Odds of predicting winners of all World Cup games?

My company is thinking of launching a World Cup bracket contest for folks in the Middle East to fill out brackets to win a money prize. We have had some discussion about what the users should choose and came up with winners and final score of each match. My questions:

What are the odds that someone could predict the winner of every World Cup match?

What are the odds that someone could predict the winner AND the final score of each match? I know this is essentially impossible, but are the odds easy to figure?

I tried to sort the math but I know I am not close to the real answer. For the first question I came up with 64 the 64th power...am I close?

For the second part, predicting the scores...I'm going to assume this is outside the realm of human comprehension?

With all those headlines about World Cup of Bribery for merchandizing rights, sex for umpires, money for players ... why don't you figure out which country has the most money with which to pay bribes?

Quote: FleaStiff

With all those headlines about World Cup of Bribery for merchandizing rights, sex for umpires, money for players ... why don't you figure out which country has the most money with which to pay bribes?

That's easy...Qatar is the place to target if you want bribes.

So, 64 games.

But, there are ties possible in first 48.

My initial thought is that it would be a bit more likely to hit than Buffett's billion dollar NCAA bracket pool because there are more "sure thing" games. But ties probably negate much of that difference.

As for predicting the score of every game. Impossible. Even if you capped the number of goals to 4, that's 30 possibilities per game for opening round games.

I had the odds of predicting the scores capped at 6 goals max for either team and had 42 possible score combinations not including ties. The odds would be 42 to the power of 42 for guessing the scores? All I know is it "breaks" Excel when I enter power(42,42) as the equation so there is little chance of paying the grand prize.

From a story about the NCAA brackets, with 64 single-elimination teams, calculated by Warren Buffett's math guy:

The odds of anyone picking a perfect bracket are subject to debate. The 9.2 quintillion estimate assumes each team has a 50% chance of winning every game, which past practice suggest is not the case. Others have put the odds at 7.4 billion to 1, or 42 times worse than your chance of winning the Powerball lottery.

"There is no perfect math...There are no true odds, no one really knows," Buffett told CNN in January when the challenge was announced.

CNN story

Thank you for the info. I based my initial estimates off this article, substituting the 64 teams with 32 and then figuring a 3 game round robin schedule. I am not a mathematician (or even particularly proficient with math) so this stuff scrambles my brain.

Quote: abugabby

My company is thinking of launching a World Cup bracket contest for folks in the Middle East to fill out brackets to win a money prize. We have had some discussion about what the users should choose and came up with winners and final score of each match. My questions:

What are the odds that someone could predict the winner of every World Cup match?

What are the odds that someone could predict the winner AND the final score of each match? I know this is essentially impossible, but are the odds easy to figure?

I tried to sort the math but I know I am not close to the real answer. For the first question I came up with 64 the 64th power...am I close?

For the second part, predicting the scores...I'm going to assume this is outside the realm of human comprehension?

If I understand correctly, the round of 16 matchups and subsequent matchups are predetermined by position within group (i.e. 2nd in group G will play 1st in group H.) So it is entirely possible to arrange a pool to determine the winners of all 48 first-round games, their appropriate spots in the knockout rounds, and the subsequent knockout round winners.

The biggest problem, of course, that a World Cup pool would pose that an NCAA tournament pool would not is the fact that ties are a possibility in 75% of games - and the format of the first round is different. How exactly would one award points for a tie? The obvious answer for picking a winner is to award the picker the same number of points the team he/she picks earns, so if I pick Germany to beat Portugal and they do, you would give me 3 points, but if they tie, I would earn only 1 point.

So do you allow people to pick ties, thus making it like the baccarat side bet with a likely lower expected value unless you award something like 12 points for picking a tie correctly? Do you forbid picking ties, thus capping the number of points a person can earn in a tie game at one? Do you allow people a set number of points to wager on each game like a casino game, awarding the equivalent of even money for a win, an equal loss for a loss, and a push for a tie, thus creating an unusual tie-based side bet?

My proposal - instead of picking the individual winners in the first round, allow people to pick the group winner and the group runner-up. Based on that, people can then pick the winners of knockout games based on position and be awarded points for correct predictions based on the teams predicted still being alive. (Simply put, let's say that, out of Germany, Ghana, Portugal, and USA, a person picks Germany to win the group and Portugal to finish second. Award, say, 1 point for such a pick if Germany advances and an extra point if Germany advances as the group winner. Then, if the player picks Germany to advance to the Round of 8, award points if Germany reaches that round. The numbers are hypothetical, of course.)

You can do bonuses for predicting number of goals, the score of the last game, etc. But it's best to keep it simple, especially if it's new.

8 pods, 6 games per pool: 48 pool games
16 team single elimination tourney: 15 standard games in tourney plus 1 third place game

Picking the whole thing right is rather unlikely to happen.

If you asked me to put something simple and that rewards soccer knowledge, I would do the following:
Everyone picks the winner of each pool, and then 2nd place in each pool. Then once the bracket is filled in real life have everyone fill it out just like an NCAA bracket.

-1 pt for each team which is picked to make the knockout stages (1 or 2 in the pool) which does
-1 pt for each team which is picked to win their group which does (in addition to the previous point for making the knockout stages)
-1 pt for each correct pick in the round of 16
-2 pts for each correct pick in the quarter-finals
-3 pts for each correct pick in the semi-finals
-5 pts for each correctly picking the WC winner
-3 pts for each correctly picking the 3rd place game winner

You could tweak the scoring, but I think what I threw out there will work well (if anything I might raise the amount of the WC winner and 3rd place game winner). You could also split the contest between pool play and knockout play. However, I'm not sure that the pool stage would be very interesting without some sort of change in the scoring.

Thank you dealer and endermike for your input! Your suggestions are awesome and something we hadn't thought of.

We had initially planned to make the total possible points 500. We came up with 96 possible points in each stage (1 point for the winning team and one point for the score) with 20 points available as bonuses for recruiting friends who will play the bracket game, sms interactions, etc.

For 90 minutes result (Win, Lose, Tie) assuming equal probability for each of 1/3, then the odds are 3^64:1 = 3,4 x 10^30.
Even though some results will be more probable than others and say divide this figure by 3, the odds are still 10^30.
So basically an impossibility.

If you had only 2 results for win-lose for all games (which you can have for the last 16 games only) the figure becomes 2^64 = 1,8 x 10^19.
Say divide this by 3, the odds are still 6 x 10^18.

Somthing more plausible is that for first 48 games, the player choses 1X (Win or Tie) or 2X and for the last 16 games Win or Lose for the game.
That's: (2/3)^48 X (1/2)^16 and taking the inverse of it to get odds in the form xxx : 1
The answer becomes = 1,8 x 10^11.
Divide by 2 to take into account that probs are not the same for all options then is around 10^11
ie 100.000.000.000 : 1.
100 billion to 1.

Awesome answer! :)