My Math Forum Rule of succession problem

 January 19th, 2010, 06:32 AM #1 Newbie   Joined: May 2009 Posts: 25 Thanks: 0 Rule of succession problem Taken from an example in the 1870 version of Todhunter algebra. I spotted and answered this problem (correctly?) on Yahoo answers.... You have a bag of 5 equally sized balls the colour of which you have no information about. You draw two balls and both are white. What is the probability that all five of the balls are white ? I answered the problem without resorting to the succession rule.....how would you solve it ?
 January 19th, 2010, 07:11 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Rule of succession problem I have no idea how to solve the problem. I suppose that any solution must infer a reasonable Bayesian prior...? Link: Algebra for the Use of Colleges and Schools: With Numerous Examples, p. 466 ff.
January 19th, 2010, 10:36 AM   #3
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Re: Rule of succession problem

Hello, CarlPierce!

CRGreathouse is correct . . . This requires Bayes' Theorem.

Quote:
 You have a bag of 5 equally sized balls the colour of which you have no information about. You draw two balls and both are white. What is the probability that all five of the balls are white ?

We know there are at least 2 Whites among the 5 balls.

I assume that the assortment colors of the balls are equally likely.

$\text{Let: }\,W\,=\,\text{white ball, }\:A\,=\,\text{another color}$

$\text{Then: }\;P(5W)\:=\:P(4W,\,1A) \:=\:P(3W,\,2A) \:=\:P(2W,\,3A) \:=\:\frac {1}{4}$

$\text{Bayes' Theorem: }\:P(5W\,|\,2W\text{drawn}) \;=\;\frac{P(5W\,\wedge\,2W\text{drawn})}{P(2W\tex t{drawn})}\;\;[1]$

$\text{Numerator:}$
[color=beige]. . [/color]$P(5W\,\wedge\,2W\text{drawn}) \;=\;\left(\frac{1}{4}\right)(1) \;=\;\frac{1}{4}\;\;[2]$

$\text{Denominator: }\;P(2W\text{drawn})$

[color=beige]. . [/color]$P(5W\,\wedge\,2W\text{drawn})\;=\;\left(\frac{1}{4 }\right)(1) \;=\;\frac{1}{4}$

[color=beige]. . [/color]$P(4W,1A\,\wedge\,2W\text{drawn}) \;=\;\left(\frac{1}{4}\right)\left(\frac{{4\choose 2}{1\choose1}}{{5\choose2}}\right) \;=\;\left(\frac{1}{4}\right)\left(\frac{3}{5}\rig ht) \;=\;\frac{3}{20}$

[color=beige]. . [/color]$P(3W,2A\,\wedge\,2W\text{drawn}) \;=\;\left(\frac{1}{4}\right)\left(\frac{{3\choose 2}{2\choose0}}{{5\choose2}}\right) \;=\;\left(\frac{1}{4}\right)\left(\frac{3}{10}\ri ght) \;=\;\frac{3}{40}$

[color=beige]. . [/color]$P(2W,3A\,\wedge\,2W\text{drawn}) \;=\;\left(\frac{1}{4}\right)\left(\frac{{2\choose 2}{3\choose0}}{{5\choose2}}\right) \;=\;\left(\frac{1}{4}\right)\left(\frac{1}{10}\ri ght) \;=\;\frac{1}{40}$

$\text{Hence: }\:P(2W\text{drawn}) \;=\;\frac{1}{4}\,+\,\frac{3}{20}\,+\,\frac{3}{40} \,+\,\frac{1}{40} \;=\;\frac{30}{40} \;=\;\frac{3}{4}\;\;[3]$

$\text{Substitute [2] and [3] into [1]: }\;P(5W\,|\,2W\text{drawn}) \;=\;\frac{\frac{1}{4}}{\frac{3}{4}} \;=\;\frac{1}{3}$

January 19th, 2010, 12:51 PM   #4
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Re: Rule of succession problem

Quote:
 Originally Posted by soroban $\text{Then: }\;P(5W)\:=\:P(4W,\,1A) \:=\:P(3W,\,2A) \:=\:P(2W,\,3A) \:=\:\frac {1}{4}$
This strikes me as a questionable assumption, though. Why not assume, say, that the prior probability that a ball is white is uniformly distributed on [0, 1]? That is, there is a universe with Bp white balls and B(1-p) nonwhite balls for B very large, and then the five balls are chosen with that probability based on your universe. This seems more natural to me, and probably gives a different result.

