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#11
11-22-2007, 12:55 AM
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Re: Math puzzle: Breaking the camel\'s back.
[ QUOTE ]
[ QUOTE ] Part 1: Your camel's back can hold 1 unit. Straws have weights which are uniformly distributed from 0 to 1 unit, and independent of each other. You add one straw at a time. If the weights are 0.7, 0.1, 0.8..., then your camel's back breaks on the 3rd straw. On average, how many straws does it take to break your camel's back? [/ QUOTE ] There are many solutions. In white, I have written up the one I found without pencil or paper when I was 13 after an uncle posed the problem at a family Thanksgiving dinner, and at which blah_blah hinted. <font color="white"> The expected value of a random variable X taking positive integer values is the sum of the probabilities P(X>=1)+P(X>=2) + P(X>=3) + ... This useful trick is actually a change of the order of summation in something that looks like a single sum, but can be expanded: E(X) = P(X=1) + 2 P(X=2) + 3 P(X=3) + ... = P(X=1) + P(X=2) + P(X=3) + ... + _________P(X=2) + P(X=3) + ... + _______________ + P(X=3) + ... Summing rows, this is P(X>=1)+ P(X>=2)+ P(X>=3)+... Anyway, P(X>=n) = P(the first n-1 straws add up to less than 1, so with k=n-1 we want the sum of the probabilities that the first k straws add up to less than 1 for k=0,1,... The plane x1+...+xk =1 cuts off a simplex of volumn 1/k! from the unit cube, so the expected value is 1/0! + 1/1! + 1/2! + ... = e. </font> Also at that dinner, we discussed the geometric mean of the numbers in [0,1], which is again related to e. I'll post my solutions to Parts 2 and 3 later. I used a method which at least looks different from the suggestions of blah_black and bigpooch. However, I would not be surprised if they were all similar, from some perspective. [/ QUOTE ] Was your uncle the anomaly, or is your entire family filled with mathematicians? 
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