Probability Problem -- pulling colored balls from a box randomly

[kent davidge]

Homework Statement

Bob had a box containing 9 green balls and 1 white ball. A Bob's friend removed one ball from the box without telling Bob what was its color. Now what's the probability for Bob picking up one green ball from the box?

Homework Equations

[/B]
Not aware of "formulas", I just use those I think are useful for the problem in question.

The Attempt at a Solution

If Bob's friend removed the white ball, then the chance for Bob picking up a green ball is 9/9 = 100%.
If Bob's friend removed a green ball, then there's 8/9 chance.

But I would like to get one total probability, not two conditions like these.

 

[.Scott]

Since it is not specified, we have to assume that Bob's friend picked a ball at random.

What if Bob's friend removed 9 balls without telling Bob the result?
What would Bob's chances be then?

 

[kent davidge]

0 if the 9 balls were green
100% if 8 green and 1 white

Sorry, I still don't see a pattern

 

[.Scott]

0 if the 9 balls were green
100% if 8 green and 1 white

Sorry, I still don't see a pattern

Okay. Then let's go back to the original problem.
Assuming the friend picked at random, what is the chance that he picked a white ball and what is the chance the he picked the green one.
We'll call those pFW and pFG.
As you said originally:
The chance of Bob pick green given that Friend picked White pBGFW= 1/9
The chance of Bob pick green given that Friend picked Green pBGFG= 0
So, pBG will be a weighted sum of pBGFW and pBGFG - with the weights based on the friend's chances (pFW and pFG).

 

[kent davidge]

I think I got what you said. The problem is that if we sum we get more than 100%. For instance, if Friend picked the white ball, Bob has 100% chance of picking up a green ball. If Friend picked a green ball, Bob has a 8/9 chance of picking up a green ball. If we add these two we get 1 + 8/9. How do you normalize it to 1?

 

[Ray Vickson]

I think I got what you said. The problem is that if we sum we get more than 100%. For instance, if Friend picked the white ball, Bob has 100% chance of picking up a green ball. If Friend picked a green ball, Bob has a 8/9 chance of picking up a green ball. If we add these two we get 1 + 8/9. How do you normalize it to 1?

If you don't want an instant solution using Bayes' Theorem, you can get to the solution using a more intuitive counting method.

Imagine doing the experiment 100 times, with random choices each time. The experiments fall into two groups:
Group I (90 experiments): player 1 picks green.
Group II (10 experiments): player 1 picks white.

In the 90 group-I experiments, player 2 faces an urn having 8 green and 1 white ball, so in how many of these 90 cases will he pick green?

In the 10 group-II experiments, player 2 faces an urn having all 9 balls the color green, so in how many cases will he pick green?

How many times in 100 would player 2 pick green?

 

[kent davidge]

In the 90 group-I experiments, player 2 faces an urn having 8 green and 1 white ball, so in how many of these 90 cases will he pick green?

$90 ( 8/9 ) = 80$ times?

In the 10 group-II experiments, player 2 faces an urn having all 9 balls the color green, so in how many cases will he pick green?

$10 (9/9) = 10$ times?

How many times in 100 would player 2 pick green?

$80 + 10 = 90$ times, which in $100$ times gives $90 \%$?

But why did you choose doing exactly 90 experiments of the group 1 and exactly 10 experiments of the group 2?

 

[Ray Vickson]

$90 ( 8/9 ) = 80$ times?

$10 (9/9) = 10$ times?

$80 + 10 = 90$ times, which in $100$ times gives $90 \%$?

But why did you choose doing exactly 90 experiments of the group 1 and exactly 10 experiments of the group 2?

That is a heuristic argument; the actual number of experiments in group I would be binomial with parameters $n = 100$ and $p = 9/10$, for a mean value of 90. Of course, that means that in any actual group of 100 experiments we will not always get 90 in group I, but averaged out over many different sets of 100 we will get 90. The number in group II will be exactly equal to 100 minus the number in group I. And, of course, the same can be said about how many greens are chosen by player 2 in the group I experiments.

You would get a more accurate accounting if you did 100 billion experiments, and said that group I had 90 billion members, and player 2 picks a green ball 80 billion times withing group I. Or, if that is not accurate enough for you, do 100 trillion experiments, etc. When the numbers become large enough the (relative) errors go to zero, so the situation ceases to be "probabilistic".

The point is that the argument is supposed to be intuitive, not rigorous. For beginners who may have trouble relating to Bayes' rule, that type of counting argument can be very helpful in clarifying the basic concepts. To be sure, it can all be done exactly in about 2 lines using Bayes' rule; see, eg.
http://www.eecs.qmul.ac.uk/~norman/BBNs/Bayes_rule.htm

 

[kent davidge]

Oh, I finally figuered it.

