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.
The below is apparently based on misinterpretation of the procedure.
It seems to be random with a mean of 1.727, approximately:
totct=0;counts=zeros(1,21);
hourglass0=[repmat('r',1,40) ...
repmat('w',1,40) repmat('b',1,40) ...
repmat('y',1,40) repmat('g',1,40) ];
for i=1:1000000
cup='';
hourglass=hourglass0(randperm(200));
while ~isempty(hourglass)
if length(hourglass)<=20
ct=sum((hourglass=='b'));
hourglass='';
continue
else
cup=hourglass(end-19:end);
hourglass=hourglass(1:end-20);
end
cup(find(cup=='b'))='';
refill=20-length(cup);
if length(hourglass)<=refill
ct=sum((hourglass=='b'));
cup=[cup hourglass];
hourglass='';
else
cup=[cup hourglass(end-(refill-1):end)];
hourglass=hourglass(1:end-refill);
end
end
totct=totct+ct;
counts(ct+1)=counts(ct+1)+1;
end
fprintf('%9d/%d %10.8f
',totct, i, totct/i)
counts'
finds and average count of 1.727:
1726547/1000000 1.72654700
distributed as:
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
Edited on December 12, 2023, 7:46 pm
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Posted by Charlie
on 2023-12-12 14:32:28 |
simulation
