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A three ball problem

BobK

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Apr 8, 2003
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You have three balls in a sack, randomly colored either solid white or solid black, so that the combination of colors is unknown.

Pull out one ball and note it's color and replace in sack.

Repeat two more times.

If all three pulls are the same color ball, what is the probability that the balls in the sack are all one color?
 
BobK said:
You have three balls in a sack, randomly colored either solid white or solid black, so that the combination of colors is unknown.

Pull out one ball and note it's color and replace in sack.

Repeat two more times.

If all three pulls are the same color ball, what is the probability that the balls in the sack are all one color?
Click to expand...
To make it easier for the person with the right answer to correct my error I will separate my answer.

Assumption 1

Choices in the bag

(a) 3 black
(b) 3 White
(c) 1 black 2 white
(d) 2 black 1 white

Assumption 2

Each of the above equally likely ie 0.25 chance

Assumption 3

Chance of 3 the same if

(a) 1
(b) 1
(c) (3 blacks) 1/3 x 1/3 x 1/3 + (3 whites) 2/3 x 2/3 x2/3 = 1/3
(d) 1/3 (as above)

Assumption 4

Chance of all three being the same colour given 3 the same come out


(a) + (b)
----------------------
(a) + (b) + (c) + (d)

= 2 / 2.6 = .75

Edited to expand on answer
 
Both Lothian and Walter Wayne missed.

Lothian check your assumptions.

I'll post the solution later if no one solves it.:)
 
BobK said:
Both Lothian and Walter Wayne missed.

Lothian check your assumptions.

I'll post the solution later if no one solves it.:)
Click to expand...
Ok.

Say assumtion 2 is wrong and the odds are

All black 1/8
2 black 3/8
1 black 3/8
0 black 1/8

Then Overall odds

All black 1/8 x 1 = 1/8
2 black 3/8 x 1/3 = 1/8
1 black 3/8 x 1/3 = 1/8
0 black 1/8 x1 =1/8

Total chance of three the same coming out= 1/2.

Of the above all three the same in the bag is 1/4 therefore the chances of all three being the same colour when each time the same colour is drawn out is a quarter over a half or 0.5
 
I'm going to delurk and take a shot. Here's how I worked this out.

The possibilities are as follows, all with a 1/8 chance:
WWW
WWB
WBW
WBB
BWW
BWB
BBW
BBB

Condensed:
3W 0B = 1/8 chance
2W 1B = 3/8 chance
1W 2B = 3/8 chance
0W 3B = 1/8 chance

Now, for the odds of drawing 3 balls of the same color in each of these scenarios:
3W 0B
- 3 white -> 3/3 x 3/3 x 3/3 = 1/1
- 3 black -> 0/3 x 0/3 x 0/3 = 0/1
- Chance of drawing three of the same color = 1/1 + 0/1 = 1/1

2 W 1 B
- 3 white -> 2/3 x 2/3 x 2/3 = 8/27
- 3 black -> 1/3 x 1/3 x 1/3 = 1/27
- Chance of drawing three of same color = 8/27 + 1/27 = 1/3

1 W 2 B - as for 1W 2B, with W and B reversed

0W 3B - as for 3W 0B, with W and B reversed


So, multiply the odds of each scenario occuring times the odds of drawing 3 balls of the same color in each:
3W 0B = 1/8 chance x 1/1 chance of three of same color = 1/8
2W 1B = 3/8 chance x 1/3 chance of three of same color = 1/8
1W 2B = 3/8 chance x 1/3 chance of three of same color = 1/8
0W 3B = 1/8 chance x 1/1 chance of three of same color = 1/8

Add these all together:
1/8 + 1/8 + 1/8 + 1/8 = 1/2

So, my answer is that there's a 50/50 chance.

(And I now see someone beat me to it!)
 
Say 3 balls are A, B, and C
Possibilities are

A B C
1 B B B
2 B B W
3 B W B
4 B W W
5 W B B
6 W B W
7 W W B
8 W W W

So
All Black 1/8
All White 1/8
2 Black 1 White 3/8
2 White 1 Black 3/8

I don't remember enough of my stats to decide the impact of pulling the balls out and putting them back in. I'll take a guess that it's irrelevant and the answer is 1/4 (1/8 chance they are all black and 1/8 chance they are all white).
 
Aren't you guys forgetting that you have to go back an account for the fact that if you pull out even one black ball, let alone all three times, you must remove the possibility of 3 white balls?
 
Andonyx said:
Aren't you guys forgetting that you have to go back an account for the fact that if you pull out even one black ball, let alone all three times, you must remove the possibility of 3 white balls?
Click to expand...

I think I accounted for that, unless I'm completely confused.

That's why I attacked this problem from the direction of calculating the odds of drawing three of the same color, black or white, for each of the 8 possibilities of what is in the bag (each with a 1/8 chance of being the bag contents).

For instance, if there are 3 black balls in the bag, the odds of drawing 3 white balls in a row are 0. The odds of 3 black balls is 1. The chance of drawing three balls, same color, is the sum of the chance of 3 white plus the chance of 3 black, in this case, is 0 + 1 = 1.

Do this for each of the eight possibilities for what is in the bag, multiply each result by 1/8, add them all together (see my earlier post for the calculaltions, slightly condensed), for a total chance of 1/2.


