Ye've Lost Yer Marbles!

[kevinatilusa]

I think the post made initially by Zutalors can be quickly fashioned into an induction proof if you want rigor.

[ZutAlors!]

Speeder: To me, it's easier to think about this problem starting at the top (see my first post; reply 3).

Suppose you start a "game" with *any* non-zero number of white marbles, and *any* non-zero number of black marbles. Then, define a "round" as: keep picking marbles until you get a different color, then replace that last color (for example, a round could be "black, white, replace", or it could be "white, white, white, white, replace").

So what could happen on any given round? Three thing: 1) You could pick and discard *all* the white marbles, which guarantees you'll end with black, 2) You could pick and discard *all* the black marbles, which guarantees you'll end with white, or 3) Neither of these occurs, in which case you have to start a new round.

Each game then consists of some number of rounds. Now, and here's the key: all you need to do is determine the relative probability for the three cases above for each round. It turns out (see my post above) that the probability of cases (1) and (2) are exactly equal, for *any* number of black or white marbles. Thus, for any game, you have an equal probability of going "out" with black or white on every round, and thus you have an equal probability of going "out" with black or white for the entire game.

[ETA: Although, as kevin says, I think you could fashion it into an induction proof if you want. I'm not a rigorous mathematical thinker, though, so I'll leave it for someone else.]

[This message has been edited by ZutAlors! (edited 05-17-2004 07:50 AM).]

[12345]

i think there may be a 50/50 chance......if you discard the marble which does not follow the same one before.but the board says that no one has come up wit the right answer?

[michigan]

For the case where we start with 2 black and 1 white:

The probability of ending with a black one is 5/9.

Here are the possibilities:

W; stop; prob=1/3
B B; stop; prob= (2/3)* (1/2)^(1)
B W W; stop; prob = (2/3) * (1/2)^(2)
B W B B; stop; prob = (2/3) * (1/2)^(3)
....

The odd iterations result in ending with a black marble.
The even iterations result in ending with a white marble.

[ZutAlors!]

michigan: you've an error in your probablities. The probability of the last case is (1/3(1/2)(1/2) = 1/6. If you add the two cases where you end with a black, you'll see that the total probability is indeed 1/2.

Deathstream: I think you may have not accounted for the fact that marbles are thrown back into the bag after being picked, in some circumstances.

[Speeder]

ZutAlors! I like your original explanation in your first post. I'm still puzzled over some things though. I can't decide if your demonstration shows that the sum of the probabilities for each completed game ending in white necessarily equals the sum of probabilities of games ending in black.

Also, looking at the example 1 W and 2 B you get these combinations:
(1) WB,BB
(2) BW,WB
(3) BW,BW
(4) BBW

When you say 'discard all' is this (1) and (4) equalling prob 1/3 each? If so, this leaves BW as the only other remaining combination for the first round of the rest of the probabilities???
Alternatively...
If we say (1),(2),(3) have first rounds with equal probabilities for BW and WB then we are left with (4) on its own as the 'discard all' option. This doesn't seem right either.

I still believe in the 50% answer, but the more I think about why, the more confused I get.

The other point that is puzzling me is (as you pointed out) the probabilities are different for each combination even though there are clearly the same amount of combinations with blacks/whites at the end of drawing. If I could just demonstrate that those probabilities would always add up to 1/2 each for all combinations...

Speeder.



[This message has been edited by Speeder (edited 05-19-2004 09:44 AM).]

[ZutAlors!]

[Deathstream]

I got two answers one of which I am pretty sure is right. I think my second answer has a better chance. I got either 20.4% or 60%. I think the answer is 60% more than I do the 20.4%.
1st answer
The answer is 11 out of 54 times or about 20.4%
Why? Because there's 3 black and 1 white after reducing.
Chart displays the ratio of when white is picked 1st, 2nd, or 3rd. For information on how I got the numbers email me.
Pick# 1st 2nd 3rd
Black 3/4 8/9 3/4
White 1/4 1/9 1/4
((1/4)+(1/9)+(1/4))/3= 11/54. If you do it for black it will be 43/54 to save you work.
So white will get picked 1st, 2nd, or 3rd 11 out of 54 times (to me that does quite make sense but i think it's right so email me if you think I am right and know how to tell me why it makes sense) so 43 out of 54 times white will be picked last. Therefore black will be picked last 11 out of 54 times.
2nd answer
If that's wrong then 60% is right. Actually I got a total of 150 combos (you probably don't believe it but just trust me cause it takes a very advanced mind to see it on your own). 90 times black comes last and 60 times white is last. I think 60% is right, at least more than I think 20.4% is.
Site creators make sure you send me an email if i'm right or post it on the site because nobody else has this answer and i'm only 15 so that will be cool if I got it first.

Dear ZutAlors,
I did consider it I just messed up on the 2nd drawing of marbles.
If you can figure out something else that is wrong tell me.
Plus it can't be 50/50 because the probability of the 2nd to last ball drawn will depend on what the 3rd to last ball drawn is making one ball have the advantage. Plus I got 50% on my first try and I doubt I could have done it right the 1st time but 60% is like the 9th time around.
Consider marbles being put back if you want to see the combos on your own.




[This message has been edited by Deathstream (edited 05-19-2004 08:17 PM).]

[C. Postir]

Random post. (The lightbulb was getting on my nerves.)

C. Krett Postir

[thesmartguy]

the answer is a BLACK marble.

Even though there is more chance of drawing a black marble, it would soon change to the white marble and it will be going down 50-50

[thesupersmartguy]

The answer is a BLACK marble, i made a mistake about the explaination on the last post

Even though there is more black marbles, eventually the black marbles will be discarded and white and black would be about even.
Counting down from 729, it will even out, then it would be whites turn to go. White would go more than black since a simple sum solved this. but there is a possibility of 66.1/3% of drawing a black marble last since the white would have already gone by the time there are 2 marbles left

[Deathstream]

okay my final answer is 57.9% for the last marble being black.
I got 76 combos black finishes 44 times and white 32 times

[This message has been edited by Deathstream (edited 05-21-2004 02:31 PM).]

[BonelessConnus]

Ok, my last and final answer fially came to be 50/50.

You can automatically go to the different scenarios for the last two marbles. The following scenarios are:

--You can end up with 2 whites and the last one you discarded was white. So you will no matter what, you will end up with white.
--Another scenario, you can end up with 2 whites and the last on you discarded was black. Still, you will get white.
--Now, you can get 2 blacks and the last one you discarded was black and end up with the last one being black.
--Or you can have 2 black, the last one you discarded was white, still, you will end up with black.
Now, with those four scenarios, you can get a 50/50 probability. Or 1:1 Ratio, correct?
But there are still two more scenarios . . .
--You can have one of each, with the last color discarded a black. One possibility is you can choose black and it will all be over with, you are left with white. But! If you pick white, you start over again and the discarded marble no longer matters. So, now you have a 50/50 chance of it being either black or white. So, you have two chances to end up getting black, but chance of getting white. So, the ratio is 2:1, correct?
--Of course, you have one of each again, but with the last color discarded a white. You still have a 2:1 ratio.

Alright, so now we have three separate ratios- 1:1, 2:1, and 2:1

Now, the 1:1 ratio is equal to 1/2 chance of getting black and 1/2 chance of getting black. And one of the 2:1 ratios is equal to 1/3 chance of getting black and a 2/3 chance of getting white. And the other 2:1 ratio is vice versa. 1/3 chance of getting white and 2/3 chances of getting black.

Not to get ahead, we'll work this out all mathematically like.

So the chances of getting black are 1/2 * 1/3 * 2/3 = 2/18 = 1/9

The chances of getting white are also 1/2 * 1/3 * 2/3 = 2/18 = 1/9

So..they both have an equal chance of being the last marble left. So, the chances of black being the last one left is 50/50.

[Deathstream]

that's only true if there are an equal chance of those senarios happening

[BonelessConnus]

Well . . . true. Do you not think that the white marbles have more of a chance to be the last one given that there are less of them? If there are less white marbles, wouldn't it lessen the chance of picking them up, causing you to end up picking black marbles more often until there are an equal amount of both? Then each marble would have an equal, 50/50 chance of being picked out of the bag down to the last two. So, the probability of the last marble being a particular color is 50/50.

Suppose you have 100 Black and 99 white. Since you have more black, you are more more likely to pick out the black marble. If you DO pick out the black marble, you now have 99 white and 99 black. Since they're now equal in number, you have an equal chance of picking one out. Suppose you pick out a white one this time, you now have 98 white and 99 black. Again, you have more black, so the next time you pick one out, it is more likely going to be black. You will more than likely pick out the black on and once again reach equilibrium. This process will go on until the last ones, in which there will be and equal chance of either color being picked out.

Actually, I think I have just realized something. According to what I just said, the last scenario will end up being one color of each with the last one being discarded black. This will make the ratio 2:1. With 1/3 being the chance of white being the last one and the last one being black will be 2/3. I think I just changed my answer to 2/3.

[Deathstream]

well the way i did it came out almost 50/50 so i think your answer is too high. maybe not. try and c if u can find then combos i did. it's tricky

[Muzaktaz]

You only have one marble left, no matter what, and you only had 2 kinds of marbles in the bag, black or white...so the chance that it is black is 1 out of 2, 50/50 chance of it being black.

[BonelessConnus]

Actually, my last one was wrong. It's not 2/3 chance of being black with the scenario is two different colors with the last one being discard black. You have a 1/3 chance of the ladt marble being black and 2/3 being white. So, 1/3.

[wylarion]

Hi,

The answer is 50%.

Common reasoning tells us that if there are more black marbles, they will be picked more often and discarded (it's important that if we first pick e.g. 5 blacks and then 1 white, the white one is not discarded since it only breaks the loop). Thus the ratio of black and white marbles converges to 50/50 regardless of the initial ratio.

The mathematical proof by IQ221 seems correct.

Here's a Java code to simulate the puzzle. With 1000 iterations it gives very close 50/50 ratio with all test executions.


public class Marbles
{
static int BLACK = 0;
static int WHITE = 1;

public static void main(String[] args)
{
new Marbles();
}

public Marbles()
{
int blacks = 0;
int whites = 0;
int rounds = 1000;

for(int i = 0; i < rounds; i++)
{
int color = simulate();
if( color == BLACK )
blacks++;
else
whites++;
}
System.out.println("blacks: "+(blacks * 100.0/rounds) + " whites: "+(whites * 100.0/rounds));
}

public int simulate()
{
Vector bag = new Vector();
for(int i = 0; i < 729; i++)
{
bag.add(new Marble(Marbles.BLACK));
}
for(int i = 0; i < 243; i++)
{
bag.add(new Marble(Marbles.WHITE));
}

int prevColor = -1;
Marble m;

while(true)
{
if( bag.size() == 1 )
break;

int pos = random(bag.size());
m = (Marble)bag.get(pos);

if( prevColor == -1 || m.color == prevColor )
{
bag.remove(pos);
prevColor = m.color;
}
else
prevColor = -1;
}
m = (Marble)bag.get(0);
return(m.color);
}

private int random(int size)
{
int value = (int)(Math.random() * size);
if( value == size )
value--;
return(value);
}


public class Marble
{
int color;

Marble(int color)
{
this.color = color;
}
}
}

[Sumudu2]

It seems some new ppl are under the impression that the 'solver' of the puzzle will get credited etc. and that puzzles are on the board only until they are solved. This is not the case, the puzzle discussion thread is here simply for discussion, and it's not a race to see who can solve it first (well, not officially)...so don't assume that the correct solution hasn't been found just because the official solution isn't up.

It would appear that the 50% answer is correct, just take a look at several of the posts in this thread explaining it. Of course, if you think there's something wrong with it, you should definitely post your ideas


...also sorry if that sounds a little high-handed. anyway, that information can also be found in the sticky post at the top of this forum

[This message has been edited by Sumudu2 (edited 05-24-2004 01:44 PM).]

[Deathstream]

do you know when the answer will come out?

[Deathstream]

okay i had an error the last time i did my calculations. The answer is 7/11 or about 63.6%
For the people who do not think the origonal numbers count then why did they give us numbers that simplied so much.

[This message has been edited by Deathstream (edited 05-24-2004 03:21 PM).]

[BonelessConnus]

Deathstream, I did it the way you did it. The number of black marbles/number of total marbles and simplified down, correct? I got 7/12. But wouldn't that only count if you pick a marble out randomly and discarded it despite what color it was? I would think the pattern of the marble picking would have some effect on the answer. I think the numbers were there just to mislead the person trying to solve the puzzle. Maybe it was a coincidence.

[Sumudu2]

Deathstream, could you explain again how you're doing it, I'm not sure I understand how you're getting that number.

[Deathstream]

um, well, when i look at my paper, um, it's kind of all scribbilies so i'm not to sure which part of the paper my answer came from. Hrm.....
I think it's is 50/50 but something tells me those numbers that they gave use to start of with are what they are for a reason. If you have a black and a white marble the odds are 50/50. If one marble is added then i got it to be 17:3 but i did that quick and that seems wrong. Oh well.

[Mr Lettuce]

This problem reminds me of a linear algebra question. I took linear algebra about 4 years ago...so...yeah, I don't remember how to do anything related to linear algebra. However, I do remember this problem.

The problem is a model of a Markov process/Markov chain. A frog starts on one lily pad in a pond and them starts hopping around. The question asks, what is the probability of the frog being on lily pad "x" after "y" number of hops. The idea is that each lily pad has a probability associated with it after a certain amount of time (in this case time is measured in hops). The solution -being a linear algebra problem- involves a lot of multiplication with matrices.

Maybe it helps. But I think you might still end up with an answer of 50/50.

Cheers.

[michigan]

What exactly is the rule for discarding or shuffling? My understanding is you only discard if it is the first turn or the the draw immediately preceding is the same color as the current draw. Is that correct?


If so, for the case of 3 black and 1 white why isn't B,W,B,W,B,W,B,W,B,W,B,W,B,W,B,B,B,W
listed as one of the possible combinations

Thanks.

[Lucky Wizard]

You put the marble back if it's a different color from the last one, and you did not put the last one back. If you put the last one back, or your current marble's the same color as the last one, then you discard it. Putting the last one back is considered to be starting over.

Hence, in the scenario you give, a black marble would be taken out and discarded; then a white marble would be taken out and put back; then a black one would be taken out and discarded; the white marble would be taken out and put back again; and then the black one would be taken out and discarded. This results in the BW,BW,BW,W one which does appear in the possible combinations.

[mathgeek]

Ummm, what's the official answer? I know the correct one is 1/2, i just want to know the official one!

also, I'd like to challenge IQ's post: