Ye've Lost Yer Marbles!

[Lepton]

Puzzle
by the way, thanks to The Dogpack for the title of this one.

[Mgm]

729+243=972 marbles in total.
If you draw a black first, you have to discard all the black marbles in subsequent turns.
Also, if you draw a white first, you have to discard all the white marbles in subsequent turns.
So it all comes down to the chance of picking the first marble.
You have a 75% chance of picking a black one, discarding all the other blacks in subseqent turns leaving a white marble.

So the chance the last marble is black is 25%.

Or is there something I'm missing?

------------------
"Any mystery devised by mortal minds can be solved therewith."
“When you have excluded the impossible, whatever remains, however improbable, must be the truth.”
-- Sherlock Holmes

[Foggy]

If you draw a black marble first, you discard black marbles until you draw a white...not until the bag has one marble left.

I'm no good at the probabilty math necessary to figure this out, but i'm guessing the huge number of black marbles would result in the black and white marbles evening out, until there was roughly 1 and 1 left. So I'd say 50%.

------------------
I wasn't lying, I was writing fiction with my mouth. -- Homer Simpson

[ZutAlors!]

I'm thinking I agree with Foggy. In order for a black marble to be the only one left, you've got to discard all the white marbles first. Which means that, the turn before, you must have begun with picking a white marble, and picked all the white marbles available that turn.

What are the chances of doing this?

Suppose you have n total marbles, x of which are white. The chances of picking all the whites in a row are S[i=0 to x-1](x-i)/(n-i) which equals x!(n-x)!/n! Likewise, the chances of picking all black marbles in a row are S[i=0 to n-x-1](n-x-i)/(n-i) which equals (n-x)!x!/n! Same thing. Any other circumstance (and thus the rest of the probability) results in a new round, with a new number of marbles.

So the chances of picking all the marbles of the same color in one round are the same for both colors, no matter how many marbles are in the bag. And thus, the chances of having the oppposite color be the last one left (on the next turn) are the same. And thus the chances of having a black marble as the last one is precisely 50%. No matter what the starting count (presuming at least one of each color, obviously).

[This message has been edited by ZutAlors! (edited 05-17-2004 07:37 AM).]

[Dr. Borodog]

Yeah, I got 50% by inspection, although I devoted no effort to trying to prove it mathematically or logically.

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You will respect my philosophai.

[Faust]

There are 972 marbles. There are 2 different colors, black and white. They are in a 3:1 ratio. There is a 75% chance that the first marble drawn is black. After the first marble is drawn and discarded, the ratio, and the probability of the next marble being black is reduced slightly. If you were to conduct this experiment, you would slowly level out the ratios of marbles, since you are more likely to draw subsequent black marbles then white ones. The rules governing discard and re-shuffling act to balance the portions of marbles. I Hypothesize that by the time only a few marbles are left, the ratios would be very close, leaving you with a 50/50 chance on the color of the final margin.

[Faust]

correction...marble, not margin.

[Jack Daniels]

The answer is 50%, and the ratio between black and white marbles or the number of marbles is irrelevant, as long as there is at least one of each color (this seems wrong, but try it with any number of marbles of any color and you'll reach the same conclusion). This reminds me of another similar puzzle with a surprising answer: If a certain country decided that each family could have as many children as they could until they had a boy and then they had to stop, and the chances for having a boy are the same as having a girl, without twins, what would the ratio between boys and girls be in a few generations (what would the ratio converge to)? This means that if a family has a boy the first time around, they stop. If they have a girl, they try again. If this time they have a boy, they stop, and so on... I think China had a law like this once. The answer is of course, 50% girls and 50% boys. The reason for this is the same as in the marble puzzle.

[debashis]

The answer is .75. Because the balls are being treated in same manner, so it becomes a symmetric problem. let's the probablity of the last ball being black is p , so probablity for white ball is (1-p). now because the ratio of black and white is 3 , p/(1-p)=3.. solve it u will get p=.75

[Pablo]

Another way to approach this is to assume a much simpler case, such as 2 blacks and 1 white. If you explore the small number of possibilities, you find quickly that there are 6, 3 of which lead to white and 2 of which lead to black. If it works for a small sample, I would think it would be the same for a larger sample.

