A three ball problem

[BobK]

The probability must be greater than 0.5. Here is why.

Having three balls in a bag of which there are two possible colours represents on these following possible colour combinations:

BBB
WWW
WWB
BBW

In plain english, you can only have either all white, all black, two white and one black or two black and one white. Each of these four combinations have an equal 0.25 probability of occuring.

Therefore 50% of the time, all three balls will be the same colour and 50% of the time they will be a two/one colour combination (this is all before you start drawing them out). Therefore, the minimum probability for the solution is 0.5.

Now, we have to add the probability of drawing three consecutive on one colour when there are two of one colour and one of other. I believe I have done that previously.

Sorry Mike, but...
Possible combinations prior to 1st draw...
WWW WWB WBW WBB BBW BWB BWW BBB

Although it appears that you can call mixed combinations such as WWB and BWW one combination, they are really two different combinations.

The correct answer really is 50%

See my post above.

 

[BobK]

woolery,

See my reply above to Miketemp and my solution back about 5-6 posts.

 

[BobK]

Ladewig,

Being rather new here and believing this to be a math problem, I wasn't sure which of the two forums to put it in. Someone else mentioned it also. Now I know.

I believe my solution should also work with 4 balls each.

 

[69dodge]

If you draw a black ball you immediately negate the possibility of the balls all being white, leaving only three options - 3B, 2B/1W, 1B/2W. Since all we know is that all three drawn balls are then black we cannot differentiate further between these possibilities, so they are equally likely.

We can't conclusively rule out any of the three possibilities, but that doesn't mean they're equally likely. (If I toss a coin ten times, getting ten heads isn't impossible but it certainly is less likely than getting five heads and five tails.)

If 3B is the truth, we're guaranteed to draw three blacks; if 1B/2W is the truth, we're unlikely to draw three blacks. This implies that if we actually do draw three blacks, the relative probabilities of 3B vs. 1B/2W will shift in favor of 3B.

 

[miketemp]

Sorry Mike, but...
Possible combinations prior to 1st draw...
WWW WWB WBW WBB BBW BWB BWW BBB

Although it appears that you can call mixed combinations such as WWB and BWW one combination, they are really two different combinations.

The correct answer really is 50%

See my post above.

Sorry to belabour this, but I am still not clear. Your eight combinations are all the possible ways three balls could be drawn. I am talking about what is in the bag before you starting picking. Thus order of B and W is not important, only content.

Thus, BWW and WWB and WBW are three different orders that can be picked when there are two white balls and one black ball (along with WWW or BBB, of course).

The content of the bag before any picking is done is not related to any order. I only see four different content combinations.

 

[LW]

Second. For those who believe they have the right method, ask yourself, if there were four balls drawn instead of three would my answer produce a greater or smaller probability. If the answer is smaller, then you have a flaw in your set-up.

In case someone is interested, you can get the probability of having three identical balls in the bag after n same-colored balls (n > 0) have been drawn using the formula:

P = 1 / (1 + a)

where

a = (2^n + 1) / 3^(n-1)

This is derived from the formula I posted above by noting that in the case there are two different colors of balls in the bag, the probability of getting n same-colored balls is (2/3)^n + (1/3)^n.

This gives the following values:

P(1) = 0.25
P(2) = 0.375
P(3) = 0.5
P(4) ~= 0.61
P(5) ~= 0.71
P(6) ~= 0.79
P(7) ~= 0.85
P(8) ~= 0.89
P(9) ~= 0.93
P(10) ~= 0.95

 

[RSLancastr]

If there are three balls in a bag, and they can be either black or white, then there are four combinations of these two colors:

BBB
WWW
BBW
WWB

Since we don't care black from white, then there are only two possible mixtures of the two colours:

All the same colour; and,
Two of one colour and one of the other.

There is a 0.5 probability that they are all the same colour before we even pick any out to observe.

Mike, althought the order is not important, the number of possible combinations IS.

Try dropping three pennies onto a table (all at the same time). Do this many times, and see how often all three come up the same (all heads or all tails).

