Date: Tue, 9 Mar 2000 15:40:07 +0000

From: David L. Arnett

Being somewhat foolish in nature, I tackled this one by hand. I started out by methodically coming up with all possible ways the problem could occur, beginning with k=10. This was easy, as you had to have either xxxxxxxxxx or yyyyyyyyyy. This gives us a probability of 2*(1/2)^10. Next was k=9. Still easy, but more possibilities: xxxxxxxxxyx, xxxxxxxxyxx,... yxxxxxxxxxx. The same goes for reversing the x's and y's. This gives us a probability of 20*(1/2)^11 (the extra 1/2 comes into play as there are now 11 sticks of gum being drawn). Next was k=8. Somewhat harder (don't worry, I found a pattern I'll get to soon): xxxxxxxxxyyx, xxxxxxxxyxyx, xxxxxxxxyyxx, xxxxxxxyxxyx, xxxxxxxyxyxx,... yyxxxxxxxxxx. Once again, the same goes for reversing the x's and y's. This gives us a probability of (add it up, but don't use your fingers; you don't have enough) 110*(1/2)^12. For k=7, the same work provided 440*(1/2)^13. OK, I'm tired of reversing the x's and y's. Let's drop the factor of 2 by simultaneously lowering the exponent of the (1/2) terms by 1. Think about it, it works. This gives us (1/2)^9, 10*(1/2)^10, 55*(1/2)^11, and 220*(1/2)^12 for k=10, 9, 8, and 7, respectively. As I went through the possibilities, I also noticed the pattern I was using could be duplicated mathematically by determining the multiplier as follows: For k=8: Sum(a=1,10){a} = 1+2+3+...+10 = 55 For k=7: Sum(a=1,10){Sum(b=1,a){b}} = (1)+(1+2)+(1+2+3)+...+(1+2+3+...+10) = 220 This technique can be continued through k=1. As for the exponents, they simply increase by 1 for each value of k. This gives us final results of: k=10: 1*(1/2)^9 = 0.001953125 k=9: 10*(1/2)^10 = 0.009765625 k=8: 55*(1/2)^11 = 0.02685546875 k=7: 220*(1/2)^12 = 0.0537109375 k=6: 715*(1/2)^13 = 0.0872802734375 k=5: 2002*(1/2)^14 = 0.122192382812 k=4: 5005*(1/2)^15 = 0.152740478516 k=3: 11,440*(1/2)^16 = 0.174560546875 k=2: 24,310*(1/2)^17 = 0.185470581055 k=1: 48,620*(1/2)^18 = 0.185470581055 --------------- (Check) Total = 1 An interesting note: it is possible to come up with the sums for the multipliers easily by hand in the following manner (trust me, I didn't just plug through that many nested sums): for k=9, we have (for each possible order): 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 10 Now, start with 1 and add the corresponding number from the k=9 sequence to obtain the next number in the k=8 sequence (in this case, all the numbers added are one): 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55 For k=7, we start with 1, add 2, then add 3, etc. (numbers from the k=8 sequence): 1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 + 55 = 220 For k=6, we get: 1 + 4 + 10 + 20 + ... and so on. Why does the probability of k=1 equal the probability of k=2? Because the multiplier for k=1 is twice the multiplier for k=2, and the exponent for the (1/2) term is raised by one, cancelling each other out. Why is that? I have no idea. Anyone?

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