[This message has been edited by Pablo (edited 05-12-2004 07:52 PM).]

[Pablo]

Originally posted by Faust:

---

I Hypothesize that by the time only a few marbles are left, the ratios would be very close, leaving you with a 50/50 chance on the color of the final margin.

---

Bad hypothesis. By the time you get down to 3, the closest ratio you could have would be 2:1.

[This message has been edited by Pablo (edited 05-11-2004 05:33 PM).]

[PolarBoy]

Ok let's run with that last comment. You have 2 black marbles and 1 white. On your first pick you have a 2/3 chance drawing black. Drawing black leaves a 1/2 chance of either ball. Drawing white guarantees black being the last ball. That gives us 3 outcomes. In one of which white is last. In the other 2 it's black. Hmm...

4 balls: 3 black, 1 white. so we start with 3/4 of drawing black. If we draw white, it's over, and black is the last ball. If we draw black, we have a 2/3 chance of drawing black. drawing black a second time gives us a 1/2 chance of either color. Drawing black, then white gives us the same situation as above. With 1/3 white and 2/3 black. 2/3*1/3+1/4+1/2*2/3*3/4=6/864=.7222222222222... odds of ending on black. Close to our original ratio. I think the answer may actually be very close to 3/4. Dunno.

[Quailman]

[PolarBoy]

I was assuming the marbles had just been shuffled.

[marcus210]

Actually the marbles are not being treated the same. Whichever color you draw most often is discarded most often. This leads quickly to a 50-50 split. I worked it out in detail with only four marbles, 3 black and 1 white. With this few the odds are already 54% black and 46% white. The key is that the "different" marble is put back in.

[Smerk]

50/50
With 2 black and 1 white the math is easy.
1/3 of the time white is first so black is last.
TOTAL SO FAR: 1/3 BLACK

2/3 of the time black is first so let's see about marble #2:
1/2 of the time black is picked leaving white last so 1/2 of 2/3 = 1/3 white.

TOTAL SO FAR: 1/3 BLACK, 1/3 WHITE
Back to marble #2

The other 1/2 of the time white would be chosen and go back in the bag leaving 1 white and 1 black in the bag and a fresh start. Each marble now has an even 50/50 to get chosen. If Black is next, then white is last and vice-versa. Therefore, half of the other half of the 2/3 would go to white (1/6) and the other half of the half of the 2/3 would go to black (1/6).

TOTAL: 1/3 BLACK, 1/3 WHITE, 1/6 BLACK, 1/6 WHITE = 1/2 BLACK, 1/2 WHITE

[Pablo]

So if we can show that starting with 2 blacks and 1 white yields a 50-50 situation, can we extrapolate that to say, 20 and 10? 21 and 10? x and y?

[Foggy]

I tried writing a program in PERL (below), as my knowledge of probability is pretty limited (see above).

I believe that the program is written correctly.

code:

---

#!perl -w





print "Testing\n";





my $result = 0;





for ($j=0; $j<500; $j++) {


		$result += pick_em();


}





print $result;





sub pick_em() {


	my @marbles = ();


	my $i = 0;


	my $total_marbles = 972;


	my $black_marbles = $total_marbles * 0.75;


	my $white_marbles = $total_marbles - $black_marbles;





	my $black = -1; my $white = 1;





	for ($i=0; $i < $black_marbles; $i++) {


			push (@marbles, $black);


	}





	for ($i=$black_marbles; $i < $total_marbles; $i++) {


			push (@marbles, $white);


	}





	my $current = 0;





	while ($#marbles > 0) {


		$index = rand @marbles;


		if ($current+$marbles[$index] != 0) {


				$current = $marbles[$index];


				splice (@marbles, $index, 1);


		} else {


				$current = 0;


		}


}





	return $marbles[0];


}

---

Running this several times shows a slight edge for black.


------------------
I wasn't lying, I was writing fiction with my mouth. -- Homer Simpson

[This message has been edited by Foggy (edited 05-13-2004 08:50 AM).]

