OT - New probability puzzle

QuikSand · 12-04-2003, 08:30 AM

Okay, since we've been dabbling with probability a while, I'll offer up another one that gets into the same kind of thing. (I love this stuff)

- - - - -

You are shown four opaque containers, each of which contains some marbles, as follows:

-one contains three black marbles
-one contains two black and one white
-one contains one black and two white
-one contains three white marbles

The containers are placed randomly, and you cannot tell which is which. You reach into one random container, and withdraw a black marble, then replace it.

What is the probability that, if you withdraw another marble from the same container, the marble will also be black?

Bee · 12-04-2003, 08:32 AM

mckerney · 12-04-2003, 08:32 AM

50% chance?

QuikSand · 12-04-2003, 08:33 AM

Quote:

Originally posted by Bee
2/3

You know, I thought about rewriting the puzzle to ensure that this was exactly right (I'm sure I could have)... but that would have just made this a meta-puzzling exercise rather than a real probabilistic challenge.

Bee · 12-04-2003, 08:34 AM

how about 6 out of 9 then?

mckerney · 12-04-2003, 08:35 AM

My reasoning:

If you there is one black marble in it, that means it is one of three containers (we can obviously throw the 4th one out).

Therefore, with these containers there would be 2 marbles left. In one two black (100% chance), one white and one black (50% chance) and two white (0% chance)

The math with this then:

((1/3) x 1) + ((1/3) x .5) + ((1/3) x 0) = .5

QuikSand · 12-04-2003, 08:37 AM

Well, I've admitted that I diddn't fix it to be 2/3... but you certianly know me well enough that 1/2 has to be a trap.

Bee · 12-04-2003, 08:38 AM

Quote:

Originally posted by mckerney
My reasoning:

If you there is one black marble in it, that means it is one of three containers (we can obviously throw the 4th one out).

Therefore, with these containers there would be 2 marbles left. In one two black (100% chance), one white and one black (50% chance) and two white (0% chance)

The math with this then:

((1/3) x 1) + ((1/3) x .5) + ((1/3) x 0) = .5

you replace the black marble that you pulled out initially.

mckerney · 12-04-2003, 08:42 AM

Quote:

Originally posted by Bee
you replace the black marble that you pulled out initially.

Missed that for some reason. Maybe I should actually read things, though that's usually too much effort....

Bee · 12-04-2003, 08:44 AM

How about 7 outta 9? That's gotta be it.

Bad-example · 12-04-2003, 08:45 AM

72.3%?

Bad-example · 12-04-2003, 08:56 AM

dola

I love playing with puzzles like these but admittedly lack the skills to properly solve them in most cases.

After looking at it some more I retract my above answer and regret posting in this thread at all.

That is all.

Bee · 12-04-2003, 08:59 AM

I stand by my 7 out of 9 answer and have no regrets posting in this thread at all.

Bad-example · 12-04-2003, 09:06 AM

Quote:

Originally posted by Bee
I stand by my 7 out of 9 answer and have no regrets posting in this thread at all.

Well clearly that is the difference between us...I had the good sense to retract my incorrect answer.

cthomer5000 · 12-04-2003, 09:08 AM

2/3 if you can pick the same marble up again.

EDIT: correcting my part 2 answer

If there is some sort of grammar trick here and your are indeed picking up "another" (not the same) marble, then you have a 62.5% chance of picking a black marble.

Bee · 12-04-2003, 09:09 AM

Quote:

Originally posted by Bad-example
Well clearly that is the difference between us...I had the good sense to retract my incorrect answer.

Now see...you're going to look silly when Quiksand says I'm right.

lynchjm24 · 12-04-2003, 09:15 AM

I think there is an angle as to how you select the container.

If you randomly select the containers until you pull a black marble, you are going to have the 3 black marble container more often - this will lead to a higher percentage of pulling another black marble.

So if you randomly select a container 120 times you have each container 30 times.

