I made a convincing argument, but I am curious what kind of persuasive MATLAB-based ‘proofs’ can be made. I am offering a MATLAB t-shirt to the most persuasive entry. Use published MATLAB files, GUI’s, MATLAB script or function, whatever you think will be most convincing. Contest ends this time next week.
Assumptions: The Frisbees (or coins) are either ‘heads’ or ‘tails’. The coins may or may not be ‘fair’ (i.e. they might be heads 80% of the time!), however the odds of heads versus tails for the two coins is the same for both coins. Send entries to hull@MathWorks.com (no .rar files please!)
(Here is the result of this now closed contest)
Let p be the probability of “heads”, so (1 - p) is the probability of “tails”. The probability of same is therefore (p^2 + (1-p)^2) and the probability of different is (2*p*(1-p)). We can plot those over p = [0,1]:
p = (0:0.001:1)'; plot(p, [p.^2 + (1-p).^2, 2*p.*(1-p)]); legend Same Different
From this you’ll see that the probability of same is always greater than or equal to the probability of different, so it’s the better call to make.
Is better than I don´t say to much with me happend the same… lol and I love to play too !
Let p be the probability of “heads”, so (1 - p) is the probability of “tails”. The probability of same is therefore (p^2 + (1-p)^2) and the probability of different is (2*p*(1-p)). We can plot those over p = [0,1]:
From this you’ll see that the probability of same is always greater than or equal to the probability of different, so it’s the better call to make.
Sorry, Doug. I only just noticed your “email me with solutions” request. I hope I haven’t spoilt it for anyone else.