Probability for player 1 picking up a ball = $P_1$
... for player 2 picking up a green ball = $P_2$
... for player 1 picking up a ball given that player two has already picked a green ball = $1/9 + 8/9$.
Let's assume player 1 goes to the urn and picks a ball at random. He can pick one white ball or one green ball. So $P_1 = 1/10 + 9/10$.
Player 2 picking up a ball without interference of player 1 is $P_2 = 9/10$. Using Bayers' rule: $$P(P_2 | P_1) = \frac{(\frac{1}{9} + \frac{8}{9}) \frac{9}{10}}{\frac{1}{10} + \frac{9}{10}} = \frac{9}{10} = 90 \%$$

Also, from the explanation in the page you linked to, it seems quite obvious why Bayers' rule has the form it has.

 

[Ray Vickson]

Oh, I finally figuered it.

Probability for player 1 picking up a ball = $P_1$
... for player 2 picking up a green ball = $P_2$
... for player 1 picking up a ball given that player two has already picked a green ball = $1/9 + 8/9$.
Let's assume player 1 goes to the urn and picks a ball at random. He can pick one white ball or one green ball. So $P_1 = 1/10 + 9/10$.
Player 2 picking up a ball without interference of player 1 is $P_2 = 9/10$. Using Bayers' rule: $$P(P_2 | P_1) = \frac{(\frac{1}{9} + \frac{8}{9}) \frac{9}{10}}{\frac{1}{10} + \frac{9}{10}} = \frac{9}{10} = 90 \%$$

Also, from the explanation in the page you linked to, it seems quite obvious why Bayers' rule has the form it has.

You've got it. It would be clearer to use notation like G1 = {player 1 picks green} and G2 = {player 2 picks green} (with W1 and W2 similarly defined). Then we have P(G1) = 9/10, P(W1) = 1/10, P(G2|G1) = 8/9, P(G2|W1) = 1, so Bayes' rule says
$$P(G2) = P(G2|G1) P(G1) + P(G2|W1) P(W1) = (9/10)(8/9) + (1/10)(1) = 9/10.$$
Notice that the probability is 0.9 for both the first and second balls (provided that we don't know the identity of the first ball). Would it surprise you to find out that the probability of any last-picked ball is 0.9? If player 1 takes 4 balls (without showing them to player 2), the probability that player 2 picks green is still exactly 0.9. Even if player 1 takes 9 balls (types unknown to player 2) and so player 2 faces just one single ball, the chance that it is green is still 0.9!

I did not want to bring this up at first, when you were still struggling with the issues, but we can talk some more about it if you want.

 

[kent davidge]

Oh I see

Would it surprise you to find out that the probability of any last-picked ball is 0.9?

This is very interesting. Is it because we know for sure that player 1 picked a certain amount of balls? I mean, let's take the most interesting case, that player 1 removes 9 balls. Following your early notation, any refers to any type of balls removed by player 1,

$P(G2 | \text{any}1) = P( \text{any}1 | G2) P(G2) / P( \text{any} 1)$ in which both $P( \text{any}1 | G2)$ and $P( \text{any} 1)$ would be 1 because what matters is that player 1 removes 9 balls and we know that this has happened.

 

[Ray Vickson]

Oh I see

This is very interesting. Is it because we know for sure that player 1 picked a certain amount of balls? I mean, let's take the most interesting case, that player 1 removes 9 balls. Following your early notation, any refers to any type of balls removed by player 1,

$P(G2 | \text{any}1) = P( \text{any}1 | G2) P(G2) / P( \text{any} 1)$ in which both $P( \text{any}1 | G2)$ and $P( \text{any} 1)$ would be 1 because what matters is that player 1 removes 9 balls and we know that this has happened.

For the more general problem (more than 2 balls drawn) the easiest way is to avoid using conditional probabilities and Bayes' rule, but instead to go back to sample-space fundamentals.

Say player 1 draws (k-1) balls (without showing the results to player 2) and player 2 draws ball k. You can view this as a permutation problem: number the balls from 1-10; balls 1-9 are green and ball 10 is white. Now any drawing of balls corresponds to a permutation of the numbers from 1-10. Basically, we lay out the entire permutation (of which there are 10! equally-likely candidates) and just ignore the remaining entries. So, when drawing just 1 ball we lay out all 10! complete permutations and just ignore the outcomes 2--10, (That is, we include all the drawings we could have made, but didn't.) The number of permutations with a green ball in position #1 is 9*9!, because any of the 9 green balls can be first and then the other 9 balls can have any permutation. Thus, the probability that the first ball is green is P = 9*9!/10! = 9/10.

