Russian Roulette l

Probability Level 2

The cylinder of the gun, with six chambers, is all empty. Now I put a single bullet in the gun. I close the cylinder and spin it, and pull the trigger. Click. Its a blank!

Now I’m going to pull the trigger one more time without spinning the cylinder. What is the probability (in percentage) of the second shot also being a blank?

The answer is 80.

11 solutions

Satyen Nabar
May 19, 2014

Consider the odds if the cylinder is not spun. As the trigger was already pulled, there are five possible chambers remaining. Additionally, one of these chambers contains the bullet. That leaves four empty or safe chambers out of five. Thus the probability of survival is $\frac45$ , or 80 percent .

Please explain how a gun works :P. I thought that it never moves, and it still is on the same chamber, thus making my answer 100.

Kevin Mo - 7 years ago

Also, you would need to pull the hammer back.

Adam Quinn - 5 years, 9 months ago

For some guns, pulling the trigger will force the hammer backward and then release it.

Micah Wood - 5 years ago

Yeah I think this question relies too heavily on understanding how a gun operates. I got 100 too

A Former Brilliant Member - 5 years, 6 months ago

Basically pulling the trigger will rotate the cylinder then fire the next bullet it appear to.

You can learn more about it here .

Micah Wood - 5 years ago

yeaa the same

Muhammad Saleh Alatas - 1 year ago

Viswanath Mukundala
May 25, 2014

Please Correct me if my logic is wrong. I think the answer should be 66.67%. If I have to survive two trigger pulls, I should be lucky enough to get two adjacent empty chambers. There are 4 such combinations of adjacent empty chambers. The chamber with bullet makes up the remaining 2 (one in which the very first pull will blast my head. And then the one in which the second trigger pull will). So 6 possible combinations, 4 desired expectations. Probability of surviving a second bullet = 4/6 = 66.67%

Your answer would have been right if the question had asked for the probability of surviving both the shots. But the question only asks for the probability of surviving the second shot...

Ramachandra Guha - 7 years ago

i think its 4/5, because you already used one shot from the start. Meaning, you should also subtract the remaining shots from the chamber, since it did not spin. So, I think its 80 percent! they are right. I guess it helps....

Darwin Clutario - 7 years ago

Yeah u r right.. They should have taken variable change into consideration. Have you seen the movie 21? In that movie they have proceeded to solve the very first problem in your way.

Sayan Kumar - 7 years ago

your logic is correct if there is an uncertainty, regarding the first shot. But, its has been already clarified that he had survived the first shot. End of the day question was asked from the second shot. Confusion is spinning the cylinder. Its also totally uncertain, it may depend on the speed u spin the cylinder and the movement of the bullet position inside the cylinder.Then in that case we can only go according to the basic process. There might be another solution, provided if some condition given on spun of the cylinder...

Rajesh Atmakuri - 7 years ago

However, we already know that the first shot was a blank, getting rid of the first undesired outcome. This makes the possible outcome 5, and the number of desired outcomes still remains 4. We are left with 4/5 = 80%.

Keegan Musaalo - 5 years, 9 months ago

Viswanath Mukundala I agree with what you are saying. The chance of surviving the last shot would be 0.16 rather than 1/2 based on nearly everyone's logic. That's just like saying the chance of throwing a six on a dice is still 1/6 even though your 100 previous throws have also been a six. The chance is a lot smaller.

Sebastian Adler - 5 years, 9 months ago

Kawaii Kuma
May 25, 2014

Just to be fooled by the first shot and count that shot into your calculation. Just 1 bullet and 5 holes. So 4/5 chance for you to survive!

Souvik Nandi
May 23, 2014

A trigger is already pulled. So remaining 6-1=5. Amongst them, 1 is having the bullet. So 4 chances of survival. Percentage of probability: 4/5 * 100%=80%.

Ok! Better Soluction

Heder Oliveira Dias - 7 years ago

short n sweet

PUSHPESH KUMAR - 7 years ago

Tom Engelsman
May 1, 2021

Using conditional probability, let $A$ be the event the first shot is blank ( $P(A) = \frac{5}{6})$ and $B$ the second shot is also blank. If $A$ and $B$ are independent, then we can compute via:

$\Large P(B|A) = \frac{P(B \cap A)}{P(A)} = \frac{(5/6)(4/5)}{5/6} = \boxed{\frac{4}{5}},$ or $80\%.$

Rhishikesh Dongre
Jun 4, 2014

It is given that he pulls the trigger once and we survived.

=> There are 5 chamber left before reload. One chamber has bullet so 5 - 1 = 4 chambers are left.

By using the formula to calculate Probability

P(E) = No of possible out comes/ all Outcomes ....where E is a event.

=> P(E) = 4/5 = 0.8 = 0.8 x 100 = 80 per cent.

( Problem solvers are going to kill this person)

A Former Brilliant Member
Jun 1, 2014

now the the chance of saving is 4 and total chance is 5 so probability is 4/5 and its percentage is 80%

Stefano Tairi
Jun 1, 2014

80%, it's an application of conditional probability

Aliki Patsalidou
May 26, 2014

there areonly 5 empty chambers ,one of them gone so they remain only 4 out of5 , so 4/5of 100% is 80%

Krishna Garg
May 24, 2014

Since there are 5 slots left with one slot having bullet in it,so chances of survival are4/5 as probability that is 80% Answer.

K.K.GARG,India

Sk Riasat
May 23, 2014

6 chamber. 1 filled 6-1=5-1 blank fire=4 blank ### probability 4+1 bullet =5. ### 4/5x100=80%.

Probability of general survival is 5/6 . Since 1st shot has been survived, it serves as a probability without replacement since d cylinder wont b spun 4 d nxt shot. So there shd b a 1/6 subtraction 4rm d survival probability which gves 4/6. % of dat gves 66.67%

Saviour Kufoalor - 7 years ago

Yes, but the number of possible outcomes also decreases by 1.

Keegan Musaalo - 5 years, 9 months ago

Russian Roulette l

The answer is 80.

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