The Crazy Gangster Puzzle || Russian Roulette Puzzle
ฝัง
- เผยแพร่เมื่อ 26 เม.ย. 2019
- You are kidnapped by a crazy gangster who has a 6-shot revolver.
Initially all six chambers of the cylinder are empty.
He puts two bullets into adjacent chambers (in front of you)...spins it and closes the cylinder...
He now puts the gun to your head & pulls the trigger...
Luckily you survive....
He is now going to pull the trigger again... but before he proceeds he asks you for your preference whether:
1 - He should re-spin the cylinder and then pull the trigger... or
2 - He should pull the trigger right away (without spinning)...
(If you survive this shot as well... then you will be allowed to go home.)
So, which option would you prefer to maximize your chances of survival...
This puzzle is derived from a crazy lethal game called Russian Roulette in which a player places a single bullet in a revolver, spins the cylinder, places the muzzle against their head, and pulls the trigger.
You are most welcome to share puzzle, math problems or any topics for upcoming videos.
Gmail : logicreloaded@gmail.com
Facebook(message) : / mohammmedammar
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That gangster was Gabbar. 😂
That's What inspector Roy Did in Dhamaal😂😂
Mout ka khel😂
😂😂😂
2:28 , make sense given that the bullets were adjacent (next to each other), if bullets were not adjacent, then spin again would be best chance of surviving.
Ya because in case 1 you can still get the same blank slot but in case 2 you have one less blank chamber, which you can get..
Do we know that it spins anticlock wise or clockwise??
@@manusarda You mean if the bullets were randomly distributed? Adjacent bullets: new spin 2/3; no spin 3/4, as the puzzle solution says here. Random position: New spin 2/3 (66,67%), no spin 3/5 (60%). That's what Chiro Ghosh said correctly above.
Mathematical answer: Fire right away. The first shot was one of four empty chambers. Of those four chambers, three have a subsequent chamber that's empty. The fourth's is loaded. Therefore, there's a 75% chance of not getting a bullet on the next shot if we take the next chamber. However, spinning the cylinder resets the odds, returning us to a 66% chance of survival.
Physical answer: Depends on how he spins the cylinder, whether he interferes with the spin or doesn't set the cylinder until it stops spinning by itself, the weight of the bullets, and the design of the revolver. By loading only two chambers and not setting the bullets in opposite chambers, the cylinder's center of gravity is no longer aligned with the cylinder's axis of rotation. If he points the revolver downwards while the cylinder spins, this offset causes little change. If he holds the revolver upright so the barrel is parallel to the ground, the offset will cause a higher chance of the bullets settling in the lower chambers. The specific probability of this is dictated by the weight of the bullets used; heavier bullets mean higher probability the bullets will not reach the top chambers. Furthermore, the solution is also dependent on whether the revolver uses the bullet loaded in the highest chamber or the lowest chamber. However, all of this is negated if he interferes with the cylinder. Without these specifics, there's not enough information to discern which is more logical.
Mycroft616 you won
No one will read this long para, honestly
You are one Intelligent fella
@@bhushanbarde5041 i read it
@@bhushanbarde5041 i read it.
Solution was pretty simple. Even when I thought of it, I actually stayed paused for a while trying to think if I was missing something because it seemed too easy.
Well a very scary choice of decision to be in, though the puzzle is superb
I can't imagine doing math in such a situation
Shoot with no spin
Gangster: OK!
*empty*
Am i free to go home
*SHOOTS*
A creative puzzle
Thank you
Nice riddel. Nice graphic explenation, as always.
But!! Ok. I choose the first option of the 75% to survive.
The gangster pull the trigger, BOOM!! Im dead...
Math didn't help me this time. Ha?
Ty Ammar for nice video, I`m in heaven now. Everybody here are waiting to the next video (-:
Knowledge of probability concepts improves your luck but doesn't guarantee success.
@@manusarda
Statistics and probability were never, in my eyes, pure mathematics.
It is true that mathematical formulas are used, it is true that there are calculations and equations, but this is not really mathematics.
This may be the philosophical side of mathematics.
no more.
And I'm been a math teacher for 30 years.
Statistics and probability are not mathematics. "Increasing the chance" of any event, is a meaningless expression.
