Originally posted by eldragonfly 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.
Originally posted by Coconut Yes because (50 - 2 + 1) > 50 right?
No. If you continue to draw 2 balls of the same color, then you are going to keep adding one white ball to the initial bag of 25/25 black/white marbles, a white ball from some magical imaginary other source. Therefore it is beyond obvious that some of the rather sloppy assertions as stated here are meaningless and absurd.
Originally posted by eldragonfly No. If you continue to draw 2 balls of the same color, then you are going to keep adding one white ball to the initial bag of 25/25 black/white marbles, a white ball from some magical imaginary other source. Therefore it is beyond obvious that some of the rather sloppy assertions as stated here are meaningless and absurd.
I'm going to admit it - I can't tell if you're being serious or not. Little help?
Originally posted by eldragonfly No. If you continue to draw 2 balls of the same color, then you are going to keep adding one white ball to the initial bag of 25/25 black/white marbles, a white ball from some magical imaginary other source. Therefore it is beyond obvious that some of the rather sloppy assertions as stated here are meaningless and absurd.
The game is over after 49 draws, regardless of what colours are picked. Worst case is that you need 24 additional white balls in the 'imaginery source' - which b.t.w. could be where you keep them to play the game wit n much higher than 25.
Originally posted by Mephisto2 The game is over after 49 draws, regardless of what colours are picked. Worst case is that you need 24 additional white balls in the 'imaginery source' - which b.t.w. could be where you keep them to play the game wit n much higher than 25.
Not if your initial pick is two black balls, and you keep making the same pick. Your bag of black and white balls will keep growing ad infinitum. Nice try.
Originally posted by eldragonfly Not if your initial pick is two black balls, and you keep making the same pick. Your bag of black and white balls will keep growing ad infinitum. Nice try.
Take 2 out LEAVE THEM OUT put one in you %(*^@*&^$#@^$^#@^*&^&$#@^&^%#@*$^(#@^*%*&@#
Originally posted by eldragonfly Not if your initial pick is two black balls, and you keep making the same pick. Your bag of black and white balls will keep growing ad infinitum. Nice try.
You can pick 12 times two black balls, so no ad infinitum there. Anyway, whatever you pick, the number of balls, after picking 2 and putting one back, diminishes by one. After 49 picks, there is only one ball left, the one you put back in.
If you still want to maintain your position, then you can only be admitred for being extremely stubborn. Stubborn and stupid makes a nice alliteration too.
Originally posted by eldragonfly 1) bag with 50 balls in it - 25 black, 25 white
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 b ...[text shortened]... e what color it might be.
If this is too complicated for you to understand - let me know. 🙄
Why are you getting hung up on such irrelevant details? This whole problem is hypothetical. It is not a documentary, so yes, everything about it is imaginary, including the source of the balls that replace the balls that are drawn. There is just one bag, the original bag. The source of the balls that replace the ones that are drawn is irrelevant. It's there to make the problem interesting, and consistent. If the only source of balls were the original bag itself, then yes, there would be problem cases when you run out of white balls outside of the bag, and you start drawing paired black balls. The "Place in the bag" balls do not have to come from the same bag. They can come from anywhere as long as the condition that if you remove paired balls of the same color, you replace the pair with one white ball. Mephisto already put the reason why the drawing process cannot go on forever more elegantly, and clearly than I can, so I won't bother explaining why this process won't go on ad infinitum. I hope this clears things up.
Originally posted by Coconut Take 2 out LEAVE THEM OUT put one in..
Ahhh...now we're getting somewhere. At least you finally admit your initial definition was severely flawed. i also assumed you possibly meant to take two out and replace with a coloured ball, depending on selection, but you never made it explicit or clear. i asked you several times to clean it up, now this childish nonsense. Speaks volumes.
Originally posted by eldragonfly Ahhh...now we're getting somewhere. At least you finally admit your initial definition was severely flawed. i also assumed you possibly meant to take two out and replace with a coloured ball, depending on selection, but you never made it explicit or clear. i asked you several times to clean it up, now this childish nonsense. Speaks volumes.
now this childish nonsense
huh? You're the one being a jackass and playing 'dumb' when anyone with .15435426 of a brain would not be the slightest bit confused about what you are whining over.
Originally posted by eldragonfly Ahhh...now we're getting somewhere. At least you finally admit your initial definition was severely flawed. i also assumed you possibly meant to take two out and replace with a coloured ball, depending on selection, but you never made it explicit or clear. i asked you several times to clean it up, now this childish nonsense. Speaks volumes.
Coconut isn't even the one who posted the problem. If you're going to troll, at least direct it at the right person.
Colored balls in a bag
20 Sep '05 07:46
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