30 times you have to get a black marble in the 3 black container.
20 times you get a black marble in the 2 black container.
10 times you get a black marble in the 1 black container.

So now you have 60 times you've pulled black.

If you throw 'Another' out:

You'll select black again
30 times in the 3 black container
13.3 times in the 2 black container
3.3 times in the 1 black container

So you'll hit black again 46.6 out of 60 times.
or 77.67% of the time.

..... or not......, which I now realize is 7 out of 9.

Bee · 12-04-2003, 09:16 AM

Quote:

Originally posted by lynchjm24
which I now realize is 7 out of 9.

lynchjm24 · 12-04-2003, 09:18 AM

Dola. If we stick to 'Another'.

We've got 30 of 30 in the remaining 2 black marble container
10 of 20 in the remaining 1 black 1 white container
0 of 10 in the remaining 2 white container.

So then you've got 40 of 60..... or............ 2/3.

Bee · 12-04-2003, 09:20 AM

Quote:

Originally posted by lynchjm24
Dola. If we stick to 'Another'.

We've got 30 of 30 in the remaining 2 black marble container
10 of 20 in the remaining 1 black 1 white container
0 of 10 in the remaining 2 white container.

So then you've got 40 of 60..... or............ 2/3.

That's the way I read it at first, but then Quiksand implied it wasn't 2/3rds so I did what you did above and came up with 7 out of 9.

lynchjm24 · 12-04-2003, 09:21 AM

Quote:

Originally posted by Bee
That's the way I read it at first, but then Quiksand implied it wasn't 2/3rds so I did what you did above and came up with 7 out of 9.

I think replace makes it 7/9. How could you know which marble you had already pulled once.

cthomer5000 · 12-04-2003, 09:22 AM

hmmm 77.7% is my new answer on pulling a black marble.

balaka4 · 12-04-2003, 09:24 AM

The probability should be 25%, I believe. You have a 50% chance of randomly picking a black marble on your first try, i.e. 6/12. If you put it back, then you have a 50% chance of randomly picking a black marble again, i.e. 6/12 again. .5*.5 = .25. I don't believe the fact that there are four containers is a factor in this problem.

Bee · 12-04-2003, 09:25 AM

Quote:

Originally posted by lynchjm24
I think replace makes it 7/9. How could you know which marble you had already pulled once.

Maybe your hands are dirty?

Maple Leafs · 12-04-2003, 09:25 AM

Hold on here... should we assume all the marbles are logical actors?

lynchjm24 · 12-04-2003, 09:26 AM

Quote:

Originally posted by balaka4
The probability should be 25%, I believe. You have a 50% chance of randomly picking a black marble on your first try, i.e. 6/12. If you put it back, then you have a 50% chance of randomly picking a black marble again, i.e. 6/12 again. .5*.5 = .25. I don't believe the fact that there are four containers is a factor in this problem.

That isn't the question. The question is if you already have pulled a black marble - what is your chances of picking another.

.25 would be the answer to what are the chances of randomly picking 2 black marbles in a row, if you replace the one you pull.

Bad-example · 12-04-2003, 09:28 AM

Quote:

Originally posted by Bee
Now see...you're going to look silly when Quiksand says I'm right.

I am quite used to it by now.

balaka4 · 12-04-2003, 09:28 AM

Didn't even pick up on the fact that you need to pick from the same container... now i'm stumped.

Samdari · 12-04-2003, 10:16 AM

The answer is 14/18

If you know that you drew a black ball on the first attempt, that means there is a 1/2 chance you drew from the first bin, a 1/3 chance you drew from the second, a 1/6 chance you drew from the third and 0 chance you drew from the 4th.

For the first bin, there is a 1/2 chance you drew from it, and a 1/1 chance you would draw another black ball.

For the second bin, there is a 1/3 chance you drew from it, and a 2/3 chance you would draw another black ball.

