[Guest]

Hi,

I apologize if this one already got posted, but i haven't found it through the search function. I found

this one on the Hoeflin Power Test.

Suppose you are truthfully told that ten marbles were inserted into a box, all of them identical except that their colors were determined by the toss of an unbiased coin. When heads came up, a white marble was inserted, and when tails came up, a black one. You reach into the box, draw out a marble, inspect its color, then return it to the box. You shake the box to mix the marbles randomly, and then reach in and again select a marble at random. If you inspect ten marbles in succession in this manner and all turn out to be white, what is the probability to the nearest whole percent that all ten marbles in the box are white?

I'm not sure whether I can use the following method:

We need to find the posterior probability that when we take 10 whites, there are only whites in the box.

In order to do so we need to use Bayes' theorem: P(A/B) = P(B/A)*P(A) / P(B)

=>

- P(A) = odds that there are 10 white marbles in the box

= binominal (k = 10; n = 10, p = 1/2)

- P(B/A) = odds that one picks 10 white marbles out of 10 picks out of the box when there are 10 white marbles in the box

= binominal (k = 10; n = 10, p = 1)

- P(B) = cumulative odds of picking 10 whites out of 10 picks when not known how many whites there are in the box

= sum of i = 0..10 : binominal (k = i; n = 10, p = 1/2) * binominal (k = 10; n = 10, p = i/10)

Edited August 25, 2008 by dfhwze

[unreality]

good puzzle!

pull out one, it's white, prob of all being white:

1/(10^9)

pull out two, both are white.

9/10 of the time, prob is: 1/(10^8)

1/10 of the time, prob is: 1/(10^9)

thus average chances that all are white after pulling out two is 9.1 * 10^-9

pull out three, all three are white.

81/100 of the time, prob is 1/(10^7)

18/100 of the time, prob is 1/(10^8)

1/100 of the time, prob is 1/(10^9)

thus average chances that all are white after pulling out three is 8.281 * 10^-8

pull out four, all four are white.

729/1000 * 1/(10^6) +

... * 1/(10^7) +

... * 1/(10^8) +

1/1000 * 1/(10^9)

the numbers in between have to do with the binomial distribution, it's a pattern, I got to go now but I'll come back to this tomorrow

[Prime]

This could be another application for Bayes’ Distribution, which Chuck Rampart used for a variation on Monty Hall problem.

Anyway, my calculation from the total number of variations is as following:

Black and white marbles have the same probability to get into the box. There are ¹⁰C_n variations to have n white marbles in the box. ( ¹⁰C_n=10!/(n!*(10-n)!) )

Each of individual variation has the same probability.

If there were n white marbles out of 10 in the box – the probability to pull 10 in row as specified in the OP is (n/10)¹⁰.

There is only one variant to have all 10 marbles as white (¹⁰C₁₀). We can estimate the number of variations in which 10 white marbles were pulled in a row by adding up number of variations for each distribution multiplied by its probability to draw 10 white marbles in a row:

P = 1/(¹⁰C₀*(0/10)¹⁰ + ¹⁰C₁*(1/10)¹⁰ + ¹⁰C₂*(2/10)¹⁰ + ¹⁰C₃*(3/10)¹⁰ + ¹⁰C₄*(4/10)¹⁰ + ¹⁰C₅*(5/10)¹⁰ + ¹⁰C₆*(6/10)¹⁰ + ¹⁰C₇*(7/10)¹⁰ + ¹⁰C₉*(9/10)¹⁰ + ¹⁰C₁₀*(10/10)¹⁰)

Which I calculated in spreadsheet as 0.07019. (Just over 7%).

[Guest]

Hi,

I apologize if this one already got posted, but i haven't found it through the search function. I found

this one on the Hoeflin Power Test.

Suppose you are truthfully told that ten marbles were inserted into a box, all of them identical except that their colors were determined by the toss of an unbiased coin. When heads came up, a white marble was inserted, and when tails came up, a black one. You reach into the box, draw out a marble, inspect its color, then return it to the box. You shake the box to mix the marbles randomly, and then reach in and again select a marble at random. If you inspect ten marbles in succession in this manner and all turn out to be white, what is the probability to the nearest whole percent that all ten marbles in the box are white?

I'm not sure whether I can use the following method:

We need to find the posterior probability that when we take 10 whites, there are only whites in the box.

In order to do so we need to use Bayes' theorem: P(A/B) = P(B/A)*P(A) / P(B)

=>

- P(A) = odds that there are 10 white marbles in the box

= binominal (k = 10; n = 10, p = 1/2)

- P(B/A) = odds that one picks 10 white marbles out of 10 picks out of the box when there are 10 white marbles in the box

= binominal (k = 10; n = 10, p = 1)

- P(B) = cumulative odds of picking 10 whites out of 10 picks when not known how many whites there are in the box

= sum of i = 0..10 : binominal (k = i; n = 10, p = 1/2) * binominal (k = 10; n = 10, p = i/10)

Bayes' Theorem is the way to go, but you need to use a more complicated version. Basically, your computation only counts cases where all 10 are white - however, it's possible to pull 10 straight whites with fewer than 10 white marbles originally.

The version of Bayes' Theorem you need to use is

For clarity, when I say "pull" below I mean "pull a marble out of the bag," and by "flip" I mean "flip a coin so that the color is inserted to begin with." For example, P(flip 10)=.5^10.

P(flip 10|pull 10)=P(pull 10|flip 10)*P(flip 10) / (P(pull 10|flip 10)*P(flip 10) + P(pull 10|flip 9)*P(flip 9) + ... + P(pull 10|flip 1)*P(flip 1))

All these terms are trivial to calculate: P(flip x) = 10Cx *.5^10, and P(pull 10|flip x) = (x/10)^10. I put it all into a spreadsheet, and found P=.07019, which is also what Prime got.

I used excel to go through all that, and my answer agreed with Prime's: .07019

In case you were curious, if you pull 10 white marbles consecutively, it is most likely that you originally flipped for 8 white marbles - that probability is .339.