The math dealing with expression like "i can increase the distance"
Or: " i can increase the weight",
Or: "i can increase the area... or volume... or frequency... or length"... and so on.
But the expression: "I can increase the chance" is meaningless for math.
What is "chance" anyway?
In mathematics there is existence and meaning to expression like 5 + 7 and there is existence and meaning to expression like 10X + Y
But there is no meaning to a phrase like "a lot"
times "almost 7" equals "good chance to 95" ...
@@tamirerez2547 now you are telling something completely different.
In 7th to 11th class we were taught probability in mathematics subject and now you are saying that it doesn't belong. 😂
@@manusarda
It is belong. But not as a PURE MATH.
I was a math teacher, and never really saw statistica as math.
Just drop a dice.
The chance to get "4" is one to six, yes?
But we all know you can drop the dice 10 times and get: 1,3,3,5,6,6,1,2,3,5....
Not one "4".
What is going on?
Well, only in case you throw the dice six million times, (the rool of big numbers, remember?) Only then you will probably get "4" say, 985,457 times.
It is ALMOST 1/6 of 6,000,000.
Is this math? Pure math??
1/6 is very clear fraction! Very accurate expression!
When math say 1/6 it mean 1/6.
So simple.
Now try to answer these questions:
A. 1/6 of 120 = ?
B. 1/6 of 600 =?
C. 1/6 of 5400 =?
You probably answerd
A. 20
B. 100
C. 900
So why the question: how much is 1/6 of 6,000,000 is 985,457??
Because statistica is not pure math.thats why.
I hope you get my point.
1/6 is always 1/6.
Not almost 1/6
Not near 1/6
Not maybe 1/6
If statistica was pure math, the result of throwing a dice was look like
5,2,4,6,1,3
And you know it is not.
That is my point.
@@tamirerez2547 I get that. You are nice teacher.
I am editing my original reply
Possibility problem based puzzle
I like it
On spinning loaded chambers will settle at the bottom due to its weight while firing pin will hit the top one!
Who thinks that they are just waiting for the answer
Rather than solving it
The answer to the question would change drastically even if the question is worded slightly differently.
It was easy one... I am waiting for next version its going to be more difficult I guess
You can't think when you actually at the point of gun !😂😂
😆😆😆
😂😂😂😂
You start doing maths... asking for a pen and paper to the gangster
😂😂😂
Puzzles on this channel are awesome 😂
Took me a minute to survive 😁
But it will be interesting if you consider gravity.
bcoz everytime u spin the revolver the bullets will be at the bottom...
Yes definitely, first i thought this will be the answer..
But what if cylinder axis is vertical or gangster stop cylinder Manuely like sholey
In questions from previous comment, if safe strategy changes to spin, then we can safely say that biological phenomenon of "There is safety in numbers" gets mathematically proven.
Gr8 bro ur questions r vry tricky i would love to enjoy them😂
Great puzzle! the fact tha the bullets are in adjacent chambers is a game chamber. I got fooled.
That got me too. If the bullets were sperate, then spinning would be your best bet.
This is such an amazing variant of the Monty Hall Problem, and could be a great tool for teaching Bayes theorem ':)
Thanks Danyal.. By the way I would be posting Monty Hall Problem in a few months. Although monty hall is not very difficult to understand, but 8/10 people won't be able to agree with the solution for different reasons.
So I want to put some efforts in preparing a unique explanation.
@@LOGICALLYYOURS Great idea sir, go ahead.
next version pls, this was easy
Umm 1 feels more safer but is isnt... Well great
Woo! I got it right.
Without considering the gravity, after respin, only one bullet is now left and chance of Survyiing is 5/6 which is 83.33 % .
Hm, my theory before checking comments or watching the ending but a revolver might not really work like that, I mean how flawlessly would a revolver spin so the "heavy" side is on the bottom? But if that's the case then a new spin, if also allowed to finish the spin "naturally" would mean the two bullets would be at the bottom and recycle to one clear chamber in the worst case (as it cycles once when cocking the gun, thanks Alec Baldwin).
If it wouldn't be spun and directly fired and my spin+gravity theory is correct the gun would likely fire if it isn't spun anew, so the chances would overall be better for option 1.