In the original version of your problem we are really asking for the probability that the ball in position 2 is green. There are 9*9! such permutations, because there are 9 green balls that can be in position 2 and then the other 9 balls (in positions 1,3,4,...,10) can have any permutation. So, the probability continues to be P = 9(9!/10! = 9/10.

If player 1 draws 9 balls, the probability that player 2 draws green is again P = 9*9!/10!, because there are 9*9! permutations in which a green ball is in position 10.

It works the same way for any position k.

 

[haruspex]

Bob's friend removed one ball from the box

We are left to guess the basis on which the friend chose the ball. If it was eyes closed then, yes, the answer to the question is 0.9. But put yourself in the friend's position. Eyes open, would you pick any ball, equally likely, or any colour, equally likely?
Sack the question setter.

 

[PeroK]

@kent davidge I would look at it like this.

There are 10 balls in the bag. Bob is equally likely to get any ball. The probability it is white is, therefore, 10%.

It doesn't matter how many balls his friend removes at random before he chooses.

 

[FactChecker]

@kent davidge I would look at it like this.

There are 10 balls in the bag. Bob is equally likely to get any ball. The probability it is white is, therefore, 10%.

It doesn't matter how many balls his friend removes at random before he chooses.

True, but treacherous. One must be careful about relying on that intuition. It is the root of the argument over the "Monty Hall problem". It's often safer to beat these problems to death with a more tedious calculation of the different possibilities.

 

[Ray Vickson]

True, but treacherous. One must be careful about relying on that intuition. It is the root of the argument over the "Monty Hall problem". It's often safer to beat these problems to death with a more tedious calculation of the different possibilities.

Right, and the argument in post #12 is perfectly safe. It would have been faster and easier to look at the probability of player 2 drawing white on draw k. The number of permutations with W in position k is 9!, so the probability of drawing white is 9!/10! = 1/10.

 

[kent davidge]

We are left to guess the basis on which the friend chose the ball. If it was eyes closed then, yes, the answer to the question is 0.9. But put yourself in the friend's position. Eyes open, would you pick any ball, equally likely, or any colour, equally likely?
Sack the question setter.

Friend's is supposed to see what ball he is picking up. It's just that he doesn't tell Bob what ball it was. But as @Ray Vickson pointed out even that Friend's doesn't see what ball he is picking up Bob still has the same chances. And that, I guess, is because what matters is whether Friend's tells Bob what ball it was or not, i.e., this information given to Bob is what actually changes the probabilites for Bob. This seems to be the same thing as it happens in the Monty Hall problem mentioned by @FactChecker later on.

Does this make sense?

 

[FactChecker]

Friend's is supposed to see what ball he is picking up. It's just that he doesn't tell Bob what ball it was. But as @Ray Vickson pointed out even that Friend's doesn't see what ball he is picking up Bob still has the same chances. And that, I guess, is because what matters is whether Friend's tells Bob what ball it was or not, i.e., this information given to Bob is what actually changes the probabilites for Bob. This seems to be the same thing as it happens in the Monty Hall problem mentioned by @FactChecker later on.

Does this make sense?

It's not the same as the Monty Hall problem. CORRECTION. THE ITALICIZED STATEMENT IS MISLEADING. The key feature of the Monty Hall problem is that Monte will never open a door with the prize. That has a filtering effect which changes the probabilities. Unless Bob's friend does something intentional like that we have no new information upon which to adjust the probabilities. Whether his friend tells Bob or not does not matter. Of course a particular situation like "friend draws the white ball" would change things. In order to be clear on this, it helps to calculate probabilities after the two situations: "friend draws the white ball" and "friend draws the green ball" with the appropriate conditional probabilities and ask yourself step by step whether the friend or Bob knowing the situation would change anything. It would not. The final result would be the sum of the two disjoint situations -- Probability that Bob draws green = 9/10.

CORRECTION : I need to correct my answer. It doesn't really matter if the friend intentionally did anything. What he drew is what he drew regardless of any intention. If the friend tells Bob what he drew, that definitely changes the probabilities for Bob, given what the friend drew. So our calculated probability would change given that information. But the probability without specifying what the friend drew is not changed from the original.
In the Monty Hall problem, the result of one door changes the probabilities of the other doors. If he was not intentionally avoiding the prize, the probabilities of both other doors would remain equal. Because of his intentional filtering it changes one door probability more than your door (which is unchanged). In this problem, the result of the friend's draw (if known) changes the result of Bob's draw in a known way if the friens's draw is known. If the friend's draw is not known, there is no new information and the probabilities are unchanged.