Chances are, it will be much closer to 25% of the time than 50% of the time.

(There are eight combinations of pennies (HHH, HHT, HTH, HTT, THH, THT, TTH, TTT), only two of which (HHH, TTT) are all the same. 2/8 = 25%)

 

[miketemp]

You have three balls in a sack, randomly colored either solid white or solid black, so that the combination of colors is unknown.

So I guess the key here is what are the possible combinations from this statement.

Well I say there are two: a. three of one colour OR b. two of one colour and one of the other colour. If you open the sack and look in, and there are two colours of balls in there, how does the order of the balls matter? BBW, BWB, WBB is the same combination in a different order. Order only matters when you begin drawing balls out.

If you contend that there are eight combinations (as you have in your subsequent post) then you should have stated the following before the first sentence:

"Out of six total balls, those being three white balls and three black balls, three are chosen randomly and placed into a bag."

Then your probability of:

frequency of combinations
www = 1/8
wwb = 3/8
wbb = 3/8
bbb = 1/8

would be correct.

But you did not state that, therefore the problem begins with three unknown balls in a bag of which there are only the two possible combinations of colours.

Thus, my probability of:

frequency of combinations
www = 1/4
wwb = 1/4
wbb = 1/4
bbb = 1/4

is correct.


Maybe somebody could help me out here.....

 

[miketemp]

Try dropping three pennies onto a table (all at the same time). Do this many times, and see how often all three come up the same (all heads or all tails).

This is not analagous to the original problem. The act of dropping the pennies is analagous to stating that we are selecting three balls from a pool of six balls which contain, equally, three white and three black.

The original problem clearly states that there are already three balls in the bag which would equate to the three pennies already having been tossed and covered with a cloth.

The problem does not involve how the balls got into the bag. Who says the white and black balls have an equal chance of getting into the bag? Where does it say they are like pennies and have a 50% chance of being white or black?

 

[RSLancastr]

This is not analagous to the original problem. The act of dropping the pennies is analagous to stating that we are selecting three balls from a pool of six balls which contain, equally, three white and three black.

You're either way ahead of me, or way behind - I'm not sure which...

It is not analogous to the entire problem, but it does illustrate the fact that there are more combinations than it seems you are accounting for.

The original problem clearly states that there are already three balls in the bag which would equate to the three pennies already having been tossed and covered with a cloth.

Okay, toss them under a cloth and then remove the cloth. It does not effect the chances, from what I can tell.

Perhaps the thread should be renamed Shroedinger's Balls?

The problem does not involve how the balls got into the bag. Who says the white and black balls have an equal chance of getting into the bag? Where does it say they are like pennies and have a 50% chance of being white or black?

I thought it was implicit in the problem, and even if it were not, I'm not sure how it would effect the calculations.

Out of my depth, it seems. Not the first time!

 

[LW]

Re: Re: A three ball problem

So I guess the key here is what are the possible combinations from this statement.

OK, let's go through the problem statement once again:

You have three balls in a sack, randomly colored either solid white or solid black, so that the combination of colors is unknown.

The way I read this is that:

- we have three balls
- each ball is either white or black, randomly determined.

So, for the three balls the combinations are:

- first ball is white or black (2)
- second ball is white or black (2)
- third ball is white or black (2)

Combinations: 2 x 2 x 2 = 8.

I guess we are to assume uniform random distribution, though I haven't checked whether non-uniform distributions lead to same or different result.

 

[Larspeart]

Am I the only person who noticed that a thread that starts out with '3 balls in a sack' is just plain funny and wrong?

 

[BobK]

Hey! I knew a guy in high school with three nipples. So anything's possible.

 

[miketemp]

I'll have to defer to some of my colleagues on this one for now but let me state where I think we are with this:

One train of thought (apparently only mine) is this for the combinations in the bag:

Option A: all the same colour
Option B: not all the same colour

The probability distribution is 50% for each since there are two combinations for each option. The premise is that problem starts after the balls are in the bag (I may have to get in the bag to solve this) and there are only four different black/white combinations available.

