Fedimath Episode 3: Whack-A-Troll

Suppose that there are N instances. Steve Jail creates alt accounts while the American Society for the Prevention of Cruelty to Animals (ASPCA) purges his alts. They take turns as follows:
- First, Steve Jail creates exactly two alt accounts anywhere he likes. (They may be on the same instance or on different instances.)
- Second, the ASPCA chooses *one* instance and purges *all* of Steve's alts from that instance.

By definition, an instance reaches the *boiling point* if Steve Jail manages to have M alts on that instance existing at the same time.

Steve Jail wants to ensure that *some* instance reaches the boiling point, while the ASPCA wants to prevent it. Who wins? (The answer depends on N and M.)

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Example:

Suppose there are four instances (N = 4) and the boiling point is M = 3. Initially, Steve Jail's alt count is (0, 0, 0, 0).

Steve Jail creates alts on instances 1 and 3. Now the alt count is (1, 0, 1, 0).

The ASPCA purges instance 3. Now the alt count is (1, 0, 0, 0).

Steve Jail creates alts on instances 1 and 4. Now the alt count is (2, 0, 0, 1).

The ASPCA purges instance 1. Now the alt count is (0, 0, 0, 1).

Steve Jail creates two alts on instance 4. Now the alt count is (0, 0, 0, 3), so instance 4 has reached the boiling point. Steve Jail wins but his dog loses.

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Extra credit: How does the answer change if, on each turn, Steve Jail can create three alts?

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See Fedimath Episode 2 here: https://cawfee.club/notice/AaqqArnnuXt0I1uyMi

cc @MercurialBlack @scenesbycolleen @ceo_of_monoeye_dating @roboneko @jeffcliff @hidden @udongle @tiskaan