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A marbles game Changqing Li
A marble falls into the container. When the marble crashes the barrier in the first row, then the marble falls into the second row. When it crashes the barriers in the second row, then it falls into the third row. In this way, the marble finally down to the bottom of the container. According to the area of marble, we can win a different prize. The prize of A and G is the best, and the prize of D is the worst. The prize of A is much better than D. This can tell us that the probability of A is much better than D.
We can analyses the situation of D. The marble falls into the D has two situations: one is from the upper left and another is from upper right. So the probability of D is half of the sum of these two probabilities. Then we can get probabilities for every region.
From the result, we can easily find out that the numerator of every probability is the same as the every entry of Pascal’s Triangle.
So we can easily know that the probability of A is much greater than D according to some properties of Pascal’s Triangle. If we have 1 + C1,n + C2,n +…+ Cn-2,n +Cn-1,n +1 marbles. Then make them fall into the container and down to the n+1 regions. And the number of marbles from left to right in each region is 1; C1,n ; C2,n ;…+ Cn-2,n ;Cn-1,n ;1 This is same as the entries of (n+1)th rows in Pascal’s Triangle.
