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A three ball problem

Well, I am standing by my answer.

First we have to figure out how many different balls we have seen.
Pulled same ball all three times: 1 in 9
Pulled two balls in three tries: 6 in 9
Pulled a different ball each time: 2 in 9

Next we figure the odds that the unseen balls are identical to the one(s) we have seen.
Seen one ball: 1 in 4 chance the other two are the same colour
Seen two balls: 1 in 2
Seen three balls: 1 in 1

Calculating a weighted average
1/9*1/4= 1/36
6/9*1/2=12/36
2/9*1/1= 8/36
Sum = 21/36 or 7/12
 
miketemp said:

I can't see how this statement is true.
Click to expand...
Suppose I flip a coin fairly, no cheating or anything. Neither of us looks at it. I make the following offer: If it's heads, I'll give you a dollar; if it's tails, you give me 50 cents. Would you accept?

Now we look at the coin. It shows tails. It's the same coin. Nothing changed, except that now you have some information about it that you didn't have before. Since I'm such a nice guy, I ask you if you want to change your mind about accepting my offer. Would you change your mind?

I prefer the Bayesian approach to probability, in which probability measures degree of belief in a statement, rather than being an intrinsic property of the statement. How strongly I believe in something naturally depends on what information I have about it. Before I looked at the coin, I didn't know whether it was heads or tails. I could express this uncertainty by saying that the probability that it's heads is 1/2. After I look at it, I know it's tails. The probability that it's heads has changed to 0.

If you think that probabilities can't change in this way, how would describe why you'd accept the offer before looking at the coin but not after?
 
69dodge said:
StevenP, I still think you're making a mistake, though I'm not sure why you get the correct answer in this case. How would you solve the exact problem in my previous post, where it's given that the ball you drew was white?

If I copy your previous post, making the appropriate changes, I get this:
See what I mean?
Click to expand...

That's easy ... you've added additional information to the equation that goes beyond the presented problem.

I'm calculating for ball selection for when we know all the draws (be it one or three) of the available balls (1 or 3) are of the same color ... but we aren't given which color (and don't necessarily KNOW it).

In the one ball case, the initial states possible, with a 1/2 chance each are, which holds even if we know a ball was drawn:
W
B

However, if we KNOW the drawn ball was white, the possible states drop to
W
with a probability of 1 in 1.

So, the probability would not be divided in half in this case, where the drawn ball color is known.

Similarly, for the three ball case, each of these is 1/8 probable before we have more information:
WWW
WWB
WBW
WBB
BWW
BWB
BBW
BBB

Once we draw a ball, AND KNOW THAT IT IS WHITE, the possibilities drop to these, each of which is 1/4 probably, as we only are concerned as to what the other two balls in the bag are:
W(WW)
W(WB)
W(BW)
W(BB)

We KNOW at least one ball is WHITE, and just need to calculate for the other two draws. If draws two and three are also white,:
W(WW)
- White occurs all the time (3/3). So two draws of white occurs
(3/3) x (3/3) = 9/9. Multiple by 1/4 for the intiial possibility of this being the distribution in the bag, for 9/36.
W(WB)
- White occurs all the time (2/3). So two draws of white occurs
(2/3) x (2/3) = 4/9. Multiple by 1/4 for the intiial possibility of this being the distribution in the bag, for 4/36.
W(BW)
-Also 4/36, as above.
W(BB)
- White occurs all the time (3/3). So two draws of white occurs
(1/3) x (1/3) = 1/9. Multiple by 1/4 for the intiial possibility of this being the distribution in the bag, for 1/36.

9/36 + 4/36 + 4/36 + 1/36 = 18/36 = 1/2.

However, unlike my initial calculations in my first post, which determined that there is a 50% chance that the bag is all three OF ANY ONE COLOR, if all three draws are of the same color, this calculation shows a 50% chance of all three balls being WHITE if there are three individual draws of WHITE.


