The Game of Death (Part II)

Probability Level 5

Surprisingly, all the five agents have survived from The Game of Death Part I . The gangster leader selects an agent and forces the selected agent to play another Game of Death: Russian Roulette . The gangster leader puts two bullets in consecutive order in an empty six-round revolver, spins it, and points the muzzle at agent's head. The rules of the game are:

In each chance, the agent can choose either he wants to spin revolving cylinder again or not.
If the agent can survive from four times chance, he and the other four agents will be released. If not, the rest will be killed.

Assuming that the agent uses the best strategy . What is the probability that the agent will survive?

Note: Your answer must be in the interval $[0,1]$ .

The answer is 0.25.

4 solutions

Tunk-Fey Ariawan
Mar 7, 2014

The solution is as following:

Round 1 : Agent can choose either he want to spin the revolving cylinder again or not. The probability of agent can survive is $\frac{2}{3}$ .

Round 2 : Agent must choose to NOT spin the revolving cylinder, just pull the trigger instead. If he chooses to spin it again the probability of agent can survive is $\frac{2}{3}$ . But if agent chooses to not spin the revolving cylinder, the probability of A can survive is $\frac{3}{4}$ . Let's denote $\text{B} = \text{bullet}$ and $\text{E} = \text{empty chamber}$ in the revolving cylinder. Since the two bullets are put in the consecutive order, we can consider these two bullets as a 'single' bullet. Take a look the combination of the bullets in the revolving cylinder after round 1:

$\text{BBEEE}$
$\text{EBBEE}$
$\text{EEBBE}$
$\text{EEEBB}$

The combination of bullets that can make agent dies is only combination $(1)$ . Then it is obvious the probability of agent can survive in round 2 is $3$ out of $4$ . Of course he will choose this option (do not spin the revolving cylinder, just pull the trigger) because this option has greater probability.

Round 3 : Agent must choose to spin the revolving cylinder again although the probability of agent can survive either he spins it again or not is equal. If he chooses to spin again, the probability is $\frac{2}{3}$ and if not, the probability is also $\frac{2}{3}$ . Take a look the combination of the bullets in the revolving cylinder after round 2:

$\text{BBEE}$
$\text{EBBE}$
$\text{EEBB}$

The combination of bullets that can make agent dies is only combination $(1)$ . Then it is obvious the probability of agent can survive in round 3 is $2$ out of $3$ . He must choose to spin the revolving cylinder again because if not, the greater probability that he gets in round 4 is only $\frac{2}{3}$ (he asks to spin it again, because if not the probability is only $\frac{1}{2}$ ). If he chooses to spin it again in round 3, he can get the probability of $\frac{3}{4}$ in round 4 by choosing to not spin the revolving cylinder, just pull the trigger.

Round 4 : Agent must choose to not spin the revolving cylinder, just pull the trigger instead. The probability of agent can survive in round 4 is $\frac{3}{4}$ .

Thus, the probability that he can survive until round 4 is $\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}=\dfrac{1}{4}=\boxed{0.25}$ .

They told me to write an integer.I had it wrong the first time anyway,but I wasted my points thinking that as 0 and 1 were the only integers available,I might as well earn a few points.Shows that greed can get you down.Although it is a problem that brilliant should solve nonetheless.

A Former Brilliant Member - 7 years, 2 months ago

Sorry guys, actually the question has been edited by Brilliant (I think Calvin) and I also edited based on the Brilliant editing but unfortunately we didn't consider about changing the answer. I've already asked Calvin to fix it by email and it looks like he hasn't checked his inbox yet. The original question was asking $p+q$ where the probability was expressed by $\dfrac{p}{q}$ . I'm really sorry for my terrible mistake my friends.

Tunk-Fey Ariawan - 7 years, 2 months ago

in question it is written answer is in [0,1] so i tried 0.25 but it is not written to give in form of p+q

dp dp - 7 years, 2 months ago

This problem has two different answers it depends how literally you take the problem. If you keep the two bullets separate i.e. Keep the shots out of 6 then the agent has a 4/6 or 2/3 chance of getting an empty chamber, thus living. Then when he takes the second shot as the two bullets are separate and as we have 6 bullet chambers the chance of him getting another empty chamber is 3/5. Then for the third shot if he chooses to take the shot his chance of survival would be 2/4 which means he has an equal chance of being killed as he does living which isn't a viable tactic. Therefore if he spins the gun after the second shot has been taken then the probability essentially resets and he is once again on a 2/3 chance or 4/6 of getting an empty chamber again, and here he should just repeat the tactic. This would leave the sum being (2/3 . 3/5) + (2/3 . 3/5) = 0.79992 (roughly) Can someone tell me if they had the same problem with this question?

Henry Hollamby - 5 years, 5 months ago

Cynthia Liu
Mar 17, 2014

1st shot: Regardless of whether he spins, the agent has a 2/3 chance of surviving.

2nd shot: The agent now knows the location of one empty chamber. There are 4 remaining consecutive pairs in which the bullets can be. Only 1 pair results in the next chamber containing a bullet, thus without spinning his chance of survival is 3/4, 3/4 > 2/3.

3rd shot: The agent now knows the locations of 2 empty chambers, which leaves 3 possible pairs. Similarly, one pair results in the next chamber containing a bullet, making his probability of survival 2/3 if he does not spin, which is the same as spinning. However, if he didn't spin, the maximum probability of survival on the 4th shot is 2/3, by spinning (otherwise it is 1/2). As we have already seen, surviving a random starting position and its next move has a maximum probability of 2/3 * 3/4 > (2/3)^2, thus the agent spins on the 3rd shot. 2/3.

4th shot: The situation is exactly the same as the 2nd shot. 3/4.

Thus the answer is 2/3 * 3/4 * 2/3 * 3/4 = 1/4 = 0.25

Gabor Revesz
Mar 16, 2014

When the agent spins the cylinder there is a 1/6 chance for each of the 4 empty chambers to line up with the barrel, giving him a 2/3 chance of survival after a spin.

Assume the agent survived pulling the trigger immediately following a spin. Then the cylinder has advanced by one chamber. Thus each of the 2nd through 4th empty chambers as well as the 1st loaded chamber has a 1/4 chance of being lined up with the barrel, giving the agent a 3/4 chance of survival were he to pull the trigger without spinning. So since 3/4 > 2/3, the agent will not spin the cylinder at this juncture.

Thus the total chance that the agent makes it to the third round is (2/3)(3/4).

Assuming the agent makes it to the third round, the cylinder has advanced by another chamber. Thus each of the 3rd and 4th empty chambers as well as each of the 1st and 2nd loaded chambers has a 1/4 chance of being lined up with the barrel, giving the agent a 1/2 chance of survival were he to pull the trigger without spinning. However, since 1/2 < 2/3, the agent will spin the cylinder. Thus rounds 3 and 4 play out exactly like the 1st 2 rounds. Hence the probability of getting through all four rounds is (2/3)(3/4)(2/3)(3/4) = 1/4.

Shan Shah
Mar 16, 2014

If the agent choses among first three of the 4 empty chambers then he will surely survive the current chance and the next chance.. After surviving two consequent chance he again needs to select among those three chambers to survive next two chances. Thus the probability of his surviving is : (3/6)*(3/6) = 1/4

The Game of Death (Part II)

The answer is 0.25.

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