For the second bin, there is a 1/6 chance you drew from it, and a 1/3 chance you would draw another black ball.

Thus the probability you will draw another black ball is:

(1/2)*(1) + (1/3)*(2/3) + (1/6)*(1/3) = 1/2 + 2/9 + 1/18 = 9/18 + 4/18 + 1/18 = 14/18

Maple Leafs · 12-04-2003, 10:21 AM

Quote:

Originally posted by Samdari
The answer is 14/18

Damn, and here I was sure it was 7/9...

Samdari · 12-04-2003, 10:22 AM

Don't confuse 14/18 with 7/9. They are way different. I am clearly right and Bee is clearly wrong.

Bee · 12-04-2003, 10:25 AM

Quote:

Originally posted by Samdari
Don't confuse 14/18 with 7/9. They are way different. I am clearly right and Bee is clearly wrong.

I refuse to get into a battle of wits with an unarmed man.

QuikSand · 12-04-2003, 11:34 AM

Quote:

Originally posted by Bee
Now see...you're going to look silly when Quiksand says I'm right.

How many people read that post and chuckled?

QuikSand · 12-04-2003, 11:37 AM

By the way... you guys are ruthless on semantics.

I had no idea that "withdraw another marble" was an ambiguous phrase.

I intended that phrase to mean: "again withdraw a marble"

I did not intend: "withdraw a marble other than the one you initially withdrew"

Sorry for any confusion. (wow)

albionmoonlight · 12-04-2003, 12:01 PM

So is 7/9 right?

Samdari · 12-04-2003, 12:15 PM

Nope, its 14/18

cthomer5000 · 12-04-2003, 12:19 PM

Quote:

Originally posted by Samdari
Nope, its 14/18

wrong again, it's 77.77%

QuikSand · 12-04-2003, 12:27 PM

Cant everybody win? Why do we have to be so competitive on our society, and feel like we have to haev winners and losers. Can't we just say that everybody did a good job, and that really you're all winners?

Okay, maybe not.

It's 21/27.

QuikSand · 12-04-2003, 01:29 PM

Okay for those probability-inclined... here's a derivative puzzle for you.

Ho can I take basically this same puzzle structure, and rearrange it so the correct answer is 2/3. Ideally, I'd like to have the puzzle retain its simle symmetry in the number and design of the containers and their contents... but I don't mind varying the numbers a bit to get to 2/3.

Anyone got one that comes out to 2/3?

Samdari · 12-04-2003, 01:44 PM

Quote:

Originally posted by QuikSand
Cant everybody win? Why do we have to be so competitive on our society, and feel like we have to haev winners and losers. Can't we just say that everybody did a good job, and that really you're all winners?

Okay, maybe not.

It's 21/27.

Please post the solution or a link to the problem. I really think the 7/9 DOH! I mean 14/18, is correct.

Samdari · 12-04-2003, 01:48 PM

Quote:

Originally posted by Samdari
Please post the solution or a link to the problem. I really think the 7/9 DOH! I mean 14/18, is correct.

WOW, am I a dumbass.

AnalBumCover · 12-04-2003, 02:15 PM

i've lost my marbles.

KWhit · 12-04-2003, 02:26 PM

Quote:

Originally posted by QuikSand
By the way... you guys are ruthless on semantics.

I had no idea that "withdraw another marble" was an ambiguous phrase.

I intended that phrase to mean: "again withdraw a marble"

I did not intend: "withdraw a marble other than the one you initially withdrew"

Sorry for any confusion. (wow)

Well what do you expect after dealing with that puzzle talking about "a king or two!"

QuikSand · 12-04-2003, 02:30 PM

Quote:

Originally posted by Samdari
WOW, am I a dumbass.

I was trying to come up with a catholic way of saying so...

BishopMVP · 12-04-2003, 06:38 PM

Quote:

Originally posted by QuikSand
Anyone got one that comes out to 2/3?

