Table of Contents:
The Bridges of Konigsberg
The Value of Pi
Prime Numbers
Famous Paradoxes
The Problem of Points Proof of the Pythagorean Theorem

Let us continue our fictional story of Pascal and Fermat.
Perhaps Pascal would write:
Dear Pierre,
Let us take another look at your solution to our game. It seems to me that if we could discover a general way to compute the number of outcomes that go in your favor, then we would be well on our way to a general solution. Any outcome that featured two or more heads turning up meant a win for you. The total number of such outcomes is equivalent to the number of ways to choose two objects from four, plus the number of ways to choose three objects from four, plus the number of ways to choose four objects from four. Here the 'events' of the coin coming up in your favor become 'objects' in our counting terminology. Let us denote the number of ways to choose to choose r objects from n objects as nCr. Thus I think the likelihood that you would have won the game is given by:
 total outcomes Now, we have previously discussed [see the Dr. Math Combinations Questions Page] how many ways there are to choose r from n, not counting duplicates. We established that this number is
nCr =  r*(r1)*...*2*1 * (nr)*(nr1)*...*2*1
Now, at first it would appear that computing these quotients is also quite tedious. However, I have
noticed that they correspond to the numbers on various rows of my 'adding triangle'
[Pascal's Triangle]  that is, the figure
can be represented by
 * s (sum of entire row) Player 2 should receive
 * s (sum of entire row) Note that for our case this yields ((1+4+6)/(1+4+6+4+1))*100 = 68.75 F. I pray for your friend's swift recovery and hope that all else is well with you, Your Friend Blaise 
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