# Super Bowl Coin Toss

Is there any advantage to calling the coin toss in the Super Bowl?  It doesn’t appear so, as in the 49 games to date, tails has come up 25 times and heads has come up 24 (see http://www.docsports.com/super-bowl-coin-toss-history.html for detail).  This is not surprising, given the conventional wisdom that a tossed coin comes up heads half the time.

Suppose, however, that you are playing a game with a coin toss, and you suspect that your opponent has an unfair coin, but you don’t know whether it comes up heads more often or tails more often.  As long as you are calling the toss, you can make the game fair by calling heads half the time and tails half the time (secretly flip your own coin first to decide).  Even if the game coin comes up, say, heads, all the time, you are fine, because you will have a 50-50 shot at choosing heads.  It is easy to check that no matter what percentage of the time heads comes up on the game coin, you will have a 50 percent chance of winning if you randomize your call.

Here’s how.  Suppose the game coin comes up heads percent of the time.  Since you choose heads 50% of the time, you will win because you chose heads .5*p  percent of the time and you will win because you chose tails .5*(1-p) percent of the time.  Add these together and you get .5p+.5-.5p=.5=50%.  So you never need to worry about the fairness of your opponent’s coin as long as you call the toss.

Some people have suggested a different solution.  It is as follows: flip two coins.  The caller calls even (both heads or both tails) or odd (one heads, one tail).  Assuming the two coins both have a p chance of coming up heads, this gives the following odds.  For Even it is p^2 (both heads) + (1-p)^2 (both tails), which equals 2p^2 +1 -2p.  It is easy to see that this hits its minimum at p=.5 (ok–not so easy if you dont remember calculus, but take the derivative and set to 0 to determine that p = .5 is a min or a max and take the second derivative to determine that it is a mininum).
Thus, if you call even, your odds will be AT LEAST 50%=2(.5)^2 + 1 – 2(.5).  If p is EITHER smaller or larger than 0.5, then your odds of winning by calling even are better than 50-50.

The second method is therefore advantageous to the person calling ‘even’ or ‘odd’, because by always calling ‘even’ the caller has odds of winning 50% if the coin is fair, and more than 50% if the coin is unfair, no matter whether it comes up heads more often or tails more often.