### Brain Teasers

# Loaded Revolver

Probability
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.

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?

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?

### Answer

Henry should have Gretchen pull the trigger again without spinning.We know that the first chamber Gretchen fired was one of the four empty chambers. Since the bullets were placed in consecutive order, one of the empty chambers is followed by a bullet, and the other three empty chambers are followed by another empty chamber. So if Henry has Gretchen pull the trigger again, the probability that a bullet will be fired is 1/4.

If Gretchen spins the chamber again, the probability that she shoots Henry would be 2/6, or 1/3, since there are two possible bullets that would be in firing position out of the six possible chambers that would be in position.

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## Comments

i figured it out,but i think the other commentors simply show their stupidity.

Why don't you use logic to answer if you don't agree?

I'm not sure this problem is so cut and dry. Couldn't you figure the probability from another angle - what's the likelihood that you would spin the chamber such that it would land in a position where two successive shots are blanks. That'd be 3/6, compared to the more favorable odds of 2/6 of being shot with a respin of the chamber for the second shot. Did I do something wrong in my figuring?

GT -- It is true that the probability a priori of landing on a spot that has two successive blanks is 3/6. But there is also a chance that you land on a bullet (2/6 of the time) or that you landed on a place that is a blank and then a bullet (1/6). So when the first pull of the trigger is not a bullet, you are in either the 1/6 case or the 3/6 case. Since you are three times as likely to be in the case where you do no have a bullet in the next chamber, the probability you (actually, Henry) will be shot is 1/4. The probability of getting shot on a respin is indeed 2/6, but I would prefer a 1/4 chance to a 1/3. (So would Henry. )

i think this is just fine, it doesnt matter if you are a guy or girl, interesting thinking to come up with this though! 2 thunbs up!!!

You say that the bullets are put in succession in the chamber, and if you're using a revolver this isn't always %100 the case the bullets can be scattered thruought the chamber in a Russian roulette situation such as this. The only time bullets have to come in succession is when they're in a clip.

It says specifically in the problem that the bullets were put in succession though.

that is a good one! I was actually boggled but not to much. I was absolutely not gay!!! that was actually mean to say that!

I found this one to be very fun and interesting! I like teasers with a sense of urgency and meaning. I'm tired of figuring out what object has rubber feet. I'm going to check out your others.

Got it. "Guessed right"

AHHH!!!! it makes my brain hurt!

this was awesome . . . kinda easy, (only 50% chance of getting it wrong) but still, I liked it alot

I liked it.

ok this might seem far fetched or whatever but if it's a good gun it 'll probably be best if you take another spin

it's rather unlikely (if the chamber rolls smoothly) that it' ll stop spinning when the 2 bullets are on top

(due to gravity for the slower minds among us )

it's rather unlikely (if the chamber rolls smoothly) that it' ll stop spinning when the 2 bullets are on top

(due to gravity for the slower minds among us )

you guys have big comments for these teasers. now here is the love

Oct 17, 2005

very interesting....i think i'm jes confusing myself, but another way to look at it (lol...the way i did )could be for him to choose to spin again....why?...if she shot again, instead of having a 2/6 (33%) chance of getting shot, he now has a 2/5 chance of it being a bullet (40%)....if he chose to simply spin again, he's back to 2/6....this is all on ur individual perception of the relevance of the secession of the two bullets

i liked the teaser. but i'll have to agree w/angels. gravity. i shoot 357 revolver. matthematically the probability of getting shot is correct. but. bullets do hsve weight to them. especially 357 revolvers. angel has the right idea. if 2 bul loaded side by side, will pull a well greased chamber away from the hammer. therefore upping your odds. besides she didn't want the responsibility of shooying him. so she gave him the choice. lol.

The mathematical probability makes little sense. Especially if the act of placing the bullets in the chamber is consecutive, the odds of the convict being shot is lessen by spinning the cartridge a second time. But I am not an expert shot, so I am not absolutely certain. Yeah, Right.

