Do coins fall one way or another when you try to set them on their rounded edge?

Assume the coins are perfectly balanced and are not predisposed to fall on either side from spinning or flipping.

I think this is the answer:

We have to take some amount of coins out of the group and flip them all. Though my motivation for this idea may seem arbitrary, you will see that it will work out:

So, let's say that out of our group of 100 coins, we choose x of them and form a group.

In the group we formed, let's imagine that we picked h heads (obviously h can not be more than 10 since there are only 10 heads from the initial group to choose).

So, in our new group, since we picked x coins and h of them are heads, we have:

Heads: h
Tails: x-h

Thus, in the remaining group (consisting of 100 - x coins) we have:

Heads: 10-h
Tails: (90-[x-h])

What we really want is a thing to do that does not depend on h, the amount of heads we choose in our pile. We want to determine a value of x coins to pick that doesn't depend on the number of h heads that will end up in our pile since we have no way of determining h.

We take our group of x coins that we picked out and flip them. Hence, we will have:

Heads: x-h
Tails: h

So, then number of heads in our new group is x-h. The number of heads in the remaining group (the 100-x coins that we didn't pick out of the initial group) is 10-h. Hence, we learn that we must select x=10 coins to pick out.

So, our solution is as follows: take our initial group of 100 coins, pick at random 10 of them, flip those 10 and the two groups should have the same amount of heads.

I think this is right...

i used to be so good @ these types of things
lets see here though:
can u feel the sides of each coin to tell if its h or t?
idk

but LongLive...what if the coins dont fall the right way, u know just bc there is a 50/50 chance doesnt mean it always happens that way

LLy got it right, with a very nice method to find 10 as the number of coins needed

he said to take 10 and flip them, what he means is take any 10 coins and turn them over, so if you take 10 tails and turn them over, you would get 10 heads, if one of the 10 coins you took is a head, when you turn them over you will have 9 heads an 1 tails, but the original stack also has 9 heads since you took one of the heads.

... Huh?

edit: I should've never gone to this topic, the '3+6+8+7 = 4' thing is pissing me off.

the goal is to get 2 piles with the same number of heads.

since we know there are 10 heads flip 10 coins at random and have those be one pile and the other 90 another pile

if these 10 ARE all heads then we have 2 piles with 0 heads

if there are 9 heads we have 2 piles with 1 head each (since the tail we flipped is now a head)

if there are 8 heads we have 2 piles with 2 heads each (8 heads become tails and the 2 tails become heads)

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.
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if there are 0 heads we will now have 2 piles with 10 heads (since we just flipped 10 tails)

What's brown and sticky?

black tar heroin?

A stick.