Probability Question (non-poker related)
 

Hello,

I was talking with a friend at work about the March Madness Tournament and what the odds would be of picking all the games right. Being the mental midgets that we are we were unable to figure this out so I'm wondering if someone else can do the calculation and give me the approx. odds of this happening.

Assuming each game is a coin toss and there are 32+16+8+4+2+1 games played, what are the odds of picking every game correctly?

Thanks,

Swede

Re: Probability Question (non-poker related)
 

If every game is a coinflip...

The odds of picking one game is 1/2. The odds of picking two are (1/2)*(1/2).

Therefore, the odds of picking sixty three games is 1/2 taken to the sixty-third power, or one in 9,223,372,036,854,780,000.

I can't count that high.

Of course, they aren't coinflips.

Re: Probability Question (non-poker related)
 

[ QUOTE ]
Assuming each game is a coin toss and there are 32+16+8+4+2+1 games played

[/ QUOTE ]

Stupid nitpick... there's actually one more game than that... there is the "play-in game" between the 65th and 64th team.

BTW... The easiest way to figure out how many games are played in any single-elimination tournament is to realize that every team except one will lose exactly once, and each game must have exactly one loser. Therefore the number of games must be N-1, where N is the number of teams in the tournament. This works regardless of byes.