Total Posts:22|Showing Posts:1-22
Jump to topic:

Probability Question

Wallstreetatheist
Posts: 7,132
Add as Friend
Challenge to a Debate
Send a Message
9/24/2013 6:26:18 PM
Posted: 3 years ago
A Russian gangster kidnaps you. He puts two bullets in consecutive order in an empty six-round revolver, spins it, points it at your head and shoots. Click. You're still alive. He then asks you, do you want me to spin it again and fire or pull the trigger again. For each option, what is the probability that you'll be shot?
DRUG HARM: http://imgur.com...
Primal Diet. Lifting. Reading. Psychedelics. Cold-Approach Pickup. Music.
Bullish
Posts: 3,527
Add as Friend
Challenge to a Debate
Send a Message
9/24/2013 7:33:51 PM
Posted: 3 years ago
At 9/24/2013 6:26:18 PM, Wallstreetatheist wrote:
A Russian gangster kidnaps you. He puts two bullets in consecutive order in an empty six-round revolver, spins it, points it at your head and shoots. Click. You're still alive. He then asks you, do you want me to spin it again and fire or pull the trigger again. For each option, what is the probability that you'll be shot?

1/3; 1/4.

I'd ask for him to spin it again through, because he is likely to do the opposite.

And lets be nicer to Russians.
0x5f3759df
Poetaster
Posts: 587
Add as Friend
Challenge to a Debate
Send a Message
9/24/2013 7:50:44 PM
Posted: 3 years ago
If he spins it, the probability is 1/3.

Without a re-spin, getting shot on the second firing after the first was empty implies that the revolver was initially aligned with the only unoccupied slot before one of the loaded slots (which are consecutively placed). The probability of the gun being aligned with that particular slot is 1/6. Hence, the second method produces a 1/6 probability of being shot on the second firing.

If the bullet placements are not consecutive, then the two methods are equiprobable, I believe.
"The book you are looking for hasn't been written yet. What you are looking for you are going to have to find yourself, it's not going to be in a book..." -Sidewalker
Bullish
Posts: 3,527
Add as Friend
Challenge to a Debate
Send a Message
9/24/2013 8:03:11 PM
Posted: 3 years ago
At 9/24/2013 7:50:44 PM, Poetaster wrote:
If he spins it, the probability is 1/3.

Without a re-spin, getting shot on the second firing after the first was empty implies that the revolver was initially aligned with the only unoccupied slot before one of the loaded slots (which are consecutively placed). The probability of the gun being aligned with that particular slot is 1/6. Hence, the second method produces a 1/6 probability of being shot on the second firing.

If the bullet placements are not consecutive, then the two methods are equiprobable, I believe.

The problem with that is that the fact that the gun did NOT go off is already a KNOWN. So since the revolver must have landed on one of the empty slots, and only one of those can result in the no bullet-bullet configuration, it's 1/4.
0x5f3759df
Poetaster
Posts: 587
Add as Friend
Challenge to a Debate
Send a Message
9/24/2013 8:12:01 PM
Posted: 3 years ago
At 9/24/2013 8:03:11 PM, Bullish wrote:
At 9/24/2013 7:50:44 PM, Poetaster wrote:
If he spins it, the probability is 1/3.

Without a re-spin, getting shot on the second firing after the first was empty implies that the revolver was initially aligned with the only unoccupied slot before one of the loaded slots (which are consecutively placed). The probability of the gun being aligned with that particular slot is 1/6. Hence, the second method produces a 1/6 probability of being shot on the second firing.

If the bullet placements are not consecutive, then the two methods are equiprobable, I believe.

The problem with that is that the fact that the gun did NOT go off is already a KNOWN. So since the revolver must have landed on one of the empty slots, and only one of those can result in the no bullet-bullet configuration, it's 1/4.

The question asks about the likelihood of being shot on the 2nd trial after the first comes up empty, with no re-spin between the trials. There is only one chamber out of the six which satisfies that condition.
"The book you are looking for hasn't been written yet. What you are looking for you are going to have to find yourself, it's not going to be in a book..." -Sidewalker
AnDoctuir
Posts: 11,060
Add as Friend
Challenge to a Debate
Send a Message
9/24/2013 8:19:01 PM
Posted: 3 years ago
At 9/24/2013 8:12:01 PM, Poetaster wrote:
At 9/24/2013 8:03:11 PM, Bullish wrote:
At 9/24/2013 7:50:44 PM, Poetaster wrote:
If he spins it, the probability is 1/3.

