The answer you gave to the puzzler about the card cheat who was being executed by the sheriff was wrong. The card cheat could see the bullets in the chamber–or not–and would know whether the next chamber held a round, so when the sheriff gave him a choice whether to just fire the next round or spin the cylinder, the card cheat already knew whether there would be a bullet in that chamber because he saw this. In other words, his odds in this situation were 100%, not a percentage.
I know this because I saved my life when an intruder broke into my home and held a gun on me. When I saw he didn’t have a bullet in the next chamber, I disarmed him.
A revolver, Western-style or modern, single action or double action, depends on revolving the cylinder to position the next round into a firing position. The card cheat was just lucky on the first bullet, but he could see whether there was a bullet in the next position in the cylinder and knew it was empty, so when the sheriff gave him a choice whether to just fire again or to spin the cylinder, the card cheat knew there was no bullet in the next chamber. His choice had nothing to do with ‘playing the odds.’
Several people (including me) have posted on this, but yours is the most penetrating and clear.
Jump down peoples’s throats? Not me!
Note that TomRay specified that the rounds were in adjacent chambers in the solution, not in the original puzzler, and THAT was what the guy saw. Can you say “Never handled a gun in their lives”? What they say about NPR being excessively liberal is true.
Whether they were in adjacent chambers or not is irrelevant. You are right, the solution given is wrong.
russ’s answer was true if incomplete. Looking down the muzzle of a loaded revolver(DON’T TRY THIS AT HOME OR ANYWHERE ELSE), you can see 4 of the 6 chambers, including the next chamber in rotation. The card cheat will know the next chamber is loaded, or not. If it is loaded he will request a spin giving himself a 67% probability of survival.
I thought of the same thing.
Would it be possible to see the bullets in the chamber at 20 paces?
But never mind that question. My point is that the answer as given is wrong on its own terms, even if you assume that the chamber can’t be seen:
there’s a problem with the logic and calculation of probability.
Assuming two consecutive shots, bullets in adjacent chambers 5 and 6. Look at the follwing outcome chart:
Position chamber odds outcome
1 empty 1/6 Life
2 empty 1/6 Life
3 empty 1/6 Life
4 empty 1/6 death
5 Bullet 1/6 death
6 Bullet 1/6 death
He?s got to hope that the marshall spins to positions 1, 2 or 3. Positions 4, 5, or 6 result in death.
You can?t assume that ?If the first chamber is empty, there?s a 1/4 chance that it was any one of positions 1, 2, 3, or 4.? In fact, there was a 1/6 chance that it was either 1, 2, 3, or 4. All that the first shot has done is reveal that it wasn?t on position 5 or 6. The revelation does not actually change the original probability that its landed on any other specific position. He?s got to hope that it was position 1, 2, or 3, each of which there was only ever a 1/6 chance of landing on. Therefore, he only has a 1/2 chance total. Better to have the marshall spin again for the 4/6 chance.
Remember the Monty Hall problem:
Sethro:
Since apparently I am the only one so far who has done any “field testing” for this scenario and literally bet my life on the outcome, I took my .44 caliber 1847 Walker Colt out this afternoon, loaded one chamber, took the appropriate precautions of course, and looked down the cylinder from 20 paces (actually, since a “pace” is an inexact measurement, I did this from 40, 50, and 60 feet to accomodate different strides).
The puzzler said the execution was to take place at high noon, so lighting was not an issue. One can assume that it was clear and bright on execution day, with sunlight overhead, and, not so coincidentally, casting a shadow from overhead.
FYI, for the non-shooters out there, a .44 or a .45, two of the more popular caliber revolvers from this era, is not quite 1/2 inch in diameter. Another popular caliber was the .32, which is approximately 1/3 inch. Since I don’t own a .32 caliber revolver, I didn’t try out all possibilities, but it would admittedly be more difficult to see into the cylinder of this size revolver. However, an 1880’s .32 would be a rather undersized and inaccurate weapon for this purpose at this distance, even with a direct hit.
I can report that from 20 paces, however you measure it, you can clearly see down the cylinder of a .44 caliber revolver. A loaded chamber shows the light-colored lead bullet, which sticks out farther and catches the sunlight; an empty chamber appears black in the shadow of high noon. The difference is readily apparent.
Admittedly, it does get more difficult to see clearly with increased distance, but it is possible to see a 1/2 inch opening which has a definite color discrepancy at 20 paces–especially if one’s senses are heightened by the thought of imminent death. Trust me, the barrel of a gun looks as big as a bowling alley when you are looking down it.
I still contend the card cheat’s chances with the second shot were 100% because he KNEW the next chamber was empty. No need to do hasty mathmatical permutations on the elements of chance when facing almost certain death. I would take what I saw from direct observation, like the puzzler stated, than to calculate the odds on a second shot.
The fallacy here that too many people are pursuing is that they are over-thinking the solution.
Sethro,
You are quite simply wrong. There is no way that three chambers with two bullets can all result in firing off a live round, thus "death". Think about it.
"Positions 4, 5, or 6 result in death."please explain. Two shots are fired, without altering the position of the cylinder after the first shot. Position 4, which is empty, fires blank on the first shot, and then advances to position 5, which fires live on the second shot. Positions 5 and 6 result in death on the first shot. So who’s wrong?