I'm sure someone will point out the fatal flaw in my calculations soon.
 
25%

EDIT: I think I missed it..I change to 81.25% (13/16) hehe...

Last EDIT: I hurry to much. I agree with those who say 50% :)
(6/8)*(8/27+1/27) = 1/4 and 2/8*1=1/4 did it for me.
Never would have guessed it though, as you see above :P
 
Lothian, I agree with your second post.

StevenP, you forgot a step, although in this case it coincidentally didn't affect the answer. Read the last paragraph of Lothian's post. Or consider a bag with one ball in it, whose initial probability of being white is 1/2. You pull the ball out and it's white. Now, the probability that it's white is 1, of course, but your method would say 1/2.

Gods Advocate, pulling the balls out and replacing them doesn't by itself change the probability, but looking at them and noticing their color does: you've gained new information, and in general that changes probabilities. What if you pulled out a ball and replaced it 100 times, and each time it was white? Wouldn't you be almost certain that all three balls were white? If there were a black ball in there, it's very unlikely you'd have missed it every one of the 100 times.
 
I'd like to buy a vowel, Pat.

I'll change my answer to P=0.6125
 
69dodge said:
StevenP, you forgot a step, although in this case it coincidentally didn't affect the answer. Read the last paragraph of Lothian's post. Or consider a bag with one ball in it, whose initial probability of being white is 1/2. You pull the ball out and it's white. Now, the probability that it's white is 1, of course, but your method would say 1/2.
Click to expand...


Maybe I'm not following my own logic, but I'm confused how it would get the correct answer, then. I worked this out using the logic of my attempt at the three ball problem, but with the one ball, and it came out to a result of a 1 in 1 chnce of getting all of the one draws to match.

The possibilities are as follows, each with a 1/2 chance:
W
B

Condensed:
1W 0B = 1/2 chance
0W 1B = 1/2 chance

Now, for the odds of drawing 1 ball of the same color (obviously) in each of these scenarios:
1W 0B
- 1 white -> 1/1 = 1/1
- 1 black -> 0/1 = 0/1
- Chance of drawing one of the same color = 1/1

0W 1B
- 1 white -> 0/1 = 0/1
- 1 black -> 1/1 = 1/1
- Chance of drawing three of same color = 1/1

So, multiply the odds of each scenario occuring times the odds of drawing 3 balls of the same color in each:
1W 0B = 1/2 chance x 1/1 chance of one of same color = 1/2
0W 1B = 1/2 chance x 1/1 chance of one of same color = 1/2

Add these all together:
1/2 + 1/2 = 1

100% chance of drawing one ball of the same color (obviously, as it's only one ball, but I wanted to test the method)

I think the confusion may be in that I'm addressing this from the perspective of all three draws being the same color, BUT WITHOUT US KNOWING WHICH COLOR (black or white), while others are assuming that we know it's balck or white at that point.

I think each result comes out to 1/2, but the extra consideration of the lack of knowledge of it being black or white, only knowing it's all the same color, is adding some extra steps to my work that may seem out of place for those attacking the problem differently.

Of course, I could be completely off base, and the problem may lie elsewhere.

I'm waiting for BobK to post the solution, where he'll tell us something completely different.
 
69dodge said:

... but looking at them and noticing their color does: you've gained new information, and in general that changes probabilities.
Click to expand...
I can't see how this statement is true.
 
The middle two sentences are a red herring. The probability will not change no matter how many times you do the experiment. WITH the proviso that the first proclamation is true. "so that the combination of colors is unknown.." The author has inclusively declared that the exclusivity of either color is non-allowed.
 
TillEulenspiegel said:
The middle two sentences are a red herring. The probability will not change no matter how many times you do the experiment. WITH the proviso that the first proclamation is true. "so that the combination of colors is unknown.." The author has inclusively declared that the exclusivity of either color in non-allowed.
Click to expand...

Not if you take "the combination of colors is unknown" as you would something like "the number of participants is unknown".

Just as the number of participants can be zero, so can the combination ... there is no combination, therefore all of one color. (Or, as it is unknown, there is a combination of some sort, also unknown).

"Unknown" can include 0.

Of course, i if you're right, congratulations. I always have trouble when these math-types problems turn out to be a tricky English problem in disguise, and usually give myself a good slap in the head when the answer is revealed.
 
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:
The possibilities are as follows, each with a 1/2 chance:
W
B

Condensed:
1W 0B = 1/2 chance
0W 1B = 1/2 chance

Now, for the odds of drawing a white ball in each of these scenarios:
1W 0B
- Chance of drawing a white ball = 1/1

0W 1B
- Chance of drawing a white ball = 0/1

So, multiply the odds of each scenario occuring times the odds of drawing a white ball in each:
1W 0B = 1/2 chance x 1/1 chance of a white ball = 1/2
0W 1B = 1/2 chance x 0/1 chance of a white ball = 0/2

Add these all together:
1/2 + 0/2 = 1/2
Click to expand...
See what I mean?
 
Your "probably" right. My effort was to pare off the non-allowed states to make the logic easier, but the basic premise remains. The probability of outcomes is the same if the initial state stay's the same ( same balls in the sack ) and if the initial state of the sample universe does not change. ( i.e no balls are removed after each sample)
 
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