Quote from: MrDrowner on September 08, 2018, 01:37:11 PM

Now there are 9 places in the Trial room and all our bots (insects) are pacifists (drown with only 1% probability, but revenge for drowning themself of course). If someone come with aggressive bot (like Loyce's one) it will climb with 8 pacifists - it's a question if it can beat 8 with the same ease as 6.

I can still beat 8 of them Cheesy

I'm now up against 8 of these:
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I'm using the same settings as before.

With one player like me
The probability of them drowning anyone is 1%. I win if nobody drowns anyone. If just one bot drows another bot, I'm very likely to lose.
I hope my math is correct:
Round 1: the probability of nobody drowning anyone is (0.99^8)^8 (=0.525596) (because there are 8 bots drowning 8 others with 1% probability)
Round 2: the probability of nobody drowning anyone is (0.99^8)^8
Round 3: the probability of nobody other than me drowning anyone is (0.99^8)^8

The probability of nobody other than me drowning anyone for 3 rounds is ((0.99^8)^8)^3 (=0.145). That means there's a 14.5% chance I win the round if I drown them all on round 3. With a 1 bit buy-in and 8.91 bit pot, that gives me an average return of 1.29x my wagered amount.
There's a small bonus: sometimes I win a shared prize (say 1.11 bit), which slightly increases the returns.

With 2 players like me
If anyone were to join using the same strategy, the math changes:
Round 1: the probability of nobody drowning anyone is (0.99^7)^7 (=0.611117)

The probability of nobody other than us drowning anyone for 3 rounds is ((0.99^7)^7)^3 (=0.228). That means there's a 22.8% chance we both win if we drown them all on round 3 and have to share the pot. That gives an average return of 1.01677x the wagered amount.