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.
Apparently what's intended is that there are 10 rounds. At the end of each round 20 candies are eaten. In each round the first phase consists of one dispense in which the non-blues are rejected and sent back to the hourglass, and the second phase is another dispense, where whatever is presented into the cup is accepted.
A couple of sets of ten rounds is shown:
first dispense cup after rejects sent back cup contents that are eaten blues eaten
rrwwbgybbgrgwrgybywg bbbb bbbbwrbrwyrybbrgwygr 7
wwwwwbwrrrwgwywgygrr b byybgygywgbbbbwrbgyb 8
ywyyrrgywwwggygwrgyy wwbgywyrbrywybybwggw 4
bggbgwgwgyrwrrgwybrb bbbb bbbbywrrryyrrbrgwgbr 6
rgbyrgbwbbwywrygbwyb bbbbbb bbbbbbrggwwrwygyrrgy 6
gyygwyrywwgwwwywrrwg rbbgrygwygywwbgrrywr 3
ryybywggrgggwwbrwwrr bb bbwwwywyywyrbgryywyg 3
byggywwyggrrgrwwyyrw b brgbrgyggrrwwyywwyrg 2
gwywwrrwryyggygrwygw ggggyrgwggyrygywgrwr 0
rbrrywywwgggwwrwgwgr
1 blues eaten in last cup
In the final round, there are only 20 left in the hourglass, so ultimately they are eaten. In this first case, there's only one blue candy in this last cup eaten.
Second example:
gyyyrgbwywywwygwwgrg b bgrbygrryyrrbbggwrby 5
yyygbwrrwggwrbgyrgrw bb bbywbgyyyywbgyrwrwgy 4
grgywbgyggrrygwybryy bb bbwbrrggwwwyyrwrrwyb 4
gbgyrwgbrgggyybgygwy bbb bbbgyygyybrbrwywyryg 5
ybwwrywrbgrbwbyygwyg bbbb bbbbryggrbyrwygwrrry 5
grrggbbrbyygwywrbwyw bbbb bbbbbbgrrwggwbgwrggg 7
grbgwggrwgyrwwyrwyww b byrrgrwwrywwbgyybggw 3
gwrggywywgwrrwybwgrb bb bbgrwwyyywwwrbywgbrw 4
wgrryrbyyggygwbrwgbw bbb bbbyggywgywrrrrgrgrg 3
wwwgwwgwygrrygygwygw
0 blues eaten in last cup
In this second example, all the blue candies were eaten in rounds prior to the last round, so no blue candies were in the last cup.
totct=0;counts=zeros(1,21);
for i=1:1000000
cup='';
hourglass0=[repmat('r',1,40) ...
repmat('w',1,40) repmat('b',1,40) ...
repmat('y',1,40) repmat('g',1,40) ];
hourglass=hourglass0(randperm(200));
% disp(hourglass)
while ~isempty(hourglass)
% disp(' ')
if length(hourglass)<=20
ct=sum((hourglass=='b'));
cup=hourglass;
% fprintf('%-30s\n',cup)
hourglass='';
continue
else
cup=hourglass(end-19:end);
% fprintf('%-30s',cup)
hourglass=hourglass(1:end-20);
end
hourglass=[hourglass cup(find(cup~='b'))];
cup(find(cup~='b'))='';
% fprintf('%-30s',cup)
refill=20-length(cup);
r=randperm(length(hourglass));
hourglass=hourglass(r);
if length(hourglass)<=refill
cup=[cup hourglass];
ct=sum((cup=='b'));
% fprintf('%-30s\n',cup)
hourglass='';
else
cup=[cup hourglass(end-(refill-1):end)];
% fprintf('%-30s',cup)
% fprintf(' %d' , sum(cup=='b'))
hourglass=hourglass(1:end-refill);
end
end
% disp(ct)
totct=totct+ct;
counts(ct+1)=counts(ct+1)+1;
end
fprintf('%9d/%d %10.8f\n',totct, i, totct/i)
The sample printing was commented out, to provide statistics for a million trials.
The results was
>> candies200
486532/1000000 0.48653200
probably good to 3 places, 0.487
|
|
Posted by Charlie
on 2023-12-13 07:46:47 |
simulation with new interpretation
