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Probability Riddle Loaded Revolver

Probability Riddle Loaded Revolver - 13 june

Henry has been caught stealing cattle, and is brought into town for justice. The judge is his ex-wife Gretchen, who wants to show him some sympathy, but the law clearly calls for two shots to be taken at Henry from close range. To make things a little better for Henry, Gretchen tells him she will place two bullets into a six-chambered revolver in successive order. She will spin the chamber, close it, and take one shot. If Henry is still alive, she will then either take another shot, or spin the chamber again before shooting.

Henry is a bit incredulous that his own ex-wife would carry out the punishment, and a bit sad that she was always such a rule follower. He steels himself as Gretchen loads the chambers, spins the revolver, and pulls the trigger. Whew! It was blank. Then Gretchen asks, 'Do you want me to pull the trigger again, or should I spin the chamber a second time before pulling the trigger?'

What should Henry choose?


For Solution: Click Here

28 comments:

  1. spin the chamber : 1/6 probability

    second time pulling the trigger : 6/15 probability

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  2. Since,she is a rule follower,She'll shoot! She'll have to abide by law. N Henry,should ask her to shoot.

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  3. Shoot again, without spinning.

    Since there's two bullets in successive positions in the chamber, spinning the chamber and shooting has a 2/6 = 1/3 chance of being a bullet. There are four positions without a bullet, only one of which has a bullet immediately following. Thus shooting again immediately after an empty shot has a `1/4 chance of being a bullet.

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    Replies
    1. I disagree, I think Rob below has got the right idea.

      Since there are two bullets in the chamber there is a 2/6 = 1/3 chance of being shot or if you take the compliment then there is a 4/6 = 2/3 of surving on the first shot.

      However, out of those 4 safe chamber positions, only 3 of them will be empty the second time the gun is fired due to the fact that 1 of the 4 "safe" chamber positions is guaranteed to be followed by a bullet on the second shot.

      Thus, given that the first shot was a blank, there is only a 3/6 = 1/2 probability that you will survive the second shot. Where as if you were to spin the chamber again you would have a 4/6 = 2/3 chance of surviving. (1 - 2/3 = 1/3 probability of being shot)

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    2. Tim is absolutely correct. 1/4 chance of the bullet in the chamber without spinning...1/3 chance with a spin. So DON'T spin.

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    3. she has a double barrel shotgun you do the math

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  4. shoot again without spinning

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  5. spin cus the bullet would weigh the chamber down and would be at the bottom, thus it would definitely be a blank

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  6. Replies
    1. yeah. he should say give me a minute then say close your eyes, then RRRRRRRUUUUUUUUUNNNNNNNNNNNNN

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    2. OMG something i would do LOL !!

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  7. If theyre in successive positions, and theres a total of six possible shots, and shot one was a blank, there is a 2/5 chance of a bullet. If you spin, you could end up with the same empty chamber u just shot out of, hence restoring the possibility to 2/6...so spin >.< am i right here?

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  8. The solution is INCORRECT. spinning is the safest way to go. There may be 4 empty spots however the one before the bullet could have been the first shot which is deadly if the chamber isn't spun again. Only 3 of the empty spots are "safe" which are the 3 after the bullets. The chance of getting these "safe" slots on the first spin is .5 just the same as a deadly spot if not spun....where as landing on an empty slot after spinning again is .66....I may be wrong but.I like arguing! LolLol

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    1. Simpler : there are 3 spots that will kill u if u don't spin so its 50/50. Since we know he survived first shot, spin! You have 2/3 of surviving!

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    2. No, the answer is correct. You aren't working out the possibility of the first spin, as we know that there was no bulletin the chamber the first time. This means we only cacalculate the probability of the second shot.

