alpha to expand that one:
http://www.wolframalpha.com/input/?i=1+%2F+%282^63+*+2^63%29
Second round: 1 / 4^16 since there are 16 winners to pick and 4 choices for each. But this assumes you can pick a loser from round 1. So, it's actually 1 / 2^16 since there are 16 winners to pick and 2 choices for each.
Picking 2 rounds correct is 1 / 2^32 * 1 / 2^16 = 1 / 2^48
One thing: "this assumes you can pick a loser from round 1"
You can pick a loser from Round 1, because all selections are made before any games are played. So it looks like its:
1 / 2^32 * 1 / 4^16
(And we should probably add another 1/2 term to account for the play-in game =)
In other words, it's a straightforward application of the chain rule for probabilities:
p(A,B) = p(A) * p(B|A)
Probability of A (round 1 / selecting the right urn):
1 / 2^32 (32 games, 2 possible winners for each)
Probability of B (round 2:
1 / 4^16 (16 games, 4 possible winners for each)
Probability of A and B:
(1 / 2^32) * (1 / 4^16)
So it would appear the chain rule for probability agrees with my point. What am I missing?
The first term (.5^32) is the probability of picking all of the winners in round one (p(A)). The second term (p(B|A)), must be the probability of picking all the winners in round two, given that you picked correctly in round one. Again, that's .5^16, not .25^16.
Thanks for sticking with it.
Assuming all teams have the same probability of winning. So, it's not nearly as bad as 2^63. Probably in any given game one team has a 0.66 chance of winning. That takes it down to a much more manageable 2^63 in 3^63.
Wait . . .