Pascal's Generalization

A Math Forum Project

Table of Contents:

Famous Problems Home

The Bridges of Konigsberg
· Euler's Solution
· Solution, problem 1
· Solution, problem 2
· Solution, problem 4
· Solution, problem 5

The Value of Pi
· A Chronological Table of Values
· Squaring the Circle

Prime Numbers
· Finding Prime Numbers

Famous Paradoxes
· Zeno's Paradox
· Cantor's Infinities
· Cantor's Infinities, Page 2

The Problem of Points
· Pascal's Generalization
· Summary and Problems
· Solution, Problem 1
· Solution, Problem 2

Proof of the Pythagorean Theorem

Proof that e is Irrational

Book Reviews

References

Links

Let us continue our fictional story of Pascal and Fermat. Perhaps Pascal would write:

Dear Pierre,

    I do in fact find your solution satisfactory. Enclosed you will find the money I owe you. Inspired by your solution I have given more thought to this problem. I realized that if, by chance, we had decided upon a larger number of points as the winning amount for our game, your method of writing out all the possibilities would become tedious. Thus I seek to generalize the problem. I believe I have found a satisfactory solution.

    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:

    4C2 + 4C3 + 4C4
    ---------------
    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

    n*(n-1)*(n-2)*...*2*1
    nCr = ---------------------------------------------------
            r*(r-1)*...*2*1 * (n-r)*(n-r-1)*...*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

    Thus, let us suppose that two players are playing, and the first player needs n points to win, while the second needs m. To compute how the stakes should be divided, one computes row (n+m) of the triangle, and then adds up the first m entries. This number is corresponds to the first player's (who needs n points to win) share of the stake. The remaining entries should be added up to determine the second players share. Thus, if s represents the number of Francs wagered, player 1 should receive

    (sum of first m entries)
    --------------------------------- * s
    (sum of entire row)
    Francs.

    Player 2 should receive

    (sum of last n entries)
    -------------------------------- * s
    (sum of entire row)
    Francs.

    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

to the Problem of Points
to a Summary and Some Problems

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August, 1998