Without a re-spin, getting shot on the second firing after the first was empty implies that the revolver was initially aligned with the only unoccupied slot before one of the loaded slots (which are consecutively placed). The probability of the gun being aligned with that particular slot is 1/6. Hence, the second method produces a 1/6 probability of being shot on the second firing.

If the bullet placements are not consecutive, then the two methods are equiprobable, I believe.

The problem with that is that the fact that the gun did NOT go off is already a KNOWN. So since the revolver must have landed on one of the empty slots, and only one of those can result in the no bullet-bullet configuration, it's 1/4.

The question asks about the likelihood of being shot on the 2nd trial after the first comes up empty, with no re-spin between the trials. There is only one chamber out of the six which satisfies that condition.

Yeah, but you know that one chamber wasn't either of the chambers with the bullets in them, so 1/4.
Poetaster
Posts: 587
Add as Friend
Challenge to a Debate
Send a Message
9/24/2013 8:26:04 PM
Posted: 3 years ago
At 9/24/2013 8:19:01 PM, AnDoctuir wrote:
Yeah, but you know that one chamber wasn't either of the chambers with the bullets in them, so 1/4.

It seems I forgot that there are two bullets.
"The book you are looking for hasn't been written yet. What you are looking for you are going to have to find yourself, it's not going to be in a book..." -Sidewalker
Poetaster
Posts: 587
Add as Friend
Challenge to a Debate
Send a Message
9/24/2013 8:33:38 PM
Posted: 3 years ago
At 9/24/2013 8:26:04 PM, Poetaster wrote:
At 9/24/2013 8:19:01 PM, AnDoctuir wrote:
Yeah, but you know that one chamber wasn't either of the chambers with the bullets in them, so 1/4.

It seems I forgot that there are two bullets.

Ah, I retract that near-concession; see here: the probability of the first shot being empty is 4/6. The probability of first occupying that unique chamber out of the four empty chambers with no re-spin is 1/4, but the probability of the whole sequence occurring is:

(4/6)*(1/4)= 1/6

Hence, the second method still produces a probability of 1/6 of being shot.
"The book you are looking for hasn't been written yet. What you are looking for you are going to have to find yourself, it's not going to be in a book..." -Sidewalker
AnDoctuir
Posts: 11,060
Add as Friend
Challenge to a Debate
Send a Message
9/24/2013 9:36:35 PM
Posted: 3 years ago
At 9/24/2013 8:33:38 PM, Poetaster wrote:
At 9/24/2013 8:26:04 PM, Poetaster wrote:
At 9/24/2013 8:19:01 PM, AnDoctuir wrote:
Yeah, but you know that one chamber wasn't either of the chambers with the bullets in them, so 1/4.

It seems I forgot that there are two bullets.

Ah, I retract that near-concession; see here: the probability of the first shot being empty is 4/6. The probability of first occupying that unique chamber out of the four empty chambers with no re-spin is 1/4, but the probability of the whole sequence occurring is:

(4/6)*(1/4)= 1/6

Hence, the second method still produces a probability of 1/6 of being shot.

Yeah, the 4/6 has already come to fruition, though,and so you're left with a 1/4 vs. a 1/3, no? I mean otherwise it's 1/6 vs. ((2/3)*(1/3)+1/3) blahh .55555%

You're treating this like the Monty Hall Problem maybe where it's not, I think. The odds in that problem haven't come to fruition and so are still subject to variable change or whatever that was about. In this one you know you've achieved the 4/6 and so that's out of consideration. You're just left with the 1/4 to make up your 1/6.
Poetaster
Posts: 587
Add as Friend
Challenge to a Debate
Send a Message
9/24/2013 10:04:56 PM
Posted: 3 years ago
At 9/24/2013 9:36:35 PM, AnDoctuir wrote:
At 9/24/2013 8:33:38 PM, Poetaster wrote:
At 9/24/2013 8:26:04 PM, Poetaster wrote:
At 9/24/2013 8:19:01 PM, AnDoctuir wrote:
Yeah, but you know that one chamber wasn't either of the chambers with the bullets in them, so 1/4.

It seems I forgot that there are two bullets.

