Russian Roulette -- II

Probability Level 3

Let’s play a game of Russian roulette. I have tied you to a chair. I have a gun in my hand. I show you the cylinder of the gun, six chambers, all empty. Now you see me put two bullets in ADJACENT chambers in the gun. I close the cylinder and spin it. I put the gun to your head and pull the trigger. Click. Its a blank! You survived!

Now I’m going to pull the trigger one more time WITHOUT spinning the cylinder. What is the percentage probability of survival for the second shot?

The answer is 75.

1 solution

Satyen Nabar
May 20, 2014

Consider bullets in adjacent chambers 5 6. Since first shot is a blank, it had to be chamber 1 2 3 or 4 that was fired. So next chamber to be fired will be 2 3 4 or 5

3/4 chance of survival --- 75 %

The problems that you post are really good ones.....Keep up the good work!

Eddie The Head - 7 years ago

I agree with you

Archiet Dev - 7 years ago

I think the same.

Bernardo Sulzbach - 6 years, 11 months ago

Why have you not added the 6th chamber in the last line...

Archiet Dev - 7 years ago

I also want to know that! Please help @satyen nabar .

Swetank Raj - 6 years ago

Dear Archiet it cant be chamber 6 because the previous chamber to be fired would then be 5. That cannot be because the first shot was a blank. Same reason why chamber 1 is eliminated too ( since previous chamber to be fired wd be 6)

Satyen Nabar - 7 years ago

Tx Eddie :) feel the same about your problems too.

Satyen Nabar - 7 years ago

I agree with you, too.

Heder Oliveira Dias - 7 years ago

Russian Roulette -- II

The answer is 75.

1 solution

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