I’ll explain how this game works for anyone who wants to know. first of all let’s say there’s a game of this where it’s gotten down to four spheres. If whoever goes next takes one sphere the other player takes three and wins. If they take two spheres the person who goes next takes two and wins. If they take three spheres the person who goes next takes the last one and wins. Now let’s say there are eight to start with. If the person who goes first takes three the other player can take one and then the other person has to be the person who takes next when there are four left which I already explained loses the game. Same if the first picks two and second picks two or the first picks one and the second picks three. This can be expanded to any number of starting orbs that’s a multiple of four and assuming the second player plays perfectly the first can’t win. Let’s assume either the starting number is changed to not a multiple of four or player two isn’t playing perfectly. Then you take however many you need to get to a multiple of four and next turn your opponent will have to take first on a multiple of four allowing you to win if you always take three if your opponent takes one, two if your opponent takes two, and three if your opponent takes two.

I hope this explanation was easy enough to understand. Sorry if someone has already explained in another comment somewhere.

Or rather, make sure it’s a multiple of 4 after your turn. (As noted above: if it’s 4, then whoever goes first will leave 1, 2, or 3, which whoever goes second can just take and win. If it’s 8, then whoever goes first will leave 5, 6, or 7, which whoever goes second can reduce to 4 and thus win no mater what whoever goes first does. Recurse for each multiple of 4.)

If it starts at 4 or any multiple thereof, whoever goes second can leave it a multiple of 4.

Otherwise, whoever goes first can make it a multiple of 4 in one turn, and then just keep making it a multiple of 4.

In other words: if it’s 4, 8, 12, 16, 20, or the like, go second and you can win. If not, go first and you can win.

The unicorn is right, of course. This is a solved game. If you can guarantee all multiples of 4 you will win. Thus, whoever goes second will win every game.

I’ll explain how this game works for anyone who wants to know. first of all let’s say there’s a game of this where it’s gotten down to four spheres. If whoever goes next takes one sphere the other player takes three and wins. If they take two spheres the person who goes next takes two and wins. If they take three spheres the person who goes next takes the last one and wins. Now let’s say there are eight to start with. If the person who goes first takes three the other player can take one and then the other person has to be the person who takes next when there are four left which I already explained loses the game. Same if the first picks two and second picks two or the first picks one and the second picks three. This can be expanded to any number of starting orbs that’s a multiple of four and assuming the second player plays perfectly the first can’t win. Let’s assume either the starting number is changed to not a multiple of four or player two isn’t playing perfectly. Then you take however many you need to get to a multiple of four and next turn your opponent will have to take first on a multiple of four allowing you to win if you always take three if your opponent takes one, two if your opponent takes two, and three if your opponent takes two.

I hope this explanation was easy enough to understand. Sorry if someone has already explained in another comment somewhere.

Quite clear, and I’m sure it was helpful to some of the more challenged readers.

Not me, of course.

So, from what I understand, it’s a really good thing that Baron Mistycorn can also multiply so he can beat the house. Or rather, the future-house.

“and three if your opponent takes two”

Bit hard by the copy / paste

So to win, you say: You go first.

Or rather, make sure it’s a multiple of 4 after your turn. (As noted above: if it’s 4, then whoever goes first will leave 1, 2, or 3, which whoever goes second can just take and win. If it’s 8, then whoever goes first will leave 5, 6, or 7, which whoever goes second can reduce to 4 and thus win no mater what whoever goes first does. Recurse for each multiple of 4.)

If it starts at 4 or any multiple thereof, whoever goes second can leave it a multiple of 4.

Otherwise, whoever goes first can make it a multiple of 4 in one turn, and then just keep making it a multiple of 4.

In other words: if it’s 4, 8, 12, 16, 20, or the like, go second and you can win. If not, go first and you can win.

RandomFan: You could say that, but “let me win” is pretty good, too.

The unicorn is right, of course. This is a solved game. If you can guarantee all multiples of 4 you will win. Thus, whoever goes second will win every game.

Does a solved game count as ‘rigged’?

If the other player is forced to play as the “losing” starting position, then I’d say it is.

Gods above, below, and to all sides, do I hate Nim.

Even in a video game where I can just look at a reminder sheet, I. Hate. Nim.

Is Baron M’s horn getting…taller?

That’s what she said!

It definitely is. Unicorns must get longer horns whenever they level up.

Terry Pratchett used a similar strategy to solve the Labyrinth Puzzle.

Flash Gordon couldn’t have done it better. Great SF reference. (Yes, I know it’s a different Godian. But the cheesy helmets.)

Occam’s Razor: The simplest explanation is often the correct one.

Maxim #37: There is no “kill” or “overkill”, there is only “open fire” and “cover me while I reload”.

Obviously the Baron is playing by the rules, therefore there is no foul by this scenario.