There's a couple of scenarios, which depend on when you run out of blue balls:
1) If you draw all 50 blue balls first, for every blue ball you draw, you are adding 2 white balls. Hence, once you've drawn all the blue balls, you have replaced them with 100 white balls. This means you've drawn 50 balls total and there's now 100 white balls and 50 red balls (150 balls) in the container. Then, you continue and draw the next 50 balls. These could be any combination of white and red balls, we don't know which color is preferred when you continue drawing.
2a) If you draw all red balls in these next 50 draws, you would have 50 red balls out and 50 blue balls out. So, you would end with 100 white balls in the container.
2b) If you draw all white balls in these next 50 draws you would have 50 red balls, 50 blue balls, and 50 white balls out. So, you would end with 50 white balls and 50 red balls in the container.
2c) If you draw a mixture of red and white balls in these next 50 draws, the number leftover in the container would vary according to the specific combination drawn.
Remember, the order in which the balls are drawn matters to this question since we have a rule in place that changes the overall total (or population) of balls within the container. In each of these scenarios, you've drawn 50 times initially (all blue) and then 50 times again (in some unknown combination of red and white). You've drawn 100 times total and changed the number of white balls in the container from zero to an amount dependent on how many times you drew a white ball on your second round of 50 draws.
2b) If you draw all white balls in these next 50 draws you would have 50 red balls, 50 blue balls, and 50 white balls out. So, you would end with 50 white balls and 50 red balls in the container.
... so after removing 100 balls, I've removed 150 balls? And the 150 balls that I've removed are red, white and blue despite the fact that I removed 50 blue balls initially and then 50 white ones.
Just because it fails one test in a particular way doesn’t mean it lacks reasoning entirely. It clearly does have reasoning based on all the benchmarks it passses
You are really trying to make it not have reasoning for your own benefit
> You are really trying to make it not have reasoning for your own benefit
This whole thread really seems like it's the other way around. It's still very easy to make ChatGPT to spit out obviously wrong answers depending on the prompt. If it had actual ability to reason as opposed to just generating continuation to your prompt, the quality of the prompt wouldn't matter as much
e composition of what's left in the container.
There's a couple of scenarios, which depend on when you run out of blue balls:
1) If you draw all 50 blue balls first, for every blue ball you draw, you are adding 2 white balls. Hence, once you've drawn all the blue balls, you have replaced them with 100 white balls. This means you've drawn 50 balls total and there's now 100 white balls and 50 red balls (150 balls) in the container. Then, you continue and draw the next 50 balls. These could be any combination of white and red balls, we don't know which color is preferred when you continue drawing.
2a) If you draw all red balls in these next 50 draws, you would have 50 red balls out and 50 blue balls out. So, you would end with 100 white balls in the container.
2b) If you draw all white balls in these next 50 draws you would have 50 red balls, 50 blue balls, and 50 white balls out. So, you would end with 50 white balls and 50 red balls in the container.
2c) If you draw a mixture of red and white balls in these next 50 draws, the number leftover in the container would vary according to the specific combination drawn.
Remember, the order in which the balls are drawn matters to this question since we have a rule in place that changes the overall total (or population) of balls within the container. In each of these scenarios, you've drawn 50 times initially (all blue) and then 50 times again (in some unknown combination of red and white). You've drawn 100 times total and changed the number of white balls in the container from zero to an amount dependent on how many times you drew a white ball on your second round of 50 draws.