Ofc that only works if the theory is correct but also might depend...hey technically either option is fine if it's a single action revolver as if he just pulls the trigger (hence without cocking the gun) nothing would happen hence it's safe in both scenarios xD (also thanks Alec Baldwin for indirectly teaching me this lesson xD )
Edit: Okay guess gravity isn't even an option, I wonder if that would be the case though, or what the parameters would be, also again in a single action revolver the survival odds are 100%, unless the gun is malfunctioning of course.
Awesome sir
In the revolver top chamber fires.
Chambers with bullets are heavier and got higher chance to be on the bottom.
Nice
You can also look at this from the point of where the chamber is, in no spin only one chamber in 6 the one before the loaded bullets results in death ie 1/6 chance in the spin scenario there is 2/6 chance
I am present sir
Wouldnt it give you higher chances if you respin the barrel since bullets are loaded besides one another resulting to the center of gravity of barel to shift and chances are bullets will stablize the rotation when bullets are at the bottom part of the barrel
I thought the same well done bro
I ALSO THOUGHT IN SUCH WAY.
In next shot... If he spins, automatically the bullet goes down due to gravity... so 1 is good I think. And it is a good question.
Am also thinking gravity
Acctually respinning is half of what you showed because you must calculate what is the chance to spin twice and survive so 0.5x66.7%
First option is 100% safe
When the gangster re spin both bullet go down because of its weight
So when he fire the gun the shot will be in B or C
or the opposite is true... the cylinder spins faster when the bullets are at the bottom and slower when they are at the top. therefore the chance it halts is bigger when it will be loaded ^^
but seriously he should point to the ground while spinning - avoids such effects
But when we re-spin , I think gravity comes into picture so bullet have more chance of staying down , so increase the chance of survival.
Damn i didnt think of that. Well done.
Nope. It does not.
Nice puzzle
Its a nice one bro
Until the gun appears to be self repeating until loaded chamber...
5 chamber left without spin so mathematically 3/5*100 =60%
Puzzles only work when you are dumbed down and not be able to think outside the box. Why the heck would you answer the question, when you could just take him by surprise as he speaks his ass?
No.2 because if he was not able to kill you despite putting his gun to the head, it means a few things.
1. He has a toy gun.
2. He can't shoot. And if he can't shoot despite LITERALLY putting the gun on your head, then just run a few distances because he will 100% miss.
Another solution is not to answer the question itself. Just beat the shit outta the guy. Take him by surprise.
Think outside the box my ass.
according to an article on this question ' missing from the interview question is the assumption candidate know that a revolver cylinder advances one position after each pull of the trigger'
which approach is better if there is just one bullet?
which approach is better if 2 bullets are not exactly next to each other?
Nice thought. Think yourself to become more intelligent.
@@rkg6081 I'm more philosophy kind of person... I have put one more comment, related to philosophy.
Jaydeep.... that's really nice thinking... let me tell you two interesting things:
1 - The origin of this puzzle is based on your point 1. That's how "Russian Roulette" game is played. In that case it's better to re-spin because the new probability of survival would be 5/6 (and without re-spin, it would be 4/5 which is little lesser).
2 - The next version I will be posting is based on random placement (as you highlighted in your point 2).. but you mentioned about "non-adjacent"... But the puzzle will become more interesting if it's completely Random placement.
I would have to factor the chance of survival if I try to overtake the gangster.
Is that you mista ??
With SIX PISTOLS !!!
, Inside elevator.
Bingo got it!!!!!
Your explanations and making videos is amazing...
Many thanks Vinay :)
Logical❤❤
Yeah Ammar
Out side the box
While spin the bullet camber due to gravitational force the bullets occupy the bottom of the area of the chambers , even after n no of time we will have got a chance to survive
Pls review my comment
After the first bullet is shot, only one bullet is remaining, so the chances of surviving in the second scenario are rather 5/6 = 83% which is the best option
Lol 😂😂 see the video again!