At first I disagreed with the chances when you don't spin the chamber a second time, but when I worked it out in my own way with a few different routes of logic, It makes perfect sense to me.

wow a little on th hard side for me

For those who are trying to introduce a red herring by considering the weight of the bullets, you don't know how the gun is being held when it is spun. Maybe it's upside down. It is irrelevant!

Good puzzle btw!

Good puzzle btw!

Very good teaser! I didn't get it but enjoyed reading it.

My only question is - did he get shot the second time around? I'm concerned for the poor man's safety!

My only question is - did he get shot the second time around? I'm concerned for the poor man's safety!

Excellent Teaser Tsimkin. Well written and well thought out.

As for 'Andyman's' puerile and pathetic comments at the start, one only needs to review his contribution to the submission of teasers, as well as his own comments on his home page. Those comments suggest a rather screwed up person.

As for 'Andyman's' puerile and pathetic comments at the start, one only needs to review his contribution to the submission of teasers, as well as his own comments on his home page. Those comments suggest a rather screwed up person.

The chances are the same either way, 0% chance of death! Bullets can't leave the barrel without a propellant. If the judge inserted two cartridges, then the answer is correct.

I'm just giving a technical answer because I got it wrong!

I'm just giving a technical answer because I got it wrong!

The odds are the same if she spins again because there is still the chance of getting the same chamber. Therefore, the odds didn't change and he has the same chance if she spins again - 4/6 or 2/3 chance of not getting a bullet

I disagree with the probability of 1/4 of the survival. I guess you have the chance of 1/6 that Henry will survive.

My analysis:

Overall Henry had 50% of surival probability, because there were 3 "safe" chambers, and 3 "unsafe" ones. You have 4/6 chance that you'll survive the first round. And Henry had survived, so the bullet is in one of the 4 chambers. But that does not mean that he has 1/4 chance of getting a talk with the god. You missed the fact that getting a blank doesn't change the chances if we don't respin the barrel. And in fact there is only one dangerous chamber of the 6. So the probability of surviving in the current situation is 1/6.

By the way, the probability of dying in the first round is 2/6, in the second round is 1/6, and 2/6+1/6=3/6=1/2=50%, so this way it gives the correct probabilities of the survival in the whole action.

My analysis:

Overall Henry had 50% of surival probability, because there were 3 "safe" chambers, and 3 "unsafe" ones. You have 4/6 chance that you'll survive the first round. And Henry had survived, so the bullet is in one of the 4 chambers. But that does not mean that he has 1/4 chance of getting a talk with the god. You missed the fact that getting a blank doesn't change the chances if we don't respin the barrel. And in fact there is only one dangerous chamber of the 6. So the probability of surviving in the current situation is 1/6.

By the way, the probability of dying in the first round is 2/6, in the second round is 1/6, and 2/6+1/6=3/6=1/2=50%, so this way it gives the correct probabilities of the survival in the whole action.

Sorry, made som typos in the previous comment: the probability of survival is 5/6 and the probability of dying is 1/6. The rest of comment should be correct.

rblaster -- you said, "You missed the fact that getting a blank doesn't change the chances if we don't respin the barrel." This is incorrect. It absolutely changes the probability. This would be like saying, what is the probability that the next letter in the alphabet is "F" given that the current letter is a vowel? Initially, the probability that a randomly selected letter would be "F" is 1/26. But knowing that the alphabet has an order, this probability increases dramatically to 1/5. In the same way, the bullets in the chamber have an order (4 blank, then 2 loaded), so knowing that one was blank, we know we are in the section of 4 blanks. Only one of these has a bullet behind it, so the probability that the next pull will result in a bullet coming out is 1/4.

Yes, you are right. I got it. Also thanks for the example.

I've mixed up some stuff. In fact Henry has the chance 1/6 of dying on the second round only if we take every possible outcome. But now dying on the first shot is ruled out, so only 4 slots are remaining with equal probabilities of which only one is fatal.