 

[PeroK]

True, but treacherous. One must be careful about relying on that intuition. It is the root of the argument over the "Monty Hall problem". It's often safer to beat these problems to death with a more tedious calculation of the different possibilities.

It's not intuition. Each player has picked a single ball at random. Each is equally likely to have the white ball.

More generally, if they each get five picks, then they each have an equal chance of getting the white, regardless of the order they get picked. Etc.

To calculate that with your method you'd need 5 pages of conditional probabilities:

If Bob gets the first, second, fourth, sixth and ninth picks, then that is one laborious calculation. And you'd need a different calculation for every combination of picks for each player.

Monty Hall is a different problem, because Monty's pick is not random. He knows where the car is.

 

[FactChecker]

Monty Hall is a different problem, because Monty's pick is not random. He knows where the car is.

I realize that, but the fact remains that it is the same intuition on the part of (often) very well educated people which leads to the wrong answers in the Monty Hall problem. It's only when the conditional probabilities are specifically considered that the true answer is obvious to all. That's why I call it treacherous.

 

[Ray Vickson]

I realize that, but the fact remains that it is the same intuition on the part of (often) very well educated people which leads to the wrong answers in the Monty Hall problem. It's only when the conditional probabilities are specifically considered that the true answer is obvious to all. That's why I call it treacherous.

Treacherous, yes, but that does not mean that conditional probabilities are required; the sample-space arguments in posts #12 and #16 are perfectly safe, and do not use conditional probabilities. Also: depending on the problem's scenario, a multiple-drawing situation as described in post #19 can sometimes use straightforward conditional probability arguments without needing a lengthy analysis---not always, but sometimes.

 

[PeroK]

I realize that, but the fact remains that it is the same intuition on the part of (often) very well educated people which leads to the wrong answers in the Monty Hall problem. It's only when the conditional probabilities are specifically considered that the true answer is obvious to all. That's why I call it treacherous.

Why this obsession with Monty Hall?

My argument rests on the fact that the probability that the nth ball drawn from the bag is white is 10%.

You have applied that with $n =1$, which seems to me just as "treacherous".

 

[PeroK]

I realize that, but the fact remains that it is the same intuition on the part of (often) very well educated people which leads to the wrong answers in the Monty Hall problem. It's only when the conditional probabilities are specifically considered that the true answer is obvious to all. That's why I call it treacherous.

Okay, here's a question.

You and three friends are each dealt a card face down. What is the probability you have been dealt the Ace of Spades?

a) If you were dealt the first card?
b) If you were dealt the second card?
c) Ditto the third card?
d) Ditto the fourth card?
e) if you are not told the order the cards were dealt?

Recognising that you have the same odds, 1/52, in all cases, without indulging in complicated conditional arguments, is an important insight into probabilities.

 

[FactChecker]

The Monty Hall problem is a good example of a case where people erroneously assumption that intermediate information does not change the initial probabilities. That intuitive assumption is correct in this case, but wrong in the Monty Hall case. And that is why I mentioned it.

The number of intermediate draws is not the issue that I am concerned about -- it's the issue of whether they change the probabilities that should not be intuitively assumed without some thought. Bayes' rule addresses that issue directly.

 

[PeroK]

It's logic, not intuition. I can think of several simple, logically watertight reasons why picking a ball or being dealt a card doesn't depend on its place in the choosing sequence.

Ultimately, your Bayesian approach is equally dependent on the fundamentals of logic, even if you don't want to admit it. All mathematics is dependent on logic, in fact.

One of the things I like about probability problems is that there are often different ways to solve the problem, often by recasting it in a new light.

It seems sad to me that you feel the only way to solve such problems is by your Bayesian number crunching.

I'd also add that in the Monty Hall problem, you are equally at risk of simply crunching the wrong numbers. Solving the Monty Fall problem, in other words. Bayes does not make it impossible to misinterpret the problem.

We're both entitled to solve problems our own way, so why you are so adamant that yours is the only correct approach baffles me!

 

[FactChecker]

Sorry. I must have over-stated an earlier post. When I said that what you did was a "treacherous" approach, I didn't mean to imply that it was wrong and that the more tedious approach must be followed. Some people get both this problem and the Monty Hall problem correct immediately and intuitively. Apparently you are one of those. But a significant percent of people are misled by their intuition on the Monty Hall problem and Bayes' rule seems to work better for them.