@@kraugel7 Tell me why I should watch it again
Oh, I also thought that way, but then I realized that first shot was without bullet. So, both are still there.
easy one tho this time
like if you guessed it right!!
sorry... but i think the chance of survival was calculated like this :
if one of the empty chamber is triggered, it leaves 60% (3 empty 2 loaded) since the first shot has been taken
and if you respin, chance are 66% (4 empty 2 loaded)
i think is simple as that...
it's right I choice re-spin options that give me 6% of more change a living
Pull the trigger right away is not the correct answer, since you can never reach the 'A' position because, there was a bullet insife it, but you managed to survive from it, so it should be 66% only.
2/3 as A is impossible
Lol, I comment it on 2021 :V
What if a gangster put bullets randomly, for instance on A and C? Your chances will dramatically reduced to only 50% (D and E are safe, B and F not), so your choice should be depend on the bullets position
1 St view
1 viewer
25% to 33% of the people who played, never found out the result.
Imagine applying so much math but still get killed
How depressing
Nice problem, two questions, though: shouldn't we take into account that statistically speaking with many shots with bullets still in chambers the chance of firing a bullet gets higher? Because statistically speaking you should fire two bullets per six shots, you cannot expect magically not to fire a single bullet for 100 shots and still give it a chance of only 1/3 on the next shot. So, even on the second shot the odds should get slightly worse with none bullet fired (as well as better with one fired, if not to your head, of course).
Interesting question: if the cylinder had to be spinned after each shot and the first shot would be lucky for you as in the video, would you try your chance with the 2nd shot, or would you ask for one shot to a blank and go with the 3rd one?
If I understood your question correctly.. then after the first lucky click(blank), I would choose direct 2nd shot.... and if in this case i survive, then for the 3rd shot the probabilities of 1) shooting right away && 2) spin then shoot will be equal i.e. 2/3... so any choice would work.. But moving onward for 4th shot, it will be better to choose a re-spin.. because in case of shooting right away the probability further reduces to 1/2.
@@LOGICALLYYOURS Thanks for your answer, though what I meant was choosing between these two options, after the first shot:
a) spin, second shot to your head, done; or
b) spin, second shot to the air, spin again, third shot to your head, done - because in this case there is a possibility to get rid of one bullet in the 2nd shot, but it also means pushing your luck further if both bullets stays in
@@littleschnitzel8226 If both bullets stay in and you respin the barrel, then you're still in the same situation, with 2/3 chances to survive
@@_Ytreza_ Guess none of you heard of statistics and probability. Stating over and over the chance is 2/3 doesn't make it so. Yes, that is indeed the chance for each shot individually, but with the string of shots you have to multiply all the probabilities. The fact you already know the result of the first shot doesn't change that. Basically what you're saying is that you have the same chance of surviving one shot as well as a million as long as you spin before each one. If you cannot see why that is a nonsense, I don't know what to say to you.
@@littleschnitzel8226 You have the same chance of surviving each shot individually.
If you survive the first 999 shots, the chance to survive the 1000th one doesn't change (but it would be very impressive to survive 999 shots in a row yes)
Its wrong, in first case he's considering the empty slots(3/4), where as in 2nd case he's considering all ,the empty one as well as the filled one so the prob. decreases.
The premise of the riddle is, we already played once and survived. So the empty ones are the only possibility as to where we started.
Either way, choosing to shoot again is technically better chance to survive (if my math is right), even if you consider the first shot could still be lethal (unlike the video).
In the scenario of 2 shots where you always choose to shoot again, you have a 50% chance of surviving. (3 shots are lethal, 3 that are not. the 3 lethal shots include the empty shot that then leads up to a killing shot)
In the scenario you always choose to spin again, this means you have to weigh the probability of a 2/6 (or 1/3) chance occurring at least once, if rolled twice. Labeling all possible outcomes (using 1/3 for simplicity, math should be the same as 2/6 in the end) is a table of [1,1][1,2][1,3][2,1][2,2][2,3][3,1][3,2][3,3]. Any chosen number (that represents the bullet) occurring from a random selection from this table is 5/9, or a 55% chance. This means you have a 45% chance of surviving, which is 5% less likely than the 50% chance if we choose for the gunner to pull the trigger again.
Think logically and give the right answer
After he spins it, the chambers with bullet would be towards the lower side due to the gravitational acceleration. Hence, I'd ask him to spin it again.
