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in your multiplication: (2 + 3) 2 does not equal 2 2 + 3 2 and so ( a + b ) 2 cannot equal a 2 + b 2 ; instead, it equals:

    To calculate ( a + b ) 3 , and so on, you would need to include yet more combinations of terms:

    Does anything about the numbers in bold—the binomial coefficients—look familiar? Yes, indeed. If you want to know the coefficient of any term in the expanded form of ( a + b ) n , simply count down n rows in Pascal’s triangle (taking that first 1 you drew as row 0) and count across.
    This is a pleasant discovery, but why does it work? Because, in multiplying out your binomial, you have to multiply each term by all the others and then sort the results into groups. In the expansion of ( a + b ) 5 , how many groups with 5 a’ s are there? One; but there are five groups with 4 a’ s and 1 b and ten with 3 a’ s and 2 b’ s. So the coefficients tell you how many combinations of a and b you can make when you take them in groups of 1, 2, 3, or more.
    We are now nearly ready to return to the gaming table and claim our share of the winnings. Play has stopped at a point where you are r points short of victory and your opponent s points short. If you were to go on with neither player gaining an advantage, the longest that play could continue would be another r + s —1 plays, since those are all the points available before one of you must win. Let’s call this largest number of plays n . Now, on each one of these n attempts, you could win or your opponent could win; so the universe of all possible plays can be represented as (1 + 1) n : your potential point and your opponent’s potential point through your n remaining potential games. Given that your opponent had s points to make when the game is interrupted, how much of that universe of 2 n potential points is rightfully yours?
    (1 + 1) n has a familiar form: it’s a binomial. So if we count n rows down Pascal’s triangle (again, taking the top row as zero), we should find its expansion. And since a and b in this case both equal 1, the only significant terms in that expansion will be the binomial coefficients.

    The first term in the row is 1; this represents the single case in which you win all remaining points and the Venetian wins nothing; the second term represents the n cases in which you win all but one of the remaining points and your opponent wins one. Add this term to your total and go on, counting across and adding the coefficients until you get to the s th term of the expansion. From here on to the end, the territory belongs to your opponent: the coefficients represent the number of different ways he can win starting with s points to go. This is your point of division; by comparing the sum of the total number of ways you could have won with the total universe of points, 2 n , you get the proportion for dividing the stakes.
    Returning to the Venetian, still waiting impatiently: you were within 1 point of winning, but he needed 3. Therefore, s = 3. The total number of points that could be played ( n ) is 1 + 3 - 1 = 3. The universe of points is therefore 2 3 = 8. If you count down to the third row of the triangle, you find four coefficients: 1, 3, 3, 1. Since s = 3, you can add the first three coefficients to your total: 1 + 3 + 3 = 7. Compare this with the total universe of points, 8, and you find you have the right to 7/8 of the stakes, or 56 of the 64 pistoles— just as you found before.
    In 1654, the same year that God approached him in the form of fire, Pascal listed his accomplishments in a memorial to the Academy in Paris:
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    ... the uncertainty of fortune is so restrained by the equity of reason, that each of two players can always be assigned exactly what is rightly due. . . . By thus uniting the demonstrations of mathematics to the uncertainty of chance . . . it can take its name from both sides, and rightly claim the astonishing title: the geometry of chance.
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    The geometry of chance: