A game to play with marbles

[mathgrant]

It's simple. I have 3141 black marbles and 1413 white marbles. I will draw one at random and discard it into the fiery pit where [insert favorite video game villain here] will eat it. I will then keep drawing more random marbles and discarding them until one doesn't match the previous one. That marble will not be discarded, but put back in the bag. The marbles will be reshuffled and the process will start over.

Example: I draw a black marble and discard it.
I draw another black marble and discard it.
I draw a white marble and shuffle it back into the bag.
I draw a black marble and discard it.
I draw a white marble and shuffle it back into the bag.
I draw a white marble and discard it.
I draw another white marble and discard it.
I draw another white marble and discard it.
I draw a black marble and shuffle it back into the bag.

etc.

After eternity amount of time, eventually only one marble is left. Rounded off to 6 decimals, what is the probability of it being black?

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Sometimes life gives you a deck without any aces. Deal with it.

[Daft Dog Fred]

Eeh, lad. Ye've lost yer marbles...

[ZutAlors!]

Intuitively, I'm thinking the probability is: 0.500000
Not quite sure of the formal reasoning, but it strikes me that: for any distribution of marbles, the probability of picking all the black ones, one after another, until they're all gone is equal to to the probability of picking all the white ones, one after another. All other cases (i.e., not picking all marbles of one color) simply result in a new round with a new distribution of marble, to which the earlier observation applies.

[mith]

Looks formal enough to me

[ralphmerridew]

I get .500000 .

Oddly enough, it seems not to depend on how many of each kind of marble there is.

[This message has been edited by ralphmerridew (edited 03-19-2002 10:40 PM).]

[mathgrant]

Daft Dog Fred: lol. . .

ZutAlors!: Highlight to reveal your fate: You will eat a fortune cookie. It well call you an expert of probability. .500000 is right.

mith: You may be slower, but at least you can understand Zut's logic. Luna will be making love to you 100 times as often as before.

ralphmerridew: You resisted the temptation to highlight Zut's invisible answer. You solved it yourself. Good job. Hope you're reading this! ;]

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Sometimes life gives you a deck without any aces. Deal with it.

[ChienFou]

If you played the same game with three colours of marbles would the outcome be 1/3 each? Just curious.

[Daft Dog Fred]

Well, ye'd still lose yer marbles...

*considers going into the marble selling business*

[mikegoo]

ChienFou with his 3 color bag...my immediate instinct was no. Then I tried a simple game with 2 red, 1 white, and 1 black marble and after doing much more work than I thought I would, came up with p(red)=26/72 and the other 2 being 23/72 for each. Confirming my guess (provided my math skills are not incompetent which is up for debate sometimes), but I was wondering if anyone had a more elegant way of showing it other than the boring grind it out counter example?

anyone?

[Luna]

quote:

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mith: You may be slower, but at least you can understand Zut's logic. Luna will be making love to you 100 times as often as before.

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Well, then I'm going to get awfully tired.