 # 380 pickover/pickover.04.p

## Description

This article is from the Puzzles FAQ, by Chris Cole chris@questrel.questrel.com and Matthew Daly mwdaly@pobox.com with numerous contributions by others.

# 380 pickover/pickover.04.p

Title: Cliff Puzzle 4: Time in a Bottle
From: cliff@watson.ibm.com

If you respond to this puzzle, if possible please include your name,
address, affiliation, e-mail address. If you like, tell me a little bit
about yourself. PLEASE ALSO directly mail me a copy of your response
in addition to any responding you do in the newsgroup. I will assume it
is OK to describe your answer in any article or publication I may write
in the future, with attribution to you, unless you state otherwise.
Thanks, Cliff Pickover

* * *

Consider a chain of bottles (B) each connected to one another by a thin
tube. A marble is placed in bottle 1.
Each tube contains a one-way valve so marbles can only
go from left to right in the tubes which are symbolized with "-" marks:

```1   2   3   4
B - B - B - B -
```

The tubes are thin so it takes
1 hour of constant random shaking to get the marble from B1 to B2.
Likewise for each bottle.

I have not fully described the bottle collection. Each bottle
has a backward 1-way tube to bottle 1. I've tried to diagram these
with "*" symbols. Each time the marble enters bottle B(N) it has
a 50% probability of going back to bottle 1 via these tubes.

```****<********
*           *
***<*****   *
*       *   *
* * *   *   *
1   2   3   4
B - B - B - B -
```

Stop And Think

1. In how many hours will you expect to get the marble out of bottle 10
after placing the marble in bottle 1?

2. Is there a general formula for the amount of time
required to get the ball out of bottle N into bottle N+1 given
a probability P of backwards motion (given as 50% in this problem)?

3. In how many hours will you expect to get the marble out of bottle 10
after placing the marble in bottle 1 given two backward tubes for each
bottle instead of one backward tube?

pickover/pickover.04.s

-------------------------

Subject: Re: Cliff Puzzle 4 (SPOILER)
Newsgroups: rec.puzzles

In article <1992Sep15.205532.4172@watson.ibm.com>, Cliff writes:

> 1. In how many hours will you expect to get the marble out of bottle 10
> after placing the marble in bottle 1?

The expected amount of time to go from state n-1 to n (state 11 is an
absorbing state) is

E(n-1,n) = 1 + E(1,n)/2 for 1<n<11;

also

E(n-1,n+1) = E(n-1,n) + E(n,n+1) for 1<n<11.

If n=2 then E(1,2) = 1 + E(1,2)/2 ==> E(1,2) = 2 (it should be clear
that no E is infinite for this problem).

E(2,3) = 1 + E(1,3)/2 = 1 + E(1,2)/2 + E(2,3)/2 ==> E(2,3)/2 = 2
==> E(1,3) = 6.

I claim that in general E(1,n) = 2^n - 2 and E(n-1,n) = 2^(n-1).
Assume true for n, then E(n,n+1) = 1 + E(1,n+1)/2 = 1 + E(1,n)/2 +
E(n,n+1)/2 ==> E(n,n+1)/2 = 1 + E(1,n)/2 ==> E(n,n+1) = 2 + E(1,n)
==> E(n,n+1) = 2 + 2^n - 2 = 2^n. Furthermore E(1,n+1) = E(1,n) +

E(n,n+1) = 2^n - 2 + 2^n = 2^(n+1) - 2. Therefore by induction the
result is established.

Now E(1,11) = E(1,10) + 1 (ball can't go back to bottle 1 after
leaving bottle 10) = 2^10 - 1.

> 2. Is there a general formula for the amount of time
> required to get the ball out of bottle N into bottle N+1 given
> a probability P of backwards motion (given as 50% in this problem)?

I'd have to check my work, but I get E(n,n+1) = q^n, where q = 1/(1-p).

> 3. In how many hours will you expect to get the marble out of bottle 10
> after placing the marble in bottle 1 given two backward tubes for each
> bottle instead of one backward tube?

I get E(1,n) = (q^n - q)/(q-1), so E(1,11) = E(1,10) + 1 =
(3^10 - 3)/2 + 1.
-------------------------

In regards to your fourth problem, the following comments (marked
with a ">") should be added. I thought the answer was quite surprising!
---

The expected amount of time to go from state n-1 to n (state 11 is an
absorbing state) is

E(n-1,n) = 1 + E(1,n)/2 for 1<n<11

> since we expect it'll take an hour for the ball to leave bottle n-1,
> and it then has a probability of 1/2 of returning to the first bottle;

also

E(n-1,n+1) = E(n-1,n) + E(n,n+1) for 1<n<11

> since the only way of getting to state n+1 from n-1 is to first
> go from state n-1 to n, and then from n to n+1; the total expected
> time is the sum of the two.

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