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You really should read the puzzle before guessing.

I said assuming. So there.

Cam, the question states that it is unknown how many hats of each colour there are. Anyway, Musicman's strategy is well above 50 ice creams and if they knew how many of each colour hat there was, they could easily get all 100 ice creams by simply counting them all and seeing what was missing.

You're misunderstanding me. The kids are assuming the hats are split evenly. They are taking a chance. It's an all or nothing gambit. Alternatively, If the hats are 99 black and 1 white, they might get 99 ice cream if they called out all black. Or they might get nothing at all.

My strategy is risky, but the payout would be great.

You're saying what if something along the lines of 96 black hats to 4 white hats occured? Let's say it goes, from the end of the queue, WWWWBBBBBBBBBBBBB....... :

-The first child,with the white hats, says "Black!". He's wrong about his own.
-Second says "Black!". He's wrong about his own.
-Third says "Black!". He's wrong about his own.
-Fourth says "Black!". He's wrong about his own.

-The rest then announce "Black!", and are all correct about their own hat, bagging 96 icecreams.

Just to be clearer: The protocol I'm describing is to announce the hat colour that appears most until there's an even number of black hats to white hats. After this point, all hats can be guessed correctly.


Let's look at a more evenly divided situation, say 60 black to 40 white. The list goes:
WWWWWWWWW...(40 white hats)....BBBBBBB...(60 black hats). The first 40 will all be wrong about their hats,but the last 60 will be correct.

In a more evenly divided situation, 50 black hats to 50 white hats:
WBWBWBWBWB........

-The first child, seeing 50 black hats and 49 white hats, announces "Black!", gets own hat wrong.(even though he could've deduced that his was white and got an ice cream, but he'd rather adhere to the protocol because he's a great little lad)
-Second announces "White!", seeing 49 black hats and 49 white hats. Gets his own wrong, but from this transition, every child in the queue can now deduce that there are 98 hats left, half of them are black and half of them are white.
-The third knows that there are 49 black hats and 49 white hats now. He sees 49 black hats and only 48 white hats. He knows his is white and announces "White!", gets an icecream.
-The rest do the same and all get their icecreams.

Worst case scenario: 50 black hats, 50 white hats, going like WWWWWWWW..(50 whites)....BBBBBBB..(50 blacks)...
-Every child announces "Black!". The first 50 get their own hat wrong, the last 50 get their own hat right.

TL;DR: Guaranteed at least 50 ice creams, but on average you'll end up getting more.

Congratulations, Norroen Dyrd! The strategy you posted is the best possible strategy! Using that strategy the children would get:

The above spoiler doesn't contain the answer, just how many ice creams the children should expect if they were to use it. This spoiler below contains a slightly expanded version of the answer:

If this was the first time you'd seen this problem, you have my respect, Norroen Dyrd. That's a tough puzzle. Now it's your turn to post a puzzle. I hope you have one because I'm keen to join everyone else going crazy trying to come up with the answers.

In response to you, Lndai, I agree with everything you said. Even a worst case scenario is 50 ice creams. What I'm struggling with is figuring out how many it would get on average. I was curious as to whether it would be better than Musicman's strategy. I was actually considering writing a computer program for it and running it 100,000 times or something to see what happened haha (actually, I might still do that if you're interested in the answer?). Anyway, you should probably put that in a spoiler. It bears quite a bit of resemblance to the "real" answer.

Another couple of answers that I came up with when I was trying to figure this puzzle out:

Thank you, Moeboid!

I'd like to offer a small historically-themed logic puzzle.

Medieval Architecture: A Historical Puzzle.

Your problem sounds very Google-able. I'm trying to resist the temptation. Assuming that it's not simply for the sake of consistency, does it have something to do with most people being right-handed (or right footed)?

Not sure if the answer can be found in Google (I haven't tried)

Your turn, Velvet LeChance!

I've never been particularly good at making these puzzles up. But I resisted the urge to steal one off the internet. I stole it from a show instead.

All basic maths, really. Tricky part (I found) was setting up the equation. Tell me if I've phrased it poorly.

EDIT: Ninja'd by Archer.

I'm terrible at making these so someone can take my spot.

I have a puzzle ready.

Go ahead.

Your turn, Musicman!

I read it over. They can only say what they suspect their OWN color is. They cannot say what the others are. Just responding to Musicman even though it is kinda late. It says in fine print no communications. Where's the confusion? Who cares now anyway. Puzzle has already been solved. Just throwing that out there. Not like I have the intelligence to solve it anyway.