80% nope. u will not be lucky and u will see ur creator waiting to grab u
@@manusarda Chutiya hai kya beyy? Gravitational acceleration ki baat ho rahi hai. Gravity ki nahi.
Aur beta, tumhari tarah revolver waale nahi hain. 9mm rakhte hain. 😂 Chalaao apni takk takk. 😂
@@manusardaAre mera beta, gaali lagg rahi hai? Pogo dekhta hai na? 😂
Shukhar Mawla I got this one
I am not kidnapped by any gangster.
Problem solved. ;)
When a gun has single bullet then he will says re-spin the bullets
Last question answer is re-spin chamber gun is my choice
Live 88%
I think probability in the re-spin option would be 5/6 that is 83.33 % because now only one out of 6 chamber is filled with bullet
Since the first shot was a blank click... so both bullets are still in the cylinder... that's why probability after re-spin would be 4/6.
Different numbers, think out of the box.... No spin, second shot is a 2/5 chance of getting shot = 40%. Spin = 2/6 chance of getting shot = 33.3%. Spin that chamber!
How would you be able to get to the second bullet if you hadn't already had the first one?
@@Codisrocks Yep. I watched the solution... now understand the lower 25% chance without spinning.
@LOGICALLY YOURS what if bullets are not adjacent to each other?
Wouldn't pulling trigger right away result in a 3 of 5 probability?
No. Obviously one of the blank chambers has been taken out of contention, but so has one of the loaded chambers - the 2nd loaded chamber is no longer a possibility (without a spin, it is only possible when the previous trigger-pull results in a shot). So the possibilities are 3 blank and 1 loaded chambers.
It doesn't matter because the 2 rounds in the pistol have already been discharged.
If the choice was 1 and the loaded will go down due to the load and gravity then i will survive...
Not if the gangster points the gun down while spinning.
Hi just in dilemma that the ratio's for both are not the same
Why some people are talking about the CG.. that cylinder of gun is not 100% free ...then how we can apply the gravity on cylinder ..
Same as Monty hall puzzle.
Hmm, both puzzles (this one and the Monty Hall) are in the same territory of probability. However, psychologically, the Monty Hall problem is least accepted by a common man. I will certainly give my best to prepare a convincing solution for Monty Hall.
please discuss Frog riddle by Tedx
I solved it
This was easy. I did it by probability fraction calculation. The same thing of course.
I did it by using obviousness equations
Option A is better.
As in option A we have four chance to survive. But in option B we have only 3.
Bro !! If he Spins the cylinder , Then Due to weight of the bullet the chambers with bullet will be in the bottom !!
and chances are increased much more !! Think scientifically !!
are u flat earth guy?
Then point the gun down while spinning
when you spin the cylinder bcz of gravity bullet will go to the bottom. Hence the chance of surviving is 100%.
Is it,..
That is everytime the person will be safe, if he spin it👍
Thats not how a revolver works lol
It that was the case, one can just point the gun down when they spin it.
It was easy
मरवा कर ही मानेगा क्या भाई 😂😂😂😂😂😂😂😂😂
Math aside, both options are 100% safe as the "bullets" that were loaded were already fired (dimpled primers).
Your solution is not logical, because you claim the spent round was replaced on the second shot, (replaced the spent bullet with a live round).
You don’t mention this in the rules. It clearly states he pulls the trigger a second time, NOT that he replaces the round and pulls the trigger a second time. This omission changes everything.
Think before talking bullshit!. The first round was *not* spent, else you'd be dead already.
That was easy
Again I´m calling bullshit because 1 of the 4 empty chambers has already been used so we are left with 2 safe chambers, not 3 and since we dont have to care about the first chambers anymore its 2/3= 66.6666% = 66.7% in both scenarios. Get your shit done right dude!
If there are 50 children and 49 mangoes
What part of each mango should the 50th child get so that each child have same amount of mango
What's the answer
1/49
1/50 should be the answer
Bhiya app hindi m video banao pl.
..
What a depressing riddle.
Deer hunter movie anyone?
How do u make your videos?
Actually re-spinning the wheels doesn't reset your chances, meaning the probability to have an empty chamber second time is P(first_time)*P(seconds_time)=(4/6)^2 = 44%. Do never ask to spin.
Gangeser is lol