I have 200 candies in an hourglass dispenser. There are 40 each of red, white, blue, yellow and green. Each time I turn the hourglass over the candies get thoroughly mixed and 20 candies are poured into a cup. I like the blue candies best.
When I want candy, I turn over the dispenser and sort through the candies in the cup. I leave all of the blue ones in the cup and put the rest back. Then I turn over the hourglass a second time. The dispenser fills the cup so that I get exactly 20 candies, which I eat. I continue this way until the dispenser is empty.
How many blue candies will be in the final cupful?
Note:The dispenser dispense just the right amount so that to total will still be 20; so if 4 blues remain, it will dispense 16.
(In reply to
re: simulation by Larry)
Actually I was considering fewer than 10 times through the process. My error seems to be a misinterpretation of the second phase of the process. I interpreted the phrase "put the rest back" as to put the non-blue back into the cup rather than the dispenser. then the second phase would begin where the dispenser filled the cup only enough to replace the eaten blues.
I still wonder what is meant by "if 4 blues remain" I thought that the blues were actually eaten, so much so that I interpreted it, without even thinking, as 4 non-blues remain in the cup after the blues are eaten between the two phases of one round.
I'll note my comment pointing out my misinterpretation.
I think you misinterpreted my table:
0 199284
1 298161
2 246558
3 146525
4 69482
5 27133
6 9126
7 2753
8 784
9 147
10 38
11 8
12 1
It's not implying more than 10 dispensings. It means that in 199284 of the million trials zero blues were in the last dispense, etc. Of course, aside from this, if some dispensings are only partial, there could be more than 10 of those.
Edited on December 12, 2023, 7:45 pm
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Posted by Charlie
on 2023-12-12 19:27:00 |
re(2): simulation