[Kutti]

I did the same detail calculations as Smerk but with 3 black and 1 white. And the result is just the same:

1/2 BLACK, 1/2 WHITE

Details: There are 8 paths from the beginning (3 black, 1 white) to the end (the last marble) with the following probabilities:
(drawings, last marble, probability) = (w, b, 1/4), (bww, b, 1/12), (bwbww, b, 1/24), (bwbwb, w, 1/24), (bwbb, w, 1/12), (bbww, b, 1/8), (bbwb, w, 1/8), (bbb, w, 1/4)

[So Long]

My analysis of 3 black, 1 white yielded a different answer. I came up with a total of 8 combinations and 66 permutations:

Finishing with a White we have (UC when marble discarded, lc when sorted back in):

BBB (6 ways 3.2.1)
BBwB (6 ways 3.2.1.1)
BwBB (6 ways 3.1.2.1)
BwBwB (6 ways 3.1.2.1.1)

Finishing with Black we have:

BBwW (6 ways 3.2.1.1)
BwBwW (6 ways 3.1.2.1.1)
BwWbB (12 ways 3.1.1.2.2)
WbBB (18 ways 1.3.3.2)

So that makes: 42 (63.64%) Black, 24 (36.36%) White.

Doing a similar analysis with 6 black and 2 white, I get: 288 combinations (144 end in each colour) and 5616000 permutations (3277440 (58.36%) black and 2338560 (41.64%) white).

So based on my figures, as you get more marbles, the number would converge on 50%, but I don't believe they would ever actually get there.

So why does my analysis get a different result from other peoples?

[So Long]

The puzzle states that you keep going until there is only one marble left. Kutti effectively stopped when there was only one colour left. This is not the same thing. Even if you draw the white marble first (with the 3 black, 1 white scenario), there are still 18 permutations that black balls can come out:

b1, B1, B2, B3
b1, B1, B3, B2
b1, B2, B1, B3
b1, B2, B3, B1
b1, B3, B1, B2
b1, B3, B2, B1
b2, B1, ...

------------------
Share and Enjoy

[Kutti]

Yes, I stop, when there is only one colour left, because the result is obvious then and we can calculate the probability (regardless of the remaining possible permutations).

I am doubtful about 'So Long's result, because the (66) permutations don't have the same probability (=1/66). So it's not sufficient to count the permutations with the result "finishing with black" and divide this by the total number of permutations.

[So Long]

Why don't they have the same probability?

------------------
Share and Enjoy

[Kutti]

So Long: For example, let’s have a look at the first drawing: The probability of getting the white marble is obviously 1/4, not 18/66 as in your calculation.

[Speeder]

The answer is: there is a 50% chance of ending up with either colour marble.

I suppose we all knew this intuitively but you might be interested in my javascript demonstration of this...

http://www.pygstone.pwp.blueyonder.co.uk/misc/gl/marbs.html

Speeder.



[This message has been edited by Speeder (edited 05-13-2004 10:46 AM).]

[IQ221]

The probability is 50% as shown below:

[
Let black.P_s(b,w) denote the probability of finishing with a black marble starting with a bag with b black marbles and w white marbles (where the last action was to shuffle).
Let black.P_b(b,w) denote the probability of finishing with a black marble starting with a bag with b black marbles and w white marbles (where the last action was to discard a black marble).
Let black.P_w(b,w) denote the probability of finishing with a black marble starting with a bag with b black marbles and w white marbles (where the last action was to discard a white marble).
Let white.P_s(b,w) denote the probability of finishing with a white marble starting with a bag with b black marbles and w white marbles (where the last action was to shuffle).
Let white.P_b(b,w) denote the probability of finishing with a white marble starting with a bag with b black marbles and w white marbles (where the last action was to discard a black marble).
Let white.P_w(b,w) denote the probability of finishing with a white marble starting with a bag with b black marbles and w white marbles (where the last action was to discard a white marble).
Then,
(1) black.P_s(b,w)=b/(b+w)*black.P_b(b-1,w)+w/(b+w)*black.P_w(b,w-1)
(2) black.P_b(b,w)=b/(b+w)*black.P_b(b-1,w)+w/(b+w)*black.P_s(b,w)
(3) black.P_w(b,w)=b/(b+w)*black.P_s(b,w)+w/(b+w)*black.P_w(b,w-1)
By subtracting (2) from (1) and (3) from (1) and adding the results we obtain:
(4) black.P_s(b,w)=(1/2)*(black.P_b(b,w)+black.P_w(b,w))
By symmetry, we can obtain:
(5) white.P_s(b,w)=(1/2)*(white.P_b(b,w)+white.P_w(b,w))
From (5) we obtain:
(6) black.P_s(b,w)=1-(1/2)*(white.P_b(b,w)+white.P_w(b,w))
Adding (4) and (6) we get:
(7) 2 * black.P_s(b,w)=1
Thus,
(8) black.P_s(b,w) = 50%.
]

