# 10 green bottles

wolfgang59
Posers and Puzzles 26 Jun '09 18:40
1. wolfgang59
26 Jun '09 18:40
If one green bottle should accidently fall there would be 9 green bottles sitting on the wall.
etc.
etc.

The probabilty of a bottle falling in 1 minute is 50%

How long before it is likely (>50ðŸ˜µ that there will bee no bottles left?
2. forkedknight
Defend the Universe
26 Jun '09 19:17
After 4 minutes, there is a .0625 probability that each individual bottle has NOT fallen. For all ten bottles, that provides a probability of .524 that all bottles have fallen.
3. wolfgang59
27 Jun '09 09:05
Sorry - original post not clear.

Within the first minute there is a 50% chance that A bottle will fall.

It is not a 50% chance of an individual bottle falling or not.
4. forkedknight
Defend the Universe
27 Jun '09 16:54
Originally posted by wolfgang59
Sorry - original post not clear.

Within the first minute there is a 50% chance that A bottle will fall.

It is not a 50% chance of an individual bottle falling or not.
I thought that seemed a little too easy ðŸ™‚

After n minutes, there will always be a completely normal distribution of bottles from 10 to 10-n, and the probabilities will be the coefficients of a simple polynominal expansion of (a+b)^n

This is true even after 10 minutes if you consider all negative numbers to be zero. Since the distribution is normal, it would follow that after 20 minutes, half of the distribution would fall below zero, and half above.

So after 20 minutes, there is a 50% chance that all the bottles have fallen.
5. forkedknight
Defend the Universe
27 Jun '09 16:57
Originally posted by forkedknight
I thought that seemed a little too easy ðŸ™‚

After n minutes, there will always be a completely normal distribution of bottles from 10 to 10-n, and the probabilities will be the coefficients of a simple polynominal expansion of (a+b)^n

This is true even after 10 minutes if you consider all negative numbers to be zero. Since the distribution is norm ...[text shortened]... and half above.

So after 20 minutes, there is a 50% chance that all the bottles have fallen.
Technically, it's after 19 minutes, since the distribution starts at minute 0
6. wolfgang59
28 Jun '09 06:34
Originally posted by forkedknight
Technically, it's after 19 minutes, since the distribution starts at minute 0
Could you elaborate?

I didnt have an answer to this problem; 20 mins seems a sensible guess but when you consider the quickest time they could all fall is 10 minutes (lets say that a bottle falls or not after each 1 min period) and the longest is infinite does the average come out at 20 minutes?
7. forkedknight
Defend the Universe
28 Jun '09 18:322 edits
Originally posted by wolfgang59
Could you elaborate?

I didnt have an answer to this problem; 20 mins seems a sensible guess but when you consider the quickest time they could all fall is 10 minutes (lets say that a bottle falls or not after each 1 min period) and the longest is infinite does the average come out at 20 minutes?
In 1 minute, there is a 1/2 chance of 10 bottles left, and a 1/2 chance of 9 bottles left. (a+b)^1 = 1*a + 1*b

in two minutes, there is a 1/4 chance of 10 bottles left, a 2/4 chance of 9 bottles left, and a 1/4 chance of 8 bottles left. (a+b)^2 = 1*a^2 + 2*ab + 1* b^2

In three minutes: (a+b)^3 = 1*a^3 + 3*a^2b + 3*ab^2 + 1*b^3
So there is a 1/8 chance of 10 bottles
3/8 chance of 9 bottles
3/8 chance of 8 bottles
1/8 chance of 7 bottles

etc, etc,

As you can see, there will always be an equal sum of coefficients above and below the median. Also, as you'll notice, the exponent of "b" can represent the number of bottles that has fallen. so when the coeffient of b^10 is the maximum coefficient, there there is a >=50% chance that all bottles has fallen

You will notice that since this is a binary expansion, the sum of all of the coefficients will always be 2^n
http://cose.math.bas.bg/webMathematica/webComputing/PolynomialExpansion.jsp
8. forkedknight
Defend the Universe
28 Jun '09 18:48
You can also think about it this way. Starting flipping a coin. After how many flips would you expect, on average, to flip heads 10 times.

You might say 20 (like I did at first) but after 19 flips, you have exactly identical probabilities of flipping either 9 or 10 heads.
9. forkedknight
Defend the Universe
28 Jun '09 18:52
This reminds me of a funny joke:

http://www.math.ncku.edu.tw/~cfnien/expand.jpg
10. AThousandYoung
All My Soldiers...
29 Jun '09 01:14
Originally posted by wolfgang59
If one green bottle should accidently fall there would be 9 green bottles sitting on the wall.
etc.
etc.

The probabilty of a bottle falling in 1 minute is 50%

How long before it is likely (>50ðŸ˜µ that there will bee no bottles left?
The probability should change and get lower and lower as bottles fall.
11. forkedknight
Defend the Universe
29 Jun '09 06:01
Originally posted by AThousandYoung
The probability should change and get lower and lower as bottles fall.
I'm not sure what you mean.
12. forkedknight
Defend the Universe
29 Jun '09 06:09
I don't know if anyone has use Windows PowerShell ever, but I just looked up how to write a quick script for this, and it's pretty slick.

Anyway, I was pretty sure my math was correct here, but I wasn't 100%, so I wrote a script to test it out.

In a million trials, after 20 minutes:
58.6% of trials resulted in 0 bottles
16.2% in 1 bottle
12% in 2 bottles
etc, etc.

After 19 minutes
50.0% resulted in 0 bottles
17.6% in 1 bottles
14.5% in 2 bottles
etc, etc.

If you want to try it for yourself, PM me and I can send it to you.
13. Palynka
Upward Spiral
29 Jun '09 08:251 edit
Originally posted by forkedknight
I'm not sure what you mean.
I think what he means is that he sees the problem as there being an underlying exponential for each bottle such that the probability of one of the ten falling is exactly one minute.

Let t_i be the time it takes for bottle i to fall and assume equal for all i. Then the prob of no bottle falling within one minute is 1 - P(t>1)^10 = .5 and we could back out the mean of the distribution (which fully characterizes an exponential). A similar reasoning applies if he means exactly 1 falling within one minute, just modify the expression accordingly.

I'll try to get back to this when I have more time, if nobody proposed an answer.
14. Palynka
Upward Spiral
29 Jun '09 10:011 edit
Ok, so solving this. We have that

The probability that a bottles falls within x minutes is:
P(t<x) = (1-exp(-lambda*x)), where lambda is the inverse of the mean. We then have that 1-P(t<1)^10 = 0.5 (no bottles fall within one minute*), which leads to:

exp(-lambda) = 0.5^(1/10) <=>
lambda = - log(0.5^(1/10)) = (approx) 0.0693147.

The mean is then 1/lambda approx 14.42695 minutes.

The probability that all bottles have fallen after T minutes is then:
P(t<T)^10. Using our lambda, we then have to find T such that this probability equals 0.5.

(1-exp(-lambda*T))^10 = 0.5 <=>
exp(-lambda*T) = 1-0.5^0.1 <=>
T = -ln(1-0.5^0.1)/lambda = (approx) 39 minutes.

*again I assume wolfgang meant the probability that "at least" one bottle falls, not "exactly one" bottle falls within one minute.
15. forkedknight
Defend the Universe
29 Jun '09 13:531 edit
Originally posted by Palynka

*again I assume wolfgang meant the probability that "at least" one bottle falls, not "exactly one" bottle falls within one minute.
I thought wolfgang just meant that the bottles had to fall in sequence, with the first bottle in the row having a .5 probability of falling each minute.