I've mixed up some stuff. In fact Henry has the chance 1/6 of dying on the second round only if we take every possible outcome. But now dying on the first shot is ruled out, so only 4 slots are remaining with equal probabilities of which only one is fatal.

Fun to solve; I got it by drawing the symbol for a revolver, then indicating X for the loaded chambers and figuring where it would-could probably have been for the first shot.

Fun too

Fun too

Oct 15, 2008

Great puzzle. But being the geek that I am, after thinking about my chances of surviving when the revolver is loaded with 2 rounds at random, not knowing if the rounds are successive or not, Id rather spin than shoot again. The odds of the rounds by random being successive woudl be 1/15. (1/6 x 2/5 =2/30 =1/15) The odds of the two rounds not being successive would be 1/10 (1/6 x 3/5= 3/30 =1/10). In the situations where the rounds are not successive, teh odds of the 2nd shot being fatal would be 2/4 or 1/2 as opoosed to the safer 1/4 with the successive rounds. Therefore, since the odds are more in favor of the 2 rounds NOT being subsequent, I'd rather spin then pull the trigger again. Just my 2 cents.

You say: "The odds of the rounds by random being successive woudl be 1/15. (1/6 x 2/5 =2/30 =1/15) The odds of the two rounds not being successive would be 1/10 (1/6 x 3/5= 3/30 =1/10)."

You shouldn't have the 1/6's in there (the sum has to be 1, not 1/6, as it is in your calculation). That makes p(consecutive) = 2/5, and p(not consecutive) = 3/5.

With the rest of the math, I am pretty much with you, but for the sake of completeness...

If you don't spin again, p(die) = (2/5)*(1/4) + (3/5)*(1/2) = 4/10; If you do spin again, p(die) = (2/5)*(1/3) + (3/5)*(1/3) = 1/3. Since the probability of dying is lower when you spin again, I agree, you should do so.

You shouldn't have the 1/6's in there (the sum has to be 1, not 1/6, as it is in your calculation). That makes p(consecutive) = 2/5, and p(not consecutive) = 3/5.

With the rest of the math, I am pretty much with you, but for the sake of completeness...

If you don't spin again, p(die) = (2/5)*(1/4) + (3/5)*(1/2) = 4/10; If you do spin again, p(die) = (2/5)*(1/3) + (3/5)*(1/3) = 1/3. Since the probability of dying is lower when you spin again, I agree, you should do so.

Oct 17, 2008

You are absolutley correct.

Looing back at my original calculations, I should of used 1 or 6/6 as opposed to 1/6.

I understand completely where you get the 2/5 for consecutive and 3/5 for non-consecutuive.

Using 1 or 6/6,, because the first round can be placed anywhere and satisfy the condition,

consecutive condition:

1st round loaded (6/6) x 2nd round loaded (2/5) = 12/30 = 2/5

Non-consecutive condition:

1st round loaded (6/6) x 2nd round loaded (3/5) = 18/30 = 3/5

I never would of thought to include both sets of conditions when figuring probability of the fatal bullet on the second shot.

Very clever.

Thanks for breaking that down.

Looing back at my original calculations, I should of used 1 or 6/6 as opposed to 1/6.

I understand completely where you get the 2/5 for consecutive and 3/5 for non-consecutuive.

Using 1 or 6/6,, because the first round can be placed anywhere and satisfy the condition,

consecutive condition:

1st round loaded (6/6) x 2nd round loaded (2/5) = 12/30 = 2/5

Non-consecutive condition:

1st round loaded (6/6) x 2nd round loaded (3/5) = 18/30 = 3/5

I never would of thought to include both sets of conditions when figuring probability of the fatal bullet on the second shot.

Very clever.

Thanks for breaking that down.

I thought of illustrating all the possibilities in a sheet of paper given that the revolver was not spun, consequently, I got the odds of the bullet leaving the gun as 1/4, doesn't really matter whether the revolver would rotate clockwise or anti-clockwise.

I made a quick gif to explain my reasoning: http://i.imgur.com/M6wTtFq.gifv

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