[Speeder]

An interesting answer IQ221 but I can't help but think that even if your whole argument is falacious, you might still come up with the 50-50 result anyway because of the inherant symmetry in the way you have defined your argument.

Unfortunately I'm too boggled by your calculations to verify your answer. Can anyone else?

I think I've lost my marbles.

Speeder.

P.S. Is your IQ really 221? Mine is only 150.



[This message has been edited by Speeder (edited 05-13-2004 12:06 PM).]

[So Long]

Kutti, Your point is very valid. So revisiting my permutations I get

For the White:
BBB (3/4 * 2/3 * 1/2 = 1/4)
BBwB (3/4 * 2/3 * 1/2 * 1/2 = 1/8)
BwBB (3/4 * 1/3 * 2/3 * 1/2 = 1/12)
BwBwB (3/4 * 1/3 * 2/3 * 1/2 * 1/2 = 1/24)
Total: 12/24 = 1/2

For the White:
BBwW (3/4 * 2/3 * 1/2 * 1/2 = 1/8)
BwBwW (3/4 * 1/3 * 2/3 * 1/2 * 1/2 = 1/24)
BwWbB (3/4 * 1/3 * 1/3 = 1/12)
WbBB (1/4)
Total: 12/24 = 1/2

So I have come around to the exactly 50-50 answer. It is interesting that the combinations are also exactly 50-50.

[Kutti]

Speeder: IQ221’s approach is symmetric. And from that we could conclude, that the solution is 50-50, _IF_ we could assume, that a solution independent from the number of marbles (b, w) exists. But we couldn’t. IQ221’s elegant solution just PROVES this existence.

Or in other words: The last line in IQ221’s proof shows, that the requested probability is independent from the number of marbles at the beginning of the game.

Congratulation to IQ221.

[Speeder]

Thanks Kutti, I'd like to take your word for it but how do we know that IQ221's equations (1) (2) and (3) are correct?

By the way, did anyone see my javascript demonstration I posted previously?

Speeder.

[joyce]

okay so i just had this huge brain storm of an answer then disproved it in my head reuturning to my orginal hypothesis that no matter the number of each marble the outcome is fifty-fifty, black or white. it goes to a friend of mine's theory that everyday there is a fifty percent chance of rain, simply put that each day it either rains or does not outside circumstances have no baring on her calculation.

[magnum]

if the last must be black then the first one must be white right?

so the change that your first marble is a white is (243 ncr 1 * 729 ncr 0)/972 ncr 1 = .25

so the change is 25 %

[Speeder]

Sorry Magnum but you are incorrect. The final marble colour does not directly depend on the first marble colour drawn.

You can see a demonstration of this at my site:
http://www.pygstone.pwp.blueyonder.co.uk/misc/gl/marbs.html

Doesn't anyone read previous posts? (Not just you Magnum, but we pretty much already have established the chances are 50-50. Kutti agrees, IQ221 has written a seemingly plausible proof, I've demonstrated it on my site. Several others generally agree the answer is a 50% chance for each colour. Yet, even now, someone is bound to post a new reply with a different answer.)

Speeder.

[Lambert]

Originally there are 729 black and 243 white marbles, so p(first marble is black) = 3/4.

The symmetry of the problem means that the probability of any particularly placed marble being black remains constant. Hence p(last marble is black) = 3/4.

[link]

50/50 chance is very unlikely with my analysis. If you were to pick from a bag with 3:1 ratio to black, you will probably pick black. So the chance of black occuring as the last ball left will be low. If you get what i mean. you will pick all the blacks first leaving whites to take a greater chance of being last. or you will lower the chances for the blacks.