The other train of thought (apparently everybody else) is this for the combinations in the bag:

Option A: all the same colour
Option B: not all the same colour

The probability distribution is 25% for for Option A and 75% for Option B since there are two combinations for A and six combinations for B. The premise here is that the problems starts with the distribution of the balls into the bag in the first place (i.e. the selection of the balls) in which there are eight black/white combinations available.

So far so good?

 

[RSLancastr]

Am I the only person who noticed that a thread that starts out with '3 balls in a sack' is just plain funny and wrong?

Yes, when I first saw the topic, I thought that perhaps it had to do with James Bond.

 

[Walter Wayne]

...

I had everything laid out in columns in my text editor, but it looked like crap when I loaded it. Maybe someone can tell me
how to make columns look correct when replying.

If you want fixed width fonts like a basic editor, try the code tag.

.1 .2 .3 .4
11 12 13 14

yields:

.1 .2 .3 .4
11 12 13 14

Walt

 

[Jarom]

Ok, formalization is good. Formalization is your friend.

For i = 0, 1, 2, and 3, let A_i be the event "There are i white balls in the bag (and 3 - i black balls)."

Let B be the event "All three balls drawn are white."

We wish to calculate P(A_3|B). We can do this if we calculate P(A_3), P(B|A_3), and P(B), then apply Bayes' formula.

So, P(A_3) = 1/8; that part is easy.
P(B|A_3) = 1; that part is easy too.

To calculate P(B), we'll use the fact that the events A_0, A_1, A_2, and A_3 are mutually exclusive, so we can calculate

P(B) = P(A_0)*P(B|A_0) + P(A_1)*P(B|A_1) + P(A_2)*P(B|A_2) + P(A_3)*P(B|A_3)
= (1/8 * 0) + (3/8 * 1/27) + (3/8 * 8/27) + (1/8 * 1)
= 0 + 1/72 + 1/9 + 1/8
= 1/4

Now we apply Bayes' formula, so

P(A_3|B) = [P(A_3) * P(B|A_3)] / P(B)
= [1/8 * 1] / (1/4)
= 1/2

So the answer is 1/2, as several people have correctly posted with less explanation. A couple common errors seem to be:

1) The fallacy of equal probabilities: Assuming that any set of mutually exclusive events must be assigned the same probability whenever we don't know which event occurred. Unless there's a good reason for giving them the same probability, don't!

2) Claiming (for example) that the probability that all three balls examined are different is 2/9. This would be correct if we didn't know the color of those balls, but because we know that all of those balls are white, this provides evidence in favor of these balls not being different.

-Jarom

(Yes, first post. I've been lurking for a couple of years. Nice to meet you, and all.)

 

[angard]

With 1 ball in the bag, I get 100%

2 balls, about 67% (2/3)
3 balls, 50%
4 balls, about 38%(32/85)

too boring to do more, but my guess is the chances drop. I wonder if there is a bottom limit?

EDIT: A guess with just looking at previous number of balls, would be about 29% with 5 balls.

 

[LW]

1) The fallacy of equal probabilities: Assuming that any set of mutually exclusive events must be assigned the same probability whenever we don't know which event occurred. Unless there's a good reason for giving them the same probability, don't!

Welcome to delurk.

Anyway, in case you didn't notice, you too assigned an uniform distribution for mutually exclusive events: you suppose that each ball has a 50% change of being white.

 

[Jarom]

Welcome to delurk.

Anyway, in case you didn't notice, you too assigned an uniform distribution for mutually exclusive events: you suppose that each ball has a 50% change of being white.

'Tis a fair cop. I suppose a correct answer would be "There is not enough information given to solve this problem. If we further assume that the prior probability of each ball being white is 1/2, then the answer is 1/2."

In English, on the other hand, the phrase "randomly colored either solid white or solid black" is often (but not always, hence my error) used to imply a uniform distribution.

Chalk another one up for ambiguous English, I guess.

- Jarom