Ah, I retract that near-concession; see here: the probability of the first shot being empty is 4/6. The probability of first occupying that unique chamber out of the four empty chambers with no re-spin is 1/4, but the probability of the whole sequence occurring is:

(4/6)*(1/4)= 1/6

Hence, the second method still produces a probability of 1/6 of being shot.

Yeah, the 4/6 has already come to fruition, though,and so you're left with a 1/4 vs. a 1/3, no? I mean otherwise it's 1/6 vs. ((2/3)*(1/3)+1/3) blahh .55555%

You're treating this like the Monty Hall Problem maybe where it's not, I think. The odds in that problem haven't come to fruition and so are still subject to variable change or whatever that was about. In this one you know you've achieved the 4/6 and so that's out of consideration. You're just left with the 1/4 to make up your 1/6.

Yes, I realize that I'm calculating over the whole sequence (computing the product of both events prior to empirical updating), when really the problem already fixes the first event in that sequence (it updates the sequence up to the first event).

From a prior evaluation, no re-spin between shots (with empty first chambers) does pose a 1/6 chance of firing a bullet, with re-spin posing a 2/9 chance under same condition. But I mixed up my reasoning because I was thinking of re-spins as statistically independent events (which they are individually) each with 1/3 firing chances, while considering the "no re-spin" case as a series of events with statistical dependence on the first chances of the initial condition obtaining (which was the error on my part; the problem says p=1 for that condition).
"The book you are looking for hasn't been written yet. What you are looking for you are going to have to find yourself, it's not going to be in a book..." -Sidewalker
Poetaster
Posts: 587
Add as Friend
Challenge to a Debate
Send a Message
9/24/2013 10:15:06 PM
Posted: 3 years ago
At 9/24/2013 9:59:08 PM, AnDoctuir wrote:
Wait, it's not even 1/6 then, but 50/50.

Only if we're asking: "What are the chances that I'll be shot if the gangster fires twice with no re-spins?"

The 1/6 is the answer to the question: "What are the chances that the first chamber will be empty, but the second loaded with no re-spin between shots?"

I was mistaken in interpreting the riddle as asking the second question. Not sure why I did that.
"The book you are looking for hasn't been written yet. What you are looking for you are going to have to find yourself, it's not going to be in a book..." -Sidewalker
Polaris
Posts: 1,120
Add as Friend
Challenge to a Debate
Send a Message
9/24/2013 10:39:54 PM
Posted: 3 years ago
After reading through this thread I've come to the conclusion that this is how probability should be taught in public schools... as a life or death situation.
Poetaster
Posts: 587
Add as Friend
Challenge to a Debate
Send a Message
9/24/2013 10:51:47 PM
Posted: 3 years ago
At 9/24/2013 10:11:27 PM, AnDoctuir wrote:
You had me thinking I was after falling for some elaborate ruse there for a moment.

You weren't; not an elaborate one, anyway.
"The book you are looking for hasn't been written yet. What you are looking for you are going to have to find yourself, it's not going to be in a book..." -Sidewalker
Wallstreetatheist
Posts: 7,132
Add as Friend
Challenge to a Debate
Send a Message
9/25/2013 12:16:41 AM
Posted: 3 years ago
At 9/24/2013 10:39:54 PM, Polaris wrote:
After reading through this thread I've come to the conclusion that this is how probability should be taught in public schools... as a life or death situation.
DRUG HARM: http://imgur.com...
Primal Diet. Lifting. Reading. Psychedelics. Cold-Approach Pickup. Music.
Floid
Posts: 751
Add as Friend
Challenge to a Debate
Send a Message
9/25/2013 7:46:04 AM
Posted: 3 years ago
At 9/24/2013 10:39:54 PM, Polaris wrote:
After reading through this thread I've come to the conclusion that this is how probability should be taught in public schools... as a life or death situation.

I came to the conclusion probability is not taught at all.

The fresh spin case, try to visualize it: G G G G B B

G is a good outcome (empty chamber) and B is a bad outcome (bullet). A fresh spin is obviously 1/3 (2 out of 6 chambers has a bullet).

Now what about the no spin case? A revolver acts in a deterministic fashion by progressing to the next chamber on each trigger pull. From the first trigger pull we can eliminate the two B cases from this scenario because we know we were not on a bad outcome before. So visualize the current situation:

G G G B

The two B from before have been dropped and we only have on bad case now: The chance that we are currently right before the bullets so that the next trigger pull fires a shot.

