probability challenge

[knock knock]

Ten marbles were inserted into a bag based on the tosses of an unbiased coin. Heads yielded white; Tails yielded black. Someone (unaware of the results of the coin flips) selects a marble from the bag (without looking into the bag) and records that it is white. After replacement (and mixing the marbles up) this is done nine more times, each recorded marble being white. What is the probability to the nearest percent that all ten marbles in the bag are white?

[BruceZ]

[ QUOTE ]
Ten marbles were inserted into a bag based on the tosses of an unbiased coin. Heads yielded white; Tails yielded black. Someone (unaware of the results of the coin flips) selects a marble from the bag (without looking into the bag) and records that it is white. After replacement (and mixing the marbles up) this is done nine more times, each recorded marble being white. What is the probability to the nearest percent that all ten marbles in the bag are white?

[/ QUOTE ]

Answer in white.<font color="white">
7%. </font>

[Precision1C]

The method I would use is determine the percentage chance of N white balls in the bag.

N=0 (.5)^10
N=1 [(.5)^1(.5)^9]C(10,1)
N=2 [(.5)^2(.5)^8)]C(10,2)

etc. Then figure out M(N), the chance of getting 10 white draws in a row if there are N balls in the bag.

M(0)=0
M(1)= (.1)^10
M(2)= (.2)^10

etc. Multiply M(N) by the chance of getting N white balls and sum them from N= 0 to 10. Use this number to divide the product of M(10)* the chance of getting 10 white balls this will get you the percentage chance of the bag having 10 white balls.

Too lazy to do the math. : )

[daryn]

bruce, you're the man when it comes to this stuff.. but could you describe the method please?

[Nottom]

Bruce probably just doesn't want to do the guy's homework for him. /images/graemlins/tongue.gif

[mmbt0ne]

It just looks to me like you use the Law of Total Probability on a bunch of conditional probabilities based on the chances of white going in, versus the chances of drawing all whites given what you have in the bag.

[Mano]

There are 2^10=1024 possible outcomes for 10 coin flips. Here is how the number of white balls will be distributed for all possible outcomes:

#white balls (whites) #ways to get that # of balls (ways)

0 ___________________ 1
1 ___________________ 10
2 ___________________ 45
3 ___________________ 120
4 ___________________ 210
5 ___________________ 252
6 ___________________ 210
7 ___________________ 120
8 ___________________ 45
9 ___________________ 10
10 ___________________ 1

Now calculate the probability of pulling 10 straight white balls from each bag containing N white balls, and weight it by the number of ways you can get N balls. For instance, for N=5 use P(10 whites)*252 = ((.5)^10) * 252 = .246 . Add all these together, and you come up with 14.24 (I believe this would be the expected # of times you would pull 10 white balls in a row out if you did the coin flips followed by the 10 ball selections 1024 times). The bag with 10 white balls contributed 1 of the 14.24, so the probability that the bag contained 10 white balls is 1/14.24 = .072 or 7.2% .

[knock knock]

heh. i know the answer. i'm seeing if anyone else does.

[RiverTheNuts]

you guys are good... Im smart with math, and I understand how this stuff works, I just need to take a game theory/probability class to learn all the applicable formulas

[well]

1 / Sum[(n/10)^10 * Binomial[10,n], {n,0,10}] * 100% = 781250000/111304237 %,
or about 7.019%.

[Mano]

so the probability that the bag contained 10 white balls is 1/14.24 = .072 or 7.2% .

Correction: this should be .0702 or 7.02% (made error copying number from calculator).

[phredd]

Drawing the balls from the bag is a red herring. The question is asking what is the chance that all the balls going into the bag are white. Since they were decided by random coin flip, the correct answer is 1/(2^10), or about 1/10th of 1%.

[well]

Suppose I have two cards: T /images/graemlins/spade.gif and 5 /images/graemlins/heart.gif.
I give you one at random.
You look at your card: a ten.

Following your logic, this should result in a probability of 0.5 of me having kept the five.

