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Probability of there being a bullet in the chamber

Probability Theory Question on Expected Value and Variance of Random VariableRussian roulette should a player pull the trigger or spin the cylinderColliding BulletsA probability question that I failed to answer in a job interviewProving russian roulette survival statisticsLoaded revolver puzzle.Russian Roulette variant, option to point awayCalculating probability and optimization - russian rouletteExpected value of Russian roulette at a firing range played on end with finite total shotsRussian Roulette Probability

1

$begingroup$

Suppose we have three $6$-chambered guns:

• The first has no bullets.




• The second has one bullet.




• The third has two bullets in consecutive chambers.

The cylinder advances automatically as the trigger is pulled. A man grabs a revolver at random, aims it at him, pulls the trigger $t$ times, and no shot is fired. He then aims at your head and pulls the trigger once. What is the probability you are shot?

probability conditional-probability word-problem

share|cite|improve this question

edited yesterday

Rodrigo de Azevedo

13k41960

asked yesterday

KombatWombatKombatWombat

756

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The man "grabs a revolver at random" but what is the distribution? Uniform?
  $endgroup$
  – Rodrigo de Azevedo
  yesterday
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Suppose we have three $6$-chambered guns:

• The first has no bullets.




• The second has one bullet.




• The third has two bullets in consecutive chambers.

The cylinder advances automatically as the trigger is pulled. A man grabs a revolver at random, aims it at him, pulls the trigger $t$ times, and no shot is fired. He then aims at your head and pulls the trigger once. What is the probability you are shot?

probability conditional-probability word-problem

share|cite|improve this question

edited yesterday

Rodrigo de Azevedo

13k41960

asked yesterday

KombatWombatKombatWombat

756

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The man "grabs a revolver at random" but what is the distribution? Uniform?
  $endgroup$
  – Rodrigo de Azevedo
  yesterday
  
  

  
  
  
  

add a comment | 

$begingroup$

Suppose we have three $6$-chambered guns:

• The first has no bullets.




• The second has one bullet.




• The third has two bullets in consecutive chambers.

The cylinder advances automatically as the trigger is pulled. A man grabs a revolver at random, aims it at him, pulls the trigger $t$ times, and no shot is fired. He then aims at your head and pulls the trigger once. What is the probability you are shot?

probability conditional-probability word-problem

share|cite|improve this question

edited yesterday

Rodrigo de Azevedo

13k41960

asked yesterday

KombatWombatKombatWombat

756

$endgroup$

share|cite|improve this question

edited yesterday

Rodrigo de Azevedo

13k41960

asked yesterday

KombatWombatKombatWombat

756

share|cite|improve this question

edited yesterday

Rodrigo de Azevedo

13k41960

asked yesterday

KombatWombatKombatWombat

756

edited yesterday

Rodrigo de Azevedo

13k41960

edited yesterday

Rodrigo de Azevedo

13k41960

asked yesterday

KombatWombatKombatWombat

756

asked yesterday

KombatWombatKombatWombat

756

1 Answer
1

active

oldest

votes

2

$begingroup$

Setup:

Let $X$ denote the