 January 21st, 2010, 07:37 AM #5 Newbie   Joined: May 2009 Posts: 25 Thanks: 0 Re: Rule of succession problem I solved it as follows Given 2 whites then the actual number of whites in the bag can be either 2,3,4 or 5 For each possible case the chance of drawing two whites is ... 2/20 , 6/20, 12/20 or 1 So the probability that the actual number of whites is 5 is therefore 1 / (2/20 + 6/20 + 12/20 + 1) = 0.5 The succession rule gives the same answer. namely 3/4 x 4/5 x 5/6 = 0.5 Is my reasoning valid ? Todhunters gives the answer 0.5
January 21st, 2010, 08:07 AM   #6
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Re: Rule of succession problem

Quote:
 Originally Posted by CarlPierce Is my reasoning valid ?
Yes, if you assume that the prior odds of there being 2, 3, 4, or 5 white balls are equally likely. Todhunter and soroban seem to think that this is reasonable. I don't.

First, it seems to rest on the assumption that, a priori, a ball has even odds to be white or non-white. I question this.

But given that assumption, I would expect 1/32 prior odds that all 5 are white, not 1/4 or 1/6. This would give
1/(1*1 + 5*3/5 + 10*3/10 + 10*1/10) = 1/8
rather than 1/2.

Laplace's "rule of succession" would suggest that the chance that a random ball is white is (2 + 1) / (2 + 2) = 3/4, making the chance that the three unknown balls are white 3/4 * 3/4 * 3/4. The same analysis could of course be applied here:
# white balls (prior probability)
5 (3/4)^5 * (1/4)^0 * 1 = 243/1024
4 (3/4)^4 * (1/4)^1 * 5 = 405/1024
3 (3/4)^3 * (1/4)^2 * 10 = 270/1024
2 (3/4)^2 * (1/4)^3 * 10 = 90/1024
1 (3/4)^1 * (1/4)^4 * 5 = 15/1024
0 (3/4)^0 * (1/4)^5 * 1 = 1/1024

243/(243*1 + 405*3/5 + 270*3/10 + 90*1/10) = 27/64

The answers are the same either way.

I tend to think that 27/64 is more reasonable than 1/8, but I certainly wouldn't go any higher than 27/64. In particular I'm not willing to go up to the Todhunter 32/64.

The method of Agresti & Coull seems to suggest a yet lower figure: 8/27. The Clopper-Pearson 95% interval is maddeningly wide: 1.118% to 100%.

 January 21st, 2010, 08:27 AM #7 Newbie   Joined: May 2009 Posts: 25 Thanks: 0 Re: Rule of succession problem I'm not sure I need to assume anything about the prior probabilities of a ball being white. I simple state that if 2 white balls have been drawn then the respective changes it happened given the four possible states were 2/20, 6/20, 12/20 or 1 Can't we turn the odds around to infer the chance of each state actually being true. So is it not reasonable to infer that the P(white = 5) is the ratio of these ? i.e 20/(2+6+12+20) = 0.5
January 21st, 2010, 08:36 AM   #8
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Re: Rule of succession problem

Quote:
 Originally Posted by CarlPierce I'm not sure I need to assume anything about the prior probabilities of a ball being white.
You do.

Quote:
 Originally Posted by CarlPierce I simple state that if 2 white balls have been drawn then the respective changes it happened given the four possible states were 2/20, 6/20, 12/20 or 1 Can't we turn the odds around to infer the chance of each state actually being true.
Suppose that you happened to know that someone (say, the Easter bunny; alternatively, Descartes' demon) went around the world leaving bags of five white balls everywhere. In fact the bunny leaves so many bags around that the chance you'd come across another bag is infinitesimal. In that case the probabilities would still be
2/20, 6/20, 12/20 or 1
as you describe, but the only reasonable answer would be 1.

Suppose instead that you knew the bunny left three kinds of bags: one with 5 white, one with 3 white, and one with 1 white; and that the respective probabilities were 1/4, 1/2, and 1/4. Then you would be justified in deriving a probability of 5/8.

Quote:
 Originally Posted by CarlPierce So is it not reasonable to infer that the P(white = 5) is the ratio of these ? i.e 20/(2+6+12+20) = 0.5
Only if the four possibilities were equiprobable.

January 21st, 2010, 09:40 AM   #9
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Re: Rule of succession problem

Quote:
 You have a bag of 5 equally sized balls the colour of which you have no information about.
No information is no information.

Yesterday was a weekday, and today is a weekday. What is the probability the next three days will be weekdays?

January 21st, 2010, 12:12 PM   #10
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Re: Rule of succession problem

Quote:
 Originally Posted by aswoods No information is no information. Yesterday was a weekday, and today is a weekday. What is the probability the next three days will be weekdays?
The so-called "Sunrise Problem".

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