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# Probability Question

 Posts: 7,132 Add as FriendChallenge to a DebateSend a Message 9/24/2013 6:26:18 PMPosted: 3 years agoA 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.
 Posts: 3,527 Add as FriendChallenge to a DebateSend a Message 9/24/2013 7:33:51 PMPosted: 3 years agoAt 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
 Posts: 587 Add as FriendChallenge to a DebateSend a Message 9/24/2013 7:50:44 PMPosted: 3 years agoIf 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
 Posts: 3,527 Add as FriendChallenge to a DebateSend a Message 9/24/2013 8:03:11 PMPosted: 3 years agoAt 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
 Posts: 11,060 Add as FriendChallenge to a DebateSend a Message 9/24/2013 8:05:20 PMPosted: 3 years ago1/4, Poetaster. 2 possibilities are already ruled out in your being alive.
 Posts: 587 Add as FriendChallenge to a DebateSend a Message 9/24/2013 8:12:01 PMPosted: 3 years agoAt 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
 Posts: 11,060 Add as FriendChallenge to a DebateSend a Message 9/24/2013 8:19:01 PMPosted: 3 years agoAt 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.
 Posts: 587 Add as FriendChallenge to a DebateSend a Message 9/24/2013 8:26:04 PMPosted: 3 years agoAt 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
 Posts: 587 Add as FriendChallenge to a DebateSend a Message 9/24/2013 8:33:38 PMPosted: 3 years agoAt 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)= 1/6Hence, 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
 Posts: 11,060 Add as FriendChallenge to a DebateSend a Message 9/24/2013 9:36:35 PMPosted: 3 years agoAt 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)= 1/6Hence, 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.
 Posts: 11,060 Add as FriendChallenge to a DebateSend a Message 9/24/2013 9:59:08 PMPosted: 3 years agoWait, it's not even 1/6 then, but 50/50.
 Posts: 587 Add as FriendChallenge to a DebateSend a Message 9/24/2013 10:04:56 PMPosted: 3 years agoAt 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)= 1/6Hence, 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
 Posts: 11,060 Add as FriendChallenge to a DebateSend a Message 9/24/2013 10:11:27 PMPosted: 3 years agoYou had me thinking I was after falling for some elaborate ruse there for a moment.
 Posts: 587 Add as FriendChallenge to a DebateSend a Message 9/24/2013 10:15:06 PMPosted: 3 years agoAt 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
 Posts: 1,120 Add as FriendChallenge to a DebateSend a Message 9/24/2013 10:39:54 PMPosted: 3 years agoAfter 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.
 Posts: 587 Add as FriendChallenge to a DebateSend a Message 9/24/2013 10:51:47 PMPosted: 3 years agoAt 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
 Posts: 7,132 Add as FriendChallenge to a DebateSend a Message 9/25/2013 12:16:41 AMPosted: 3 years agoAt 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.
 Posts: 751 Add as FriendChallenge to a DebateSend a Message 9/25/2013 7:46:04 AMPosted: 3 years agoAt 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 BG 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 BThe 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.
 Posts: 7,132 Add as FriendChallenge to a DebateSend a Message 9/25/2013 7:41:52 PMPosted: 3 years agoAt 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 BG 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 BThe 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! /threadDRUG HARM: http://imgur.com... Primal Diet. Lifting. Reading. Psychedelics. Cold-Approach Pickup. Music.
 Posts: 2,919 Add as FriendChallenge to a DebateSend a Message 9/26/2013 1:11:59 PMPosted: 3 years agoAt 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 BG 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 BThe 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! /threadI instinctively knew that, Of course.