Difference between Monty Hall and Russian roulette problems

Question

Please see the wikipedia for the description of the Monty Hall problem .

The Russian roulette problem is: There are two bullets placed randomly in a 6 chamber revolver. Your opponent goes first and survives the first trigger pull. You are given the option whether to spin the barrel. Should you?

The conditional probability of you losing in the next round given the opponent survived the first round is 2/5. In my opinion this is an "intuitive" answer, because there are 5 chambers left and 2 bullets in them.

On the contrary, the "intuitive" answer for switching doors in the Monty Hall is 1/2, which is incorrect. I say 1/2 because conditioned on the goat being shown, there are now two doors to choose from, one of them containing the car.

My question is: what is the difference between these two problems which makes the Monty Hall problem deceptive? Is there a way to apply Bayes Theorem to the Russian roulette problem?

Answer 1

In the Monty Hall problem there is prior knowledge (on the part of the judges) of which door the goats are behind. If the contestant on the first pick happens to pick the door with the goat. The prior knowledge of the judges will then open the door with the other goat, leaving the car still hidden. The contestant's first pick without any prior information is still only 33% chance of being correct. Thus there is 66% of the car is behind the other door. In this problem the correct answer is to switch and pick the other door.

In the Russian problem is there is no knowledge of what the next chamber is. You are correct the chance the next chamber has a bullet is 2 in 5 or 40%. If you spin the chance resets back to 2 in 6 or 33%. In this problem the correct answer is to spin again.

Hope this helps.

Answer 2

You pick door A which contains either a goat g, the car c, or the (soon to be revealed) mid-value prize r. Ag Br Cc, Ag Bc Cr, Ar Bg Cc, Ar Bc Cg, Ac Br Cg and Ac Bg Cr are your pre-pick enumerated possibilities. The instant the host reveals r, the only remaining possibilities are Ag Br Cc, Ag Bc Cr, Ac Br Cg and Ac Bg Cr since A (your pick) was not revealed ... furthermore if B was revealed that leaves Ag Br Cc, Ac Br Cg and Ac Br Cg ... so you should not switch since there is a 66% probability of Ac. Same probability if C had been revealed.

With RR possibilities where 1 is empty, 1e, a given since your opponent is alive, and a blue bullet and a red bullet are somewhere else enumerate as 20 different permutations, to wit: 2b 3r, 2b 4r, 2b 5r, 2b 6r, 2r 3b, 2r 4b, 2r 5b, 2r 6b, 3b 4r, 3b 5r, 3b 6r, 3r 4b, 3r 5b, 3r 6b, 4b 5r, 4b 6r, 4r 5b, 4r 6b, 5b 6r, 5r 6b ... tallying up we have eight 2s eight 3s eight 4s eight 5s and eight 6s ... but spinning allows for 1 to come up ... there are 20 instances of 1e ... the probability of landing on empty with a spin is 4/6 = .666. Without spinning chance of slot 2 being empty is 12/20 = 0.6, so you're better off spinning.