 Subject: Re: Random chewing
Date: Mon, 07 Sep 1998 22:34:43 +0200
From: Bernd Liebermann

Hi Alex,

here are some notes on the random chewing problem on your exchange page. Unfortunately, clicking on the pencil at the bottom of the page only had the effect that that same page was being reloaded, so you get my comments via conventional e-mail.

I assume that Tarra has already passed her exam succesful several months ago. Nevertheless, here are the answers to her questions:

Q1. "What is the probablity that k is 1? 5? 9?"
The probabilities are as follows:

p(=1) = 0,185471
p(=2) = 0,185471
p(=3) = 0,174561
p(=4) = 0,152740
p(=5) = 0,122192
p(=6) = 0,087280
p(=7) = 0,053711
p(=8) = 0,026855
p(=9) = 0,009766
p(=10) = 0,001953

Why are these the probabilities?

Try imaging what happens in the story with the cool dude by means of a decision tree; for the very simple case of n=2 sticks, the tree would look as follows:

```            (0,2)*

(1,2)
(0,1)*
(1,1)
(1,0)*
(2,2)
(0,1)*
(1,1)
(1,0)*
(2,1)

(2,0)*

1/1    1/2   1/4    1/8
```

The leafs (one of both packs is empty) are marked with a *. In the bottom line there are the probabilities of reaching each node in the respecting level above under condition of random choice. You can now add the probabilities of reaching the leafs for k=1 and k=2 which results in p=0.5 for both k=1 and k=2.

For larger n's, the tree grows rapidly, of course. There are 20 leafs for n=3, 70 for n=4 and 184756 for n=10. I have no idea, how many different ways of emptying one of the two packs there are for n=11. Does anyone of you know the answer?

What I did (as a non-mathematician) in order to find out the probabilities for n=10 was writing a very small program that marches back and forth through the whole tree, always adding the probabilities at the leafs for the different values for k. But unless the topic of Tarra's exam was programming, I assume that there is an easier way to get the probabilities in a paper/pencil mode. (??)

Q2. "What is the most likely value for k?"
It seems remarkable to me that for any n between 2 and 10, p(k=1) equals p(k=2); presumably this is also true for any other n. I would appreciate if someone could explain this to me in a few sentences.

Q3. "Ann and Betty have made a bet. If K is less than 3 Ann pays Betty \$10. If k is greater than 3 Betty pays Ann \$d. What value for d would make this game fair?"

(p(k=1) + p(k=2)) * 10 = (p(k=4) + p(k=5) + ... + p(k=10)) * d d = 3,70942 / 0,454497 = 8,16...

Best regards,
Bernd Liebermann |Reply| |Up| |Previous| |Next| |Exchange index| |Contents| |Store|