- 27 Jun '09 16:54

I thought that seemed a little too easy*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.

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. - 27 Jun '09 16:57

Technically, it's after 19 minutes, since the distribution starts at minute 0*Originally posted by forkedknight***I thought that seemed a little too easy**

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. - 28 Jun '09 06:34

Could you elaborate?*Originally posted by forkedknight***Technically, it's after 19 minutes, since the distribution starts at minute 0**

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? - 28 Jun '09 18:32 / 2 edits

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 =*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?**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

*edit* you can use this link to help you expand the polynomials.

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 - 29 Jun '09 01:14

The probability should change and get lower and lower as bottles fall.*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 (>50 that there will bee no bottles left? - 29 Jun '09 06:09I 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. - 29 Jun '09 08:25 / 1 edit

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.*Originally posted by forkedknight***I'm not sure what you mean.**

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. - 29 Jun '09 10:01 / 1 editOk, 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. - 29 Jun '09 13:53 / 1 edit

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.*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.