Now explain the solution tallanvor gave that I didn't get, please. Am I right in saying his solution doesn't make it right?
<br> I think the point is supposed to be something along the line reasoning that you're under no obligation to maintain the "12" heads in the end. Most people probably get caught on the fact that each pile has to have 6 heads in it. But, rather, somehow the flipping of the coins can even out the piles by changing the number of how many heads/tails there are. This is because the question only asks to make them even in the two piles. That being said though, I still don't entirely understand how you can be sure you have the same number in each pile... <br> Let's do this step-by-step... say we have 100 coins. 12 are heads, 88 are tails. We randomly select 12 coins to move into a separate pile. out of those, 3 are heads and 9 are tails. pile x = original pile pile y = new pile so, pile X has: 9 heads, 79 tails pile Y has: 3 heads, 9 tails I flip all the coins in pile Y pile x has: 9 heads, 79 tails pile y has: 9 heads, 3 tails
Spoiler This might have been posted already, I didn't read the whole thread. A guy is at one end of a lake, and he has a boat. He's got a baby, a dog, and a plant. He can't leave the baby with the dog because the dog will bite him. He can't take more than 1 item with him at a time, how does he get all the items to the other side safely?
Define "safely." Cuz otherwise I don't see what's wrong with the man taking the dog over first, then the plant , then the baby. Right?
Your riddle is missing a piece so it's too easy. There has to be some interaction between baby and plant too. As it is written though: Takes baby first. Comes back for dog. Leaves dog. Takes baby back. Drops baby off. Gets plant. Goes back for baby.
Yeah, I **** the bed on that one. I remember I spent days on it when my grandpa first told it to me. Maybe the plant was a cat or something, I'll try to figure it out. OK, so I googled vague parts and came up with the right one I think.
That one doesn't sound right either. The fact that he can carry two of the items, in addition to himself, makes it too easy. Take dog and baby. Drops off dog. Goes back with baby. Picks up poison. Done.
what is required: a) there are two piles b) both piles have the same number of heads up coins. Spoiler at the start there are 12 heads up coins (Hs), 88 tails up coins (Ts), and 1 pile. Step 1) move any twelve coins to a separate pile (the small pile). The small pile will have X number of (Hs). The large pile will now have 12-X number of Hs since you just removed X. Step 2) flip all the coins in the small pile. The small pile previously had X number of Hs and 12 total coins. After flipping all the coins it will now have 12-X Hs and still 12 total coins. Now the large and small pile will have 12-X Hs and both requirements are satisfied. Can't make it easier than that for you Swoly. I got it without looking a the solution (so did many posters I bet). yes, i understand it.
i guess this riddle works no matter what the total number of coins is, because your numbers add up to 102, not 100
DudeWah and tallanvor: This is what is bothering me about the riddle, or maybe it's in the way it is asked: "divide the coins into two piles" (which pile? the 88 or the 12, or the 100?) "so that each pile has the same number..." But one pile is 90 and the other is 12? Doesn't that equal 102? EDIT: dookiester is along the same lines as me. EDIT2: OK, so maybe the end piles DON'T have to be equal in quantity. Is that right? I'll be cool with that. EDIT3: dookiester, I think he meant 79 tails in pile x.
Yeah, the two piles don't have to be the same amount. And the number of heads in each pile don't have to add up to 12... just have to be the same as each other.
Swoly, The key is the number of the smaller amount of coins that are 'different'. As long as you know the exact number, it will work, no matter how many there are. Using the number 12 is just arbitrary. Think in terms of the fact that this is once again a riff on the binary nature of these riddles. There are only two choices for each coin, we'll call them A or B. Out of 100 coins, let's just randomly say that 20 of them are 'A' and the other 80 are 'B'. As long as you know ahead of time that 20 of them are 'A', then it works. Let's walk through it, step by step. 100 coins in front of you, you do not know that status of any individual coin, but you know that AMONG the total, exactly 20 are 'A', and exactly 80 are 'B'. The ONLY goal here is to ensure that in two piles, each pile contains the same exact number of 'A'-sided coins. *The actual number of 'A'-sided coins is irrelevant - only the number of 'A'-sided coins being the SAME NUMBER IN EACH PILE matter*. Ok, so, the key is, since you know that there are exactly 20 which are 'A', then you randomly select 20 coins from the pile. Odds are, at least a few of of them will be 'A'-sided coins. Let's just go with the straight odds, and assume that 4 of the 20 you randomly chose are 'A'-sided, and the other 16 are 'B'-sided. So, now you have two piles, one with 20 coins, containing 4 'A'-sided coins and 16 'B'-sided coins. The other pile now contains 80 coins, which contains 16 'A'-sided coins, and 64 'B'-sided coins. (starting to see it yet?) Now, remember, our ONLY goal is to ensure that each pile contains the SAME NUMBER of 'A'-sided coins. Now flip the coins in your first pile. What do you end up with? Before, you had 20 coins, with 4 'A'-sided coins and 16 'B'-sided coins. But you flip them, and NOW you have 16 'A'-sided coins, and 4 'B'-sided coins. Exactly the same number of 'A'-sided coins as in the bigger pile. You could have flipped all the coins in the bigger pile, it wouldn't matter, you would still end up with the same number of 'A'-sided coins in each pile. This process works, no matter how many 'A'-sided coins you draw originally, as long as you separate out the exact number of coins from the bigger pile as you know beforehand are already 'A'-sided. If you know there are 20 'A'-sided coins, and you draw 20, and NONE Of the ones you drew were 'A'-sided, it still works, because you end up with 20 in each pile once you flip them. If you happen to draw all 20, it still works, because then there would end up with zero in each pile. And it works for any number in between. Because the goal is just to make sure they both contain the same number in each pile, not the same number you started out with. It's a pretty cool little logic exercise, and really it's just the nature of exploring binary thinking.
Some of you might know this, because it's older'n dirt and I've been asking it of friends ... :grin: There are 8 stones in a pile. All of them are equally as heavy, except one which weighs more, but you don't know which one it is. You can use a balance scale only twice to find the stone that is the heaviest. How do you figure out which one of the 8 is the heaviest? Enclose the answer in Spoiler , please.
Spoiler Spoiler algorithm is to break into thirds. so 3-3-2 for the 8. put the two sets of 3 on the scale. if they are equal you are down to 2, if not you have 3 left. if two left, put on the scale and whichever is heavier wins. if three left break into thirds again (1-1-1) put 2 of the rocks on the scale. if they weigh the same then the one not on the scale is the heavy one. If they don't weigh the same than the heavier one is your rock. Had that in an interview once.