Quote:

You are shown four opaque containers, each of which contains some marbles, as follows:

-one contains three black marbles
-one contains two black and one white
-one contains one black and two white
-one contains three white marbles

The containers are placed randomly, and you cannot tell which is which. You reach into one random container, and withdraw a black marble, then replace it.

What is the probability that, if you withdraw another marble from the same container, the marble will also be black?

Change replace to set aside, or just add don't.

12-04-2003, 08:45 AM	#11
Bad-example College Starter Join Date: Oct 2000 Location: san jose CA	72.3%? __________________ Karaoke Katie drove the crowd wild Every time she'd sing they'd come in for miles Curtain came up, Katie came on Drinking like a lumberjack and singing Delta Dawn

12-04-2003, 08:56 AM	#12
Bad-example College Starter Join Date: Oct 2000 Location: san jose CA	dola I love playing with puzzles like these but admittedly lack the skills to properly solve them in most cases. After looking at it some more I retract my above answer and regret posting in this thread at all. That is all. __________________ Karaoke Katie drove the crowd wild Every time she'd sing they'd come in for miles Curtain came up, Katie came on Drinking like a lumberjack and singing Delta Dawn

12-04-2003, 09:15 AM	#17
lynchjm24 College Benchwarmer Join Date: Jan 2003 Location: Hartford	I think there is an angle as to how you select the container. If you randomly select the containers until you pull a black marble, you are going to have the 3 black marble container more often - this will lead to a higher percentage of pulling another black marble. So if you randomly select a container 120 times you have each container 30 times. 30 times you have to get a black marble in the 3 black container. 20 times you get a black marble in the 2 black container. 10 times you get a black marble in the 1 black container. So now you have 60 times you've pulled black. If you throw 'Another' out: You'll select black again 30 times in the 3 black container 13.3 times in the 2 black container 3.3 times in the 1 black container So you'll hit black again 46.6 out of 60 times. or 77.67% of the time. ..... or not......, which I now realize is 7 out of 9. Last edited by lynchjm24 : 12-04-2003 at 09:15 AM.

12-04-2003, 10:16 AM	#29
Samdari Roster Filler Join Date: Jan 2002 Location: Cicero	The answer is 14/18 If you know that you drew a black ball on the first attempt, that means there is a 1/2 chance you drew from the first bin, a 1/3 chance you drew from the second, a 1/6 chance you drew from the third and 0 chance you drew from the 4th. For the first bin, there is a 1/2 chance you drew from it, and a 1/1 chance you would draw another black ball. For the second bin, there is a 1/3 chance you drew from it, and a 2/3 chance you would draw another black ball. For the second bin, there is a 1/6 chance you drew from it, and a 1/3 chance you would draw another black ball. Thus the probability you will draw another black ball is: (1/2)(1) + (1/3)(2/3) + (1/6)*(1/3) = 1/2 + 2/9 + 1/18 = 9/18 + 4/18 + 1/18 = 14/18 __________________ http://www.nateandellie.net Now featuring twice the babies for the same low price!

12-04-2003, 10:22 AM	#31
Samdari Roster Filler Join Date: Jan 2002 Location: Cicero	Don't confuse 14/18 with 7/9. They are way different. I am clearly right and Bee is clearly wrong. __________________ http://www.nateandellie.net Now featuring twice the babies for the same low price!

12-04-2003, 08:32 AM	#2
Bee Pro Starter Join Date: Oct 2000 Location: Fairfax, VA	2/3

12-04-2003, 08:32 AM	#3
mckerney Coordinator Join Date: Oct 2000	50% chance?

12-04-2003, 08:34 AM	#5
Bee Pro Starter Join Date: Oct 2000 Location: Fairfax, VA	how about 6 out of 9 then?