[think of it as being a soccer game. if there are more better players left to pick (black), then the fewer worse players will be picked last (white).......which will (most likely)leave the worst 1 left( 1 of the whites left).]

PS you might find it difficult to understand the logic behind my explaination.

[Speeder]

Dear Lambert and link,

Wrong again!

I think what you are missing is the fact you place the marble back in the bag when you pick one that doesn't follow suit. Read the question again!

Link, taking your example of the 3:1 ratio of black to white; you get 8 combinations of drawing marbles, 4 leading to black, 4 leading to white as the final marble. See below:

4 Finishing black:
WB,BBB
BW,WB,BB
BW,BW,WB
BBW,WB

4 Finishing white:
BW,BW,BW
BW,BBW
BBW,BW
BBBW

(Note that the comma denotes the last drawn marble is put back in the bag as per the rules, then drawing continues.)

Speeder.

[ZutAlors!]

Good example, Speeder, but a little misleading since each case doesn't have the same probability. However, each of the four cases in each set match with another case from the opposite set having the same probability. Thus, the total probability for each set sums to 50%. Perhaps this outline-type flowchart would make it clearer:

code:

---

Start with 3 black marbles and 1 white one.





(P=1/4) Draw white marble


    (P=1) Next, draw black marble, replace, and draw 3 blacks = WB,BBB


(P=3/4) Draw black marble


    (P=1/3) Next, draw white marble.  Replace.


        (P=1/3) Next, draw white marble.


            (P=1) Then draw black, replace, and draw 2 blacks = BW,WB,BB


        (P=2/3) Next, draw black marble.


            (P=1/2) Then draw white marble. Replace.


                (P=1/2) Draw white, then black. Replace. Draw black. BW,BW,WB,B


                (P=1/2) Draw black, then white. Replace. Draw white. BW,BW,BW,W


            (P=1/2) Then draw black marble. 


                (P=1) Draw white; replace. Draw white. BW,BBW,W


    (P=2/3) Next, draw black marble.


        (P=1/2) Next, draw white marble.


            (P=1/2) Draw white, then black. Replace. Draw black. BBW,WB,B


            (P=1/2) Draw black, then white. Replace. Draw white. BBW,BW,W


        (P=1/2) Next, draw black marble.


             (P=1) Draw white; replace. Draw white. BBBW,W

---

This gives eight possible combinations, four ending in black:
WB,BBB P=(1/4)=6/24
BW,WB,BB P=(3/4)(1/3)(1/3)=2/24
BW,BW,WB,B P=(3/4)(1/3)(2/3)(1/2)(1/2)=1/24
BBW,WB,B P=(3/4)(2/3)(1/2)(1/2)=3/24

And four ending in white:
BW,BW,BW,W P=(3/4)(1/3)(2/3)(1/2)(1/2)=1/24
BW,BBW,W P=(3/4)(1/3)(2/3)(1/2)=2/24
BBW,BW,W P=(3/4)(2/3)(1/2)(1/2)=3/24
BBBW,W P=(3/4)(2/3)(1/2)=6/24

A little addition will confirm that the total probability in each set is 1/2.

[Sphincter]

The probability of the last ball being black is 1. I know because I asked my cat, puddy, and she can't do decimals. Puddy has a history of being right on these things.

[Lambert]

I am sorry about my first answer,I had not read the question properly.

I now agree that the answer is a half, no matter how many marbles of each colour there are to start with.

It is obvious if you start with (1,1) meaning one black and one white that it is a half in this trivial case.

Now increase the problem to (2,1) and it is easy to see it is still a half.

Now increase the problem to (3,1) and it remains a half. Indeed for all (m,n) it remains a half.

Now ain't that cute.

[Speeder]

ZutAlors! That was nicely pointed out regarding the probablities for each combination of marble drawings. Intuitively you would imagine each combination adds up to 1/2 for each side but proving it...!

Did you see IQ221's proof? I'm still not sure I understand it, probability is not my strong point. Is it correct?

Lambert, your post makes me wonder if it is possible to prove using induction.

Speeder.