Nope.

[phredd]

Incorrect. If you have a known pool of distinguishable objects, pulling one will give you vastly more information about the other objects than when you have a random pool of indistinguishable objects. A better analogy would have been if the cards were randomly selected from a pool of T /images/graemlins/spade.gif and 5 /images/graemlins/heart.gif and I drew one of them and saw it was a T /images/graemlins/spade.gif, what are the chances that all of them are /images/graemlins/spade.gifs. the answer, of course, is 50:50. My initial logic would have said 1:4, so clearly drawing does give you some information.

When drawing marbles one at a time from the bag, you can never know for certain if all of them are white - even with an infinite number of draws - but, you can be more certain than the initial random chance when you put them in. Indeed, if you proceed for long enough without drawing a black marble, you will eventually reduce the chances of encountering a black marble to vanishingly small.

Point conceded.

[well]

[ QUOTE ]
A better analogy would have been if the cards were randomly selected from a pool of T /images/graemlins/spade.gif and 5 /images/graemlins/heart.gif and I drew one of them and saw it was a T /images/graemlins/spade.gif, what are the chances that all of them are /images/graemlins/spade.gifs. the answer, of course, is 50:50.

[/ QUOTE ]

How big is the pool?
All of "them"?
50:50?

You lost me.
[ QUOTE ]

When drawing marbles one at a time from the bag, you can never know for certain if all of them are white

[/ QUOTE ]

It was never about knowing for certain.
It was about determining the probability of them all being white based on the information you have.

Initially, the probability is 2^-10.
After one draw, it already changes. Nothing is certain, it's just that you have more information to
base your predictions on.



[ QUOTE ]
Point conceded.

[/ QUOTE ]

Uhhm, let's try your proof of me being "incorrect" again?

Would be fun to deal with!

[phredd]

Your analogy was incorrect - it didn't apply to the situation at hand. I tried to improve your analogy so that it was actually applicable to the problem, but I evidently explained myself so poorly that you couldn't follow along. Oh, well.

The main point is that my initial logic was flawed, yours was better, and I conceded the point. And that, as they say, is that.

[Mano]

A very clear way to see this is wrong is that by the same logic there would be a 1/(2^10) probability that all the marbles in the bag are black, when clearly there is zero probability of that.

[knock knock]

ANSWER:

P(1) = 10/1024 x (1/10)^10
P(2) = 45/1024 x (2/10)^10
P(3) = 120/1024 x (3/10)^10
P(4) = 210/1024 x (4/10)^10
P(5) = 252/1024 x (5/10)^10
P(6) = 210/1024 x (6/10)^10
P(7) = 120/1024 x (7/10)^10
P(8) = 45/1024 x (8/10)^10
P(9) = 10/1024x (9/10)^10
P(10) = 1/1024 x (10/10)^10

SUM/100 = 1.39%

[BruceZ]

[ QUOTE ]
ANSWER:

P(1) = 10/1024 x (1/10)^10
P(2) = 45/1024 x (2/10)^10
P(3) = 120/1024 x (3/10)^10
P(4) = 210/1024 x (4/10)^10
P(5) = 252/1024 x (5/10)^10
P(6) = 210/1024 x (6/10)^10
P(7) = 120/1024 x (7/10)^10
P(8) = 45/1024 x (8/10)^10
P(9) = 10/1024x (9/10)^10
P(10) = 1/1024 x (10/10)^10

SUM/100 = 1.39%

[/ QUOTE ]

I see that you don't understand your own problem. What you have computed is just the total probability of picking 10 white balls. Your problem asked for the probability that all 10 balls are white. I hope you see the difference. We want the conditional probability that all 10 balls are white given that we picked all 10 white balls. You need to divide 1/2^10 by your total probability according to Bayes' theorem.

So the exact expression is:

1/2^10 / sum{n=1 to 10} C(10,n)*(1/2)^10*(n/10)^10

= 1/2^10 / 1.39%

= 7.01905%

knock knock only computed the sum in the denominator.