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Egg Russian Roulette:

6 eggs, 5 of which are hard boiled & 1 of which is raw. Players (1v1) each choose eggs in a turn based system whereby the egg is smashed against the forehead. The first player to smash the raw egg against his/her forehead is the loser.

Should you have the option of choosing who goes first, would going first increase or decrease your chances of winning?

Moose
  • 113

2 Answers2

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Let's list all the probabilities shall we?

If you went first there is a:
$$\frac16$$ chance of you losing on the first turn,
$$\frac56\times\frac45\times\frac14=\frac16$$ chance of you losing on the third turn, $$\frac56\times\frac45\times\frac34\times\frac23\times\frac12=\frac16$$ chance of you losing on the fifth turn,

Therefore there is a $$\frac16+\frac16+\frac16=\frac12$$ chance of you losing, thus your chances will stay the same.

Abraham Zhang
  • 894
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If you went first you'd have a $1\over6$ chance of getting a raw one and the sequence of moves would look like this. $$\text{you}={1\over6}$$$$\text{them}={1\over5}$$$$\text{you}={1\over4}$$$$\text{them}={1\over3}$$$$\text{you}={1\over2}$$$$\text{them}={1\over1}$$ Which you'd have the best odds in the beginning and if it did happen to go all the way to the end your opponent would be the one to lose.

If you went second though your opponent would go twice in the time you only go once. So just within three moves you would've had a ${1\over5}$ chance of getting the raw one, but your opponent would have had a ${1\over6}+{1\over4}={5\over 12}$ which is a much higher percentage that yours.

So I could easily justify either way. That's just my logic. Don't know if it helps or not!

If you forced me to pick, I'd say go second. Just so if you happened to make it to the end and you knew you were going to get the raw egg you could use it as a weapon against your opponent.

Fmonkey2001
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  • Adding up these probabilities gives a number greater than $1$. Instead consider: you lose first turn $\dfrac16$, they lose second turn $\dfrac56 \times \dfrac15$, you lose third turn $\dfrac56 \times \dfrac45 \times \dfrac14$, etc. – Henry Dec 03 '14 at 07:07