Quote:
Originally Posted by pdurrant
An easy one! At least, that's how it's classified where I found it over at http://www.folj.com/puzzles/easy.htm.
A pot contains 75 white beans and 150 black ones. Next to the pot is a pile of 300 black beans.
A somewhat demented chef takes the beans from the pot in the following manner:
- He takes two beans at random from the pot.
- If one of the beans is black, he places it in the pile of black beans and puts the other one (white or black) back in the pot.
- Otherwise, he throws both beans away and takes a black bean from the pile and puts it into the pot.
Note that two beans have been taken out and only one put back. He repeats the procedure, reducing the number of beans in the pot by one each time, until there is only one bean in the pot.
What colour bean is the last one in the pot?
[Oh - and please enclose answers in spoiler tags as usual. Thanks!] |
this puzzle is hurting my brain. i am not sure i am even close (since at first i thought the opposite, but upon re-reading i realised i had probably misunderstood one detail of the puzzle) but i will try it anyway...
Spoiler:
i think the last bean should be white.
the reason is, whatever the draws, the number of white beans can only be reduced by 2 at a time, but black beans are removed 1 at a time. we start with an odd number of white beans. since it's impossible to divide an odd number by 2, there will be 1 white bean left over.
the results of possible draws is this :
2 black beans : end result, 1 less black bean in the pot, same number of white beans (one put on the pile, other back in the pot)
2 white beans : end result, 2 fewer white beans in the pot but 1 more black bean (both white beans discarded, one black from the pile put in)
1 black, 1 white : end result, 1 less black bean, same number of white beans (black one on the pile, other (white) bean back in the pot)
i've changed my mind about 5 times about my answer so i am now going to go see what other people have replied.