So what are the odds here? 1/4,

Your best bet is to fire again without spinning.
Wallstreetatheist
Posts: 7,132
Add as Friend
Challenge to a Debate
Send a Message
9/25/2013 7:41:52 PM
Posted: 3 years ago
At 9/25/2013 7:46:04 AM, Floid wrote:
At 9/24/2013 10:39:54 PM, Polaris wrote:
After reading through this thread I've come to the conclusion that this is how probability should be taught in public schools... as a life or death situation.

I came to the conclusion probability is not taught at all.

The fresh spin case, try to visualize it: G G G G B B

G is a good outcome (empty chamber) and B is a bad outcome (bullet). A fresh spin is obviously 1/3 (2 out of 6 chambers has a bullet).


Now what about the no spin case? A revolver acts in a deterministic fashion by progressing to the next chamber on each trigger pull. From the first trigger pull we can eliminate the two B cases from this scenario because we know we were not on a bad outcome before. So visualize the current situation:

G G G B

The two B from before have been dropped and we only have on bad case now: The chance that we are currently right before the bullets so that the next trigger pull fires a shot.

So what are the odds here? 1/4,



Your best bet is to fire again without spinning.

Congratulations, you've won! /thread
DRUG HARM: http://imgur.com...
Primal Diet. Lifting. Reading. Psychedelics. Cold-Approach Pickup. Music.
johnlubba
Posts: 2,892
Add as Friend
Challenge to a Debate
Send a Message
9/26/2013 1:11:59 PM
Posted: 3 years ago
At 9/25/2013 7:41:52 PM, Wallstreetatheist wrote:
At 9/25/2013 7:46:04 AM, Floid wrote:
At 9/24/2013 10:39:54 PM, Polaris wrote:
After reading through this thread I've come to the conclusion that this is how probability should be taught in public schools... as a life or death situation.

I came to the conclusion probability is not taught at all.

The fresh spin case, try to visualize it: G G G G B B

G is a good outcome (empty chamber) and B is a bad outcome (bullet). A fresh spin is obviously 1/3 (2 out of 6 chambers has a bullet).


Now what about the no spin case? A revolver acts in a deterministic fashion by progressing to the next chamber on each trigger pull. From the first trigger pull we can eliminate the two B cases from this scenario because we know we were not on a bad outcome before. So visualize the current situation:

G G G B

The two B from before have been dropped and we only have on bad case now: The chance that we are currently right before the bullets so that the next trigger pull fires a shot.

So what are the odds here? 1/4,



Your best bet is to fire again without spinning.

Congratulations, you've won! /thread

I instinctively knew that, Of course.
slo1
Posts: 4,354
Add as Friend
Challenge to a Debate
Send a Message
9/30/2013 4:46:46 PM
Posted: 3 years ago
At 9/25/2013 7:41:52 PM, Wallstreetatheist wrote:
At 9/25/2013 7:46:04 AM, Floid wrote:
At 9/24/2013 10:39:54 PM, Polaris wrote:
After reading through this thread I've come to the conclusion that this is how probability should be taught in public schools... as a life or death situation.

I came to the conclusion probability is not taught at all.

The fresh spin case, try to visualize it: G G G G B B

G is a good outcome (empty chamber) and B is a bad outcome (bullet). A fresh spin is obviously 1/3 (2 out of 6 chambers has a bullet).


Now what about the no spin case? A revolver acts in a deterministic fashion by progressing to the next chamber on each trigger pull. From the first trigger pull we can eliminate the two B cases from this scenario because we know we were not on a bad outcome before. So visualize the current situation:

G G G B

The two B from before have been dropped and we only have on bad case now: The chance that we are currently right before the bullets so that the next trigger pull fires a shot.

So what are the odds here? 1/4,



Your best bet is to fire again without spinning.

Congratulations, you've won! /thread

Just to visualize it a bit better.

-Two bullets loaded right next together.
- Random spin means the bullets could be in any position, but since first attempt was a click (good), the options for the remaining 5 bullet positions are as follows.

- b,b,g,g,g
- g,b,b,g,g
- g,g,b,b,g
- g,g,g,b,b

Only one of the 4 remaining scenarios is shot through the head on the next attempt.
retroman000
Posts: 14
Add as Friend
Challenge to a Debate
Send a Message
9/30/2013 5:56:37 PM
Posted: 3 years ago
Although you have to take into account that spinning it again would cause the rounds to tend towards the bottom due to their weight. Relatively insignificant, really.