12-04-2003, 08:35 AM	#6
mckerney Coordinator Join Date: Oct 2000	My reasoning: If you there is one black marble in it, that means it is one of three containers (we can obviously throw the 4th one out). Therefore, with these containers there would be 2 marbles left. In one two black (100% chance), one white and one black (50% chance) and two white (0% chance) The math with this then: ((1/3) x 1) + ((1/3) x .5) + ((1/3) x 0) = .5

12-04-2003, 08:37 AM	#7
QuikSand lolzcat Join Date: Oct 2000 Location: Annapolis, Md	Well, I've admitted that I diddn't fix it to be 2/3... but you certianly know me well enough that 1/2 has to be a trap.

12-04-2003, 08:44 AM	#10
Bee Pro Starter Join Date: Oct 2000 Location: Fairfax, VA	How about 7 outta 9? That's gotta be it.

12-04-2003, 08:59 AM	#13
Bee Pro Starter Join Date: Oct 2000 Location: Fairfax, VA	I stand by my 7 out of 9 answer and have no regrets posting in this thread at all.

12-04-2003, 09:18 AM	#19
lynchjm24 College Benchwarmer Join Date: Jan 2003 Location: Hartford	Dola. If we stick to 'Another'. We've got 30 of 30 in the remaining 2 black marble container 10 of 20 in the remaining 1 black 1 white container 0 of 10 in the remaining 2 white container. So then you've got 40 of 60..... or............ 2/3.

12-04-2003, 09:24 AM	#23
balaka4 n00b Join Date: Nov 2002	The probability should be 25%, I believe. You have a 50% chance of randomly picking a black marble on your first try, i.e. 6/12. If you put it back, then you have a 50% chance of randomly picking a black marble again, i.e. 6/12 again. .5*.5 = .25. I don't believe the fact that there are four containers is a factor in this problem.

12-04-2003, 09:25 AM	#25
Maple Leafs Coordinator Join Date: Jan 2002	Hold on here... should we assume all the marbles are logical actors?

12-04-2003, 09:28 AM	#28
balaka4 n00b Join Date: Nov 2002	Didn't even pick up on the fact that you need to pick from the same container... now i'm stumped.

12-04-2003, 11:37 AM	#34
QuikSand lolzcat Join Date: Oct 2000 Location: Annapolis, Md	By the way... you guys are ruthless on semantics. I had no idea that "withdraw another marble" was an ambiguous phrase. I intended that phrase to mean: "again withdraw a marble" I did not intend: "withdraw a marble other than the one you initially withdrew" Sorry for any confusion. (wow)

12-04-2003, 12:01 PM	#35
albionmoonlight Head Coach Join Date: Oct 2000 Location: North Carolina	So is 7/9 right?

12-04-2003, 12:15 PM	#36
Samdari Roster Filler Join Date: Jan 2002 Location: Cicero	Nope, its 14/18 __________________ http://www.nateandellie.net Now featuring twice the babies for the same low price!

12-04-2003, 12:27 PM	#38
QuikSand lolzcat Join Date: Oct 2000 Location: Annapolis, Md	Cant everybody win? Why do we have to be so competitive on our society, and feel like we have to haev winners and losers. Can't we just say that everybody did a good job, and that really you're all winners? Okay, maybe not. It's 21/27.

12-04-2003, 01:29 PM	#39
QuikSand lolzcat Join Date: Oct 2000 Location: Annapolis, Md	Okay for those probability-inclined... here's a derivative puzzle for you. Ho can I take basically this same puzzle structure, and rearrange it so the correct answer is 2/3. Ideally, I'd like to have the puzzle retain its simle symmetry in the number and design of the containers and their contents... but I don't mind varying the numbers a bit to get to 2/3. Anyone got one that comes out to 2/3? Last edited by QuikSand : 12-04-2003 at 01:29 PM.

12-04-2003, 02:15 PM	#42
AnalBumCover College Starter Join Date: Nov 2002 Location: La Mirada, CA	i've lost my marbles. __________________ ABC's Game Giveaway list

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