I think TillEulenspiegel gets to the heart of the matter when he mentions paring off non-allowed states. I stubbornly cling to them still being allowed, but in reality, it would have to be one or the other color, so I was probably just making things too complicated for myself.
 
Walter Wayne said:
Next we figure the odds that the unseen balls are identical to the one(s) we have seen.
Seen one ball: 1 in 4 chance the other two are the same colour
Seen two balls: 1 in 2
Seen three balls: 1 in 1
Click to expand...
A guy has two children. One of them is a boy. What's the probability that the other child is a boy too? It's 1/3, not 1/2.

I think you're assuming it's 1/2.

"The unseen balls" doesn't refer to specific balls, about which we can say that each has independently a probability of 1/2 of being white, in the same way that "the other child" doesn't refer to a specific child, about which we can say that it has a probability of 1/2 of being a boy.
 
69dodge said:
A guy has two children. One of them is a boy. What's the probability that the other child is a boy too? It's 1/3, not 1/2.

I think you're assuming it's 1/2.

"The unseen balls" doesn't refer to specific balls, about which we can say that each has independently a probability of 1/2 of being white, in the same way that "the other child" doesn't refer to a specific child, about which we can say that it has a probability of 1/2 of being a boy.
Click to expand...
This is a different problem. Consider the two child problem.'

Problem 1: You are are told at least one of them is a boy, what are the odds both are boys. Answer 1/3.

Problem 2: You run into one of his children and he is a boy. What is the chance the other one is also a boy.

Seen Kid/ Unseen Kid
Girl/Girl <- eliminated
Girl/Boy <- eliminated
Boy/Girl
Boy/Boy

Here by saying the one seen is male we eliminate two possibilities instead of one. so we have a 1 in 2 chance of the other being a boy.

Problem 3: You run into one child, what is the chance that the other is the same sex as the first.

Seen Kid/ Unseen Kid
Girl/Girl -> true
Girl/Boy -> false
Boy/Girl -> false
Boy/Boy -> true

Again, 1 in 2.

Walt
 
Originally Posted By Walter Wayne
Well, I am standing by my answer.

First we have to figure out how many different balls we have seen.
Pulled same ball all three times: 1 in 9
Pulled two balls in three tries: 6 in 9
Pulled a different ball each time: 2 in 9

Next we figure the odds that the unseen balls are identical to the one(s) we have seen.
Seen one ball: 1 in 4 chance the other two are the same colour
Seen two balls: 1 in 2
Seen three balls: 1 in 1

Calculating a weighted average
1/9*1/4= 1/36
6/9*1/2=12/36
2/9*1/1= 8/36
Sum = 21/36 or 7/12
Click to expand...

I'll stand by it too. I originally got 17/32, but I incorrectly figured the probabilities of seeing one, two, or three distinct balls. After correcting for that, my answer agrees with Walter's.
 
My answer, knowing next to nothing about statistics, and having only read the first answer (7/12) is:

There are eight combinations possible to begin with:

BBB
BBW
BWB
BWW
WBB
WBW
WWB
WWW

So before you look at any of the balls, the possibility of them all being one color is 2 out of 8 (BBB and WWW).

Once you pull a ball out and see its color (say, Black), you reduced the possibilities to one (BBB) out of seven (all but WWW).

If you put the ball back, draw a ball again, and it is black, I don't see how that changes the odds (again, I don't understand much about statistics), nor do I see how repeating the process would either, so I'll stick with One Out of Seven.

Now to read on and see just how wrong I am...
 
Maybe it is time to explain my answer.

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. When this is the case, there is a 1.0 probability that you will pick out three consecutive balls of the same colour.

Therefore, we can start with a 0.5 probability (1.0 x 0.5).

The probability that there are two of one colour and one of the other colour is also 0.5 probability.

Scenario A - Two whites and one black

When the this is the scenario, and we pick out one ball at a time, there are eight possible combinations that can appear for three picks:

Probability of picking a W is 2/3.
Probability of picking a B is 1/3.

