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Subject: Re: Random chewing
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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