15 Sep 05
Originally posted by eldragonfly"1) last two balls could be black - last ball = black"
No, you are mistaken.
1) last two balls could be black - last ball = black
2) last two balls could be white - last ball = white
3) last two balls could be black and white - last ball = black.
4) clearly, there is no solution.
no. Two black balls = white
15 Sep 05
Originally posted by Mephisto2i believe it.
please elaborate. This is above my understanding level.
1) brown paper bag contains 50 marbles - 25 white & 25 black
2) draw two marbles at random
3) pick two black marbles. paperbag now contains 48 marbles
4) It is impossible to return 1 white marble as you have 2 black marbles in your hand
[If you draw two white balls, or two black balls, you place a white ball in the bag.]
5) Your experiment is flawed may never get off the ground. That is why i asked for a modification of the original problem.
17 Sep 05
Originally posted by eldragonflyThe problem says nothing of "returning". It is about replacing the last two drawn marbles with one marble. If the first draw is two black marbles, you replace them with a white marble.
i believe it.
1) brown paper bag contains 50 marbles - 25 white & 25 black
2) draw two marbles at random
3) pick two black marbles. paperbag now contains 48 marbles
4) It is impossible to return 1 white marble as you have 2 black marbles in your hand
[If you draw two white balls, or two black balls, you place a white ball in the bag.]
...[text shortened]... ed may never get off the ground. That is why i asked for a modification of the original problem.
17 Sep 05
Originally posted by rheymansYou're not paying attention. Redefine the problem or admit you have a faulty definition.
The problem says nothing of "returning". It is about replacing the last two drawn marbles with one marble. If the first draw is two black marbles, you replace them with a white marble.
19 Sep 05
Originally posted by eldragonflyYou're assuming that the marbles initially in the bag are the only marbles available.
You're not paying attention. Redefine the problem or admit you have a faulty definition.
The explanation you've been provided with shows this is a false assumption. There's only a problem with replacing the first two black balls with a white one if you assume that all the problem's white balls are in the bag!
20 Sep 05
1 edit
Originally posted by rheymans1) bag with 50 balls in it - 25 black, 25 white
You have a bag with 50 balls in it. 25 of the balls are white, and 25 are black. You draw two balls from the bag at a time. You don't know what color the balls are until you have removed them from the bag. If you draw two white balls, or two black balls, you place a white ball in the bag. If the draw is two balls of different color, you place a black ...[text shortened]... n white balls, and n black balls, when is the last ball white, and when is the last ball black?
2) random selection
3) 2 alt = black 2 same = white
4) you place a white/black ball in the bag depending on selection
So several questions come to mind.
1) Where do the balls come from that are placed "in the bag?"
2) Is it the same bag or a different bag?
3) Do these "place in the bag" balls come from the same bag or not?
These question remains unanswered. i think i know why. 😕
From your shoddy definition it appears that the bag with 50 balls will just keep growing in size, ie., ad infinitum. So there won't be any "last" ball, let alone what color it might be.
If this is too complicated for you to understand - let me know. 🙄