WWB P=4/27
WBW P=4/27
WBB P=2/27
WWW P=8/27
BBW P=2/27
BWB P=2/27
BWW P=4/27
BBB P=1/27

Therefore, the probability of picking three consecutive colours is 1/27 + 8/27=1/3. The same can applied to Scenario B which is two blacks and one white.

Summarizing using the original four possible combinations of balls in the bag (each have a 1/4 probability of being the combination in question:

BBB 0.25 x 1.0
WWW 0.25 x 1.0
BBW 0.25 x .333
WWB 0.25 x .333

Total probability of picking out three consective balls of the same colour is:

0.25 + 0.25 + 0.083333 + 0.083333 = 0.66666666666..............

This is my third different answer so would the real wizard please appear and stop the bleeding.
 
First we have to figure out how many different balls we have seen.
Pulled same ball all three times: 1 in 9
Pulled two balls in three tries: 6 in 9
Pulled a different ball each time: 2 in 9

Next we figure the odds that the unseen balls are identical to the one(s) we have seen.
Seen one ball: 1 in 4 chance the other two are the same colour
Seen two balls: 1 in 2
Seen three balls: 1 in 1

Calculating a weighted average
1/9*1/4= 1/36
6/9*1/2=12/36
2/9*1/1= 8/36
Sum = 21/36 or 7/12
Click to expand...

I agree with the first set of numbers (1in9, 6in9, 2in9) but disagree about how you carry them forward. 6 in 9 (or 18 in 27 for those considering all possibilities) is the probablity that you have drawn two different balls, but that number does not represent the probability that you drew two balls of the same color.


Therefore one has to break down the 6/9 into the four possibilities: BB, BW, WB, WW. note Each of these has a 1 in 6 chance of occurring. At this point we can only consider BB and WW. Because there is a 1/2 chance that the undrawn ball in the BB case is black, we multiply 1/6 x 1/2 to get 1/12 and the WW case with the undrawn ball being white is also 1/12.

The summation then becomes
1/9 * 1/4 = 1/36
1/6 * 1/2 = 3/36
1/6 * 1/2 = 3/36
2/9 * 1/1 = 8/36
Sum...........15/36 or 5/12

Hmmm, I was shooting for 1/2. Maybe I left out something.

Note The BB and WW have at most a 1/6 chance of occurring - it probably should be something less.
 
I think you're right about the children question being different.

However . . . I have a backup plan. :)
Walter Wayne said:
First we have to figure out how many different balls we have seen.
Pulled same ball all three times: 1 in 9
Pulled two balls in three tries: 6 in 9
Pulled a different ball each time: 2 in 9
Click to expand...
These are the correct probabilities assuming no knowledge about the colors of the balls drawn. But we know that they were all the same color, which was more likely to happen if they were all actually the same ball than if not. So the probabilities above need to be adjusted.

I think this is what Ladewig is saying too. I also couldn't get an answer of 1/2 using your approach, even after trying to adjust the probabilities. I will work on it some more.
 
Doh! Most of the answers have been calculating the probability of drawing three balls of the same color given the starting conditions. That probability is 1/2 as has been shown. The question is what is the probability of the three balls in the bag being the same color - so I'm going with 1/8 +1/8 = 1/4. (Which is what God's Advocate said at the beginning.)
 
ok math time.
The initial state is I, S is my shoe size, A is my numerical address, G is the my grandma's age and P is the probability that this is correct., so...

I*A/S/
--------- =P
G
hence the initial probability = 1/8 or 12.5 % ( just trust me on this 1 , you have already shown the initial outcomes in your posts)
after 1 sample you have reduced the possible outcomes so..
after the sample the probability = 1/4 or ..lemme me see...uh carry the 2 .. is umm 25 % but as I have never seen 1 fourth of a ball , I'm not sure that works.
The poison tincture of this problem is that all repeat the same mistake of considering samples that are identical twice
I.E BBW=WBB look at your logic tables. So the logical reduction of outcomes is reduced ( after the sample) by 50%, therefore the outcomes are ruduced and the percentage of probability increases inversly proportionatly.

"is it further to New York or by train?"
 
OK, There are eight combinations of balls.
For a combination of 3 same coloured balls, there is a 100% chance of drawing three balls the same colour.
For combinations of 2 of one colour and 1 of the other, the chance of drawing three they same colour is (2/3)<sup>3</sup>+(1/3)<sup>3</sup>=8/27+1/27=1/3

Of the combination of balls, 2 in 8 (1/4) will be all the same colour and 6 in 8 (3/4) will have two the same colour.

Weighted by probabilities
1/4*1+3/4*1/3 = 1/4+1/4=1/2

Edit to add: Up to this point it is identical to Lothians second answer. However, this is only the probability of pulling three identical balls from the sack

So this 0.5 is the probability that one will pull 3 identical balls.
0.25 is the probability that the all balls are the same colour.

So given that we sampled three identical balls the probability of 3 identical balls = 0.25/0.5 = 0.5

Edit to add: This extra step does not change the answer. However, that is just a coincidence, and one would get a different answer using the two different answers if one only pulled twice from the three ball sample.

Walt
 
I dunno. You go home for the night thinking all is solved then you come back and find it isn’t, then it is again. Can’t wait for BobK to post a different answer :D
It appears that most people agree that given the balls are randomly placed in the bag then the chances of getting three the same out are 0.5.

However (miketemp please note). That was not the question.

The question was “you have picked three out, Now what are the odds that all three were the same.”

Again there appears agreement between Walter Wayne’s method and mine that the chances of there being three the same in the bag AND them being picked out is 0.25.

It follows that the answer is

The chances of there the same in the bag and three the same picked out
--------------------------------------------------------------------------------------
The chances of there being three the same picked out


This is 0.25 / 0.5 = 0.5
 
Ahhh, I love the art of randominity... I'm going to place my odds somewhere between 1:1 and A gillion:1
 
Sorry I'm so late with this.
Here's the formula...

8 possible combinations

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

if the color is white...

calculate 3-draw probablity of www for each combination

www (3/3)^3 = 1
wwb (2/3)^3 = 8/27
wbb (1/3)^3 = 1/27
bbb (0/3)^3 = 0

multiply 3-draw probability by frequency of combination

www 1 x 1/8 = 1/8 = 9/72
wwb 8/27 x 3/8 = 24/216 = 8/72
wbb 1/27 x 3/8 = 3/216 = 1/72
bbb 0

9/72 + 8/72 + 1/72 = 18/72 = 1/4

inverse of possible combinations = 1/8

solution (1/8)/(1/4) = 1/2 = 50% probability

Note: I used ^ as notation for (raised to the power of)

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.:confused:
 
I'll pop in to add support for the 1/2 answer.

Basic events:

A - all balls are identical
B - three same-colored balls are drawn.

We want to find the probability P(A | B). "A if B is given"

By Bayes' theorem:

P(A | B) x P(B) = P(A and B) = P(B | A) x P(A), so
P(A | B) = P(A and B) / P(B)

P(A and B) is quite clearly 2 / 8 since P(A) = 2/8 and P(B | A) = 1. The probability P(B) = 1/2 as several people have already calculated: P(B) = 1/8 x (2 + 6 x 1/3) = 1/2.

So, P(A | B) = 2/8 / 1/2 = 1/2.
 
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.
 
I think this problem is far simpler than most peoples post suggest.

We know that all three draws are the same colour, but that isn't the question. The question is given that all three draws are the same colour what is the probability that all three balls in the bag are the same colour.

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. This simply leaves us with the probability that they are all black being 1/3.

Edit - the same is of course true if the first ball is white, but that doesn't alter the answer!
 
First off, why is this thread in Science instead of Puzzles?

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.
 
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