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How can the game be made fair It may be thought that I have too many red marbles in my pocket and that adding a green marble will rectify things However, examine Figure 47 where we have two green and two red marbles There are six possible samples of two marbles; two of these contain marbles of the same color while four contain marbles of different colors Adding a green marble does not change the probabilities at all!

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Why is this so Part of the explanation lies in the fact that while the number of red marbles and green marbles is certainly important, it is the number of sides and diagonals of the gure involved that is crucial It is the geometry of the situation that explains the fairness, or unfairness, of the game It is interesting to nd, in the above example, that if we have three red marbles and one green marble, then the game is fair The unfairness of having two red and one green marbles in my pocket did not arise from the presumption that I had too many red marbles in my pocket I had too few! Increasing the number of marbles in my pocket is an interesting challenge Figure 48 shows three red and three green marbles, but it is not a fair situation

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Geometric Probability Table 41 R 3 6 10 15 G 1 3 6 10 R+G 4 9 16 25

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The lines in Figure 48 show that there are 15 possible samples to be selected, 6 of which have both marbles of the same color while 9 of the samples contain marbles of different colors With a total of six marbles, there is no way in which the game can be made fair Table 41 shows some of the rst combinations of red and green marbles that produce a fair game There are no combinations of 5 through 8, 10 through 15, or 17 through 25 marbles for which the game is fair We will prove this now One might notice that the numbers of red and green marbles in Table 41 are triangular numbers, that is, they are sums of consecutive positive integers 1 + 2 = 3, 1 + 2 + 3 = 6, 1 + 2 + 3 + 4 = 10, and so on The term triangular comes from the fact that these numbers can be shown as equilateral triangles: r r r r r r r r r r

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We also note that 1 + 2 + 3 + + k = k(k + 1)/2 To see that this is so, note that the formula works for k = 2 and also note that if the formula is correct for the sum of k positive integers, then 1 + 2 + 3 + + k + (k + 1) = (k + 2)(k + 1) k(k + 1) + (k + 1) = 2 2

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which is the formula for the sum of k + 1 positive integers This proves the formula Let us see why R and G must be triangular numbers For the game to be fair, R G + 2 2 or 2R(R 1) + 2G(G 1) = (R + G)(R + G 1) = 1 R+G 2 2

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and this can easily be simpli ed to R + G = (R G)2 This is one equation in two unknowns But we also know that both R and G must be positive integers So let R G = k Then, since R + G = (R G)2 , it follows that R + G = k2 Solving these simultaneously gives 2R = k + k2 or R = k(k + 1)/2 and so G = k(k 1)/2 showing that R and G are consecutive triangular numbers

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