I have an actual issue when the solution in the intended nature of the puzzler. Tom and Ray explaining the solution say with any spin of the cylinder there is 2 bullets and 4 empty chambers meaning a 4 in 6 or 2 in 3 chance of the man walking away. Then they explain how 2 adjacent bullets would mean a 3 in 4 chance of him walking away. See the problem? They are assuming 2 bullets in the chamber when the sheriff RE-spins. But, if he re-spins after the first shot there is only ONE bullet left, meaning a 5 in 6 chance of walking away which is better than 3 in 4. No matter how the bullet are placed in the chamber, the re-spin improves his chances, even if incrementally.
“please explain. Two shots are fired, without altering the position of the cylinder after the first shot.”
Right there, you are wrong. No shots were fired in the puzzler. Pulling the trigger for the putative second shot would necessarily alter the position of the cylinder, because that’s how revolvers work. So you are wrong.
“Position 4, which is empty, fires blank on the first shot, and then advances to position 5, which fires live on the second shot. Positions 5 and 6 result in death on the first shot. So who’s wrong?”
You are wrong. Vide supra.
Try to follow along. THERE WAS NO FIRST SHOT. There was a first pull of the trigger, which resulted in no shot. There are still two live rounds in the revolver.
HaHa, good point. I guess I went into tunnel vision when I was trying to solve the puzzler and forgot the first shot was an empty chamber and not a “miss” (because he never misses) or a “blank” or whatever.
First, this puzzler is meant to be a logic or probability question. Whether a person could actually see the bullets at 20 paces, or what the differences are between 19th century revolvers and modern firearms, is irrelevant.
I take issue with Sethro’s assertion that the original probability doesn’t change after the sheriff pulls the trigger the first time.
I think Sethro’s argument would make sense if the sheriff had said, “I’m going to pull the trigger twice. I’ll give you a choice: I can pull it twice in a row, or I can pull the trigger, spin the chamber, and pull it again. Which would you prefer?” Then I’d go with the latter.
But that’s not what happened here. The sheriff pulled the trigger, and the gun didn’t fire. Then he gave the guy a choice what to do next.
Once he pulls the trigger and nothing happens, though, the probabilities have changed. Now you know for a fact that, on the first pull, the hammer landed on one of the four empty chambers. If you know the bullets are in adjacent chambers, there’s now a 3/4 chance that, if the sheriff does not spin the cylinder, the next chamber will also be empty.
No, if he doesn't spin, the chances are 3/5 that the hammer will land on an empty chamber.
If he does spin, the chances of landing on an empty chamber are 2/3.
Try to visualize this with 6 pennies and 2 dimes.
Also, no, it’s not meant to be a probability problem. It is however a logic problem: what could he possibly have seen?
True, the question is, what did he see, but my point was simply that I think you could assume that the answer did not require some particular knowledge of 19th century firearm design.
Here’s how I see it when the sheriff does not spin the cylinder between trigger pulls (“x” represents an empty chamber, “o” is a bullet):
1st…2nd
pull…pull
x…x
x…x
x…x
x…o
o…there is no second pull - the guy is dead
o…ditto
If that’s wrong, please explain what I am missing.
It would be 3/5 in the following scenario: 4 white marbles, 2 black ones. Put them in a box and take one out at random. It’s white. Don’t put it back. Reach in and take out another marble. Now it’s 3/5 that it will be white.
But that’s not an apt analogy. If the sheriff doesn’t spin the cylinder, and the chamber moves clockwise (from the sheriff’s point of view), the hammer will fall on the chamber immediately to the left of the first one. Depending on which empty chamber it landed on the first time, there’s a 3/4 chance that the next one will be empty as well. There are 4 empty chambers, and 3 of them will be followed by another empty chamber. The fourth will be followed by a loaded chamber.
“True, the question is, what did he see, but my point was simply that I think you could assume that the answer did not require some particular knowledge of 19th century firearm design.”
You are absolutely right. In the areas that matter to this Puzzler, there is no difference between 19th century and 20th-21st century firearms design. It’s a revolver, specifically a “six-shooter” (a lot of moderne revolvers hold only five or up to ten rounds). That’s all you need to know. The fact that it’s a revolver and the Puzzler hinges on something the guy saw tells you all you need to know.
Because we know the first trigger pull fell on an empty chamber, there are five remaining chambers, which include three empty chambers, hence 3/5 chance of safety. If he doesn’t spin the cylinder, it’s as if the executioner swapped for a gun which only held five rounds. Spinning the cylinder would result in a 4/6 chance of safety. If the card cheat hadn’t seen something, which I remind you yet again is the point of the Puzzler, spinning the cylinder gives a better chance of survival.
Question from a non gun user. Does a revolver turn the cylinder as the trigger is pulled or after the chamber is fired and trigger is released? If the former, then russ seems right if the card cheat has really good vision. Otherwise, Ray is right.
It turns it as the trigger is pulled.
Good question, though.
Correction: It’s the hammer going back that rotates the cylinder on a single-action revolver. Has nothing to do with the trigger. Of course the type was not specified in the Puzzler. End result is the same. It’s entirely possible that TomRay, being from their fair city, and being on NPR, don’t know the mechanical details.