I have 20 dimes on the table in front of me(I can see them, but seeing as I'm alone, probably none of you can). They are randomly placed, but exactly half of them are facing head up, and half are tails up. I will follow your instructions to manipulate the coins one at a time, but there are only three instructions I will accept:
Place coin in pile A
Place coin in pile B
Flip coin over
Please provide instructions that will sort the coins into two identical piles (the same number of coins showing heads and tails in each pile)
ok probability analysis isn't something i've ever done so have no idea on how to work this out, so will guess..
i know there's a 50/50 chance of a coin being either heads or tails so think by turning over every 2nd coin, it should be the same as the preceding one..
so place first coin in pile A, turn second coin over and place in pile B, now just keep repeating this process, 3rd coin in pile A, turn 4th coin over and place in pile B, etc.
i did it 3 times and 3 times i had the same number of heads and tails in each pile, 6 heads and 4 tails, so hope that's the right answer., even tho i don't know the exact logical reason, it seems to work for me..
anyone else want to test my answer to see if they get same result? before tomk replies.
ok thought a bit more about this problem, and am 100% positive i have the right answer and have worked out the logic.
it doesn't matter how many coins of heads or tails i put in pile a i will always have the same number in pile b.
so if i put 4 heads in pile a and because there are 10 coins in a pile i must have put 6 tails into it also,
that means i put 6 heads in pile b and 4 tails but because i turned them over pile b now contains 6 tails and 4 heads the same as pile a..
If we divide the coins into two equal piles… there will be 10 coins in each pile.
Pile A will contain X heads and 10-X tails.
If pile B is left as is… it would contain 10-X heads… so it must have X tails.
If all of the coins in pile B are turned over… Pile B will contain X heads and 10-X tails.
Therefore…. congratulations (again) terry1966 on the correct answer.
got lucky with this one, because had no idea really how to work it out, i thought the math equation would be a lot more complicated.
so my first idea just happened to give me the correct result,
then after looking at my results i realized the logic of why it always worked, and it had nothing to do with probability equations.