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  9. The solution is INCORRECT. spinning is the safest way to go. There may be 4 empty spots however the one before the bullet could have been the first shot which is deadly if the chamber isn't spun again. Only 3 of the empty spots are "safe" which are the 3 after the bullets. The chance of getting these "safe" slots on the first spin is .5 just the same as a deadly spot if not spun....where as landing on an empty slot after spinning again is .66....I may be wrong but.I like arguing! LolLol

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  10. The solution is INCORRECT. spinning is the safest way to go. There may be 4 empty spots however the one before the bullet could have been the first shot which is deadly if the chamber isn't spun again. Only 3 of the empty spots are "safe" which are the 3 after the bullets. The chance of getting these "safe" slots on the first spin is .5 just the same as a deadly spot if not spun....where as landing on an empty slot after spinning again is .66....I may be wrong but.I like arguing! LolLol

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  11. If she spins 1/3 chance that he will die if she knows how to aim if she doens's spin there will be a 1/4 chance that he will die again if she can aim so don't spin :D:P:0:B

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  12. Here's the problem with claiming there is a 50/50 chance of there being a bullet on the second fire. You are calculating based on whether the gun was spun freshly again. Look at it this way, The first spin is in fact 1/3 death. We all agree here. Should he spin again he is faced with the same odds. So up to this point no one has argued differently. Were we are failing to see eye to eye is how to fix the odds of the sequential shot.

    1st line of thought.
    Out of ALL the chambers only 3 of them will NOT be followed by a bullet. Therefore, your chances of spinning to one of those chambers on the first spin is 50/50. Spin again because 1/3 death is better odds for you.

    Problem with 1st line of thought:
    You are basing your calculations on ALL chambers on first spin. We do not care about how many bullet spaces there are. We have ALREADY spun! There could have been 200 bullets. Although it affects our first spin, it does not affect the position that the gun is now in. Therefore we can ignore the 2 bullets.

    2nd (and correct) line of thought:
    We know that we have just fired 1 of 4 empty chambers. That means that of the four empty chambers, only 1 will be followed by a bullet and thus the odds are now 1/4 of being killed. 1/4 death is much better than 1/3 death.

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  13. odds of being killed first 2/6
    second shot equals odds of 2/5
    ?

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  14. It's too simple,

    Pull the trigger without spinning, (chances are 1/4 for a bullet, 3/4 for an empty slot)

    Will explain the answer if needed

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  15. Here is my thoughts....

    1 1 l 0 0 0 l 0

    1=Bullet
    0= Blank

    % chance on spin for blanks is 2/3

    % chance that after first shot you are in the "safe zone" to re-fire is 1/2

    Because the safe zone is the 3 slots after the bullets. The unsafe zone for 2 consecutive shot are the two bullets and the balnk infront of the bullets. Leaves 3 safe and 3 unsafe chambers.

    50% change of death for 1 spin and 2 consecutive shots
    66% change of survival if you spin after the first miss.

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  16. First time has two chances of bullet...if not spun again second time only has one chance; i.e he won't die with the second bullet for sure; the fact the bullets are successive only helps when the gun is not respun

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  17. Shoot without spinning. Explanation:

    The probability for survival when spinning again is simply 4/6 = 2/3.

    So let's consider the case without spinning. there are six possible realizations (i_1, ..., i_6) for two shoots, which are uniformly distributed (first position is w.l.o.g. at front):

    #1: (1,1,0,0,0,0)
    #2: (0,1,1,0,0,0)
    #3: (0,0,1,1,0,0)
    #4: (0,0,0,1,1,0)
    #5: (0,0,0,0,1,1)
    #6: (1,0,0,0,0,1)

    We want to calculate p(i_2 = 0 | i_1 = 0), which can be done simply by counting. The condition "i_1 = 0" rules out configurations #1 and #6. Of the remaining four configurations, only #2 kills (i.e. has "i_2 = 1"), whereas for #3,#4 and #5 (which have "i_2 = 0") the prisoner survives .

    So the survival probability without spinning anew is 3/4 -- which is larger than the survival probability of 2/3 with spinning.

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  18. Same guy as before, just to confirm my previous post with a simpler explanantion.

    Consider w.l.o.g. the configuration

    (0,0,0,0,1,1)

    Spinning sets you in a random position in this configuration. Shooting advances the position by one. If the first shoot was unloaded, the possible positions after the first shoot are:

    (0,0,0,0,1,1)
    - x x x x -

    Those posisitions marked with 'x' are possible, whereas those with '-' are not possible as they would have had a loaded first shot.

    Of those positions marked with 'x', one kills (the '1') and three lead to survival (the '0's).

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