cheating on the Russian roulette

Question

This question is inspired by this one. You and your opponent play Russian roulette with two bullets. Quoting from the linked question:

You are challenged to a game of Russian roulette. Your opponent places two bullets side by side in a six chamber revolver and spins the chamber. She explains that the game is single turn-based. She then offers that you can choose who goes first.

However, since you are more motivated by survival than by fair play, both players would shoot their opponent instead of themselves if that increased their probability of surviving. To try to prevent this, a knife is placed in the table, so that if a player tries to shoot the other one and the gun doesn't fire, the other player will certainly kill him with the knife.

What's your probability of surviving the game if both players follow an optimal strategy?

Note: "Strategy" means choosing to play first or second and deciding in which turn to shoot the other player instead of their own head. Any other way of cheating is supposed to be prevented by other means.

Answer 1

😥 🅰🅱🅰🅱🅰🅱

1/6 ⚫⚫⚪⚪⚪⚪
1/6 ⚪⚫⚫⚪⚪⚪
1/6 ⚪⚪⚫⚫⚪⚪
1/6 ⚪⚪⚪⚫⚫⚪
1/6 ⚪⚪⚪⚪⚫⚫
1/6 ⚫⚪⚪⚪⚪⚫

🅰🔫
◻️ 💥➡️🅰⚰️
◻️ ❌➡️🔄
🅰🔫🅱
◻️ 💥➡️🅰⚰️
◻️ ❌➡️🅰🔪🅱➡️🅱⚰️

---

❌❌❌❌❔ (🅰)
1/1 ⚪⚪⚪⚪⚫⚫

🅰🔫
◻️ 1/1 💥➡️🅰⚰️
🅱🔫🅰
◻️ 1/1 💥➡️🅱⚰️

✔️🅱🔫🅰➡️💯🅱⚰️

---

❌❌❌❔ (🅱)
1/2 ⚪⚪⚪⚫⚫⚪
1/2 ⚪⚪⚪⚪⚫⚫

🅱🔫
◻️ 1/2 💥➡️🅱⚰️
◻️ 1/2 ❌➡️💯🅱⚰️
🅰🔫🅱
◻️ 1/2 💥➡️🅰⚰️
◻️ 1/2 ❌➡️🅰🔪🅱➡️🅱⚰️

✔️🅰🔫🅱➡️1/2🅰⚰️➕1/2🅱⚰️

---

❌❌❔ (🅰)
1/3 ⚪⚪⚫⚫⚪⚪
1/3 ⚪⚪⚪⚫⚫⚪
1/3 ⚪⚪⚪⚪⚫⚫

🅰🔫
◻️ 1/3 💥➡️🅰⚰️
◻️ 2/3 ❌➡️1/2🅰⚰️➕1/2🅱⚰️
🅱🔫🅰
◻️ 1/3 💥➡️🅱⚰️
◻️ 2/3 ❌➡️🅱🔪🅰➡️🅰⚰️

✔️🅰🔫/🅱🔫🅰➡️2/3🅰⚰️➕1/3🅱⚰️

---

❌❔ (🅱)
1/4 ⚪⚫⚫⚪⚪⚪
1/4 ⚪⚪⚫⚫⚪⚪
1/4 ⚪⚪⚪⚫⚫⚪
1/4 ⚪⚪⚪⚪⚫⚫

🅱🔫
◻️ 1/4 💥➡️🅱⚰️
◻️ 3/4 ❌➡️2/3🅰⚰️➕1/3🅱⚰️
🅰🔫🅱
◻️ 1/4 💥➡️🅰⚰️
◻️ 3/4 ❌➡️🅰🔪🅱➡️🅱⚰️

✔️🅱🔫➡️1/2🅰⚰️➕1/2🅱⚰️

---

❔ (🅰)
1/6 ⚫⚫⚪⚪⚪⚪
1/6 ⚪⚫⚫⚪⚪⚪
1/6 ⚪⚪⚫⚫⚪⚪
1/6 ⚪⚪⚪⚫⚫⚪
1/6 ⚪⚪⚪⚪⚫⚫
1/6 ⚫⚪⚪⚪⚪⚫

🅰🔫
◻️ 1/3 💥➡️🅰⚰️
◻️ 2/3 ❌➡️1/2🅰⚰️➕1/2🅱⚰️
🅱🔫🅰
◻️ 1/3 💥➡️🅱⚰️
◻️ 2/3 ❌➡️🅱🔪🅰➡️🅰⚰️

✔️🅰🔫/🅱🔫🅰➡️2/3🅰⚰️➕1/3🅱⚰️

---

❌2/3🅰⚰️
✔️1/3🅱⚰️

---

I'll call the player who goes first A and the one who goes second B.

On the fifth turn it's A's move. If the game reaches this point, A knows that a round is chambered, so he will always opt to shoot B.

On the fourth turn it's B's move. At this point the odds that a round is chambered is 1:1 even. B knows if he fires at himself he will either die now or in the next round, so he opts to fire at A, where his odds of survival are 1:1 even.

On the third turn it's A's move. At this point the odds that a round is chambered is 2:1 against. It makes no difference what A does: if he fires at himself he dies now with probability 1/3, or dies in the next round with probability 1/2, leading to an odds of survival of 2:1 against. If he fires at B his odds are the same, 2:1 against.

On the second turn it's B's move. At this point the odds that a round is chambered is 3:1 against. If B fires at himself he dies now with probability 1/4, or dies in a subsequent round with probability 1/3, leading to an odds of survival of 1:1 even. If he fires at A his odds are 3:1 against, so he chooses to fire at himself.

On the first turn it's A's move. At this point the odds that a round is chambered is 2:1 against. It makes no difference what A does: if he fires at himself he dies now with probability 1/3, or dies in a subsequent round with probability 1/2, leading to an odds of survival of 2:1 against. If he fires at B his odds are the same, 2:1 against.

Thus B has the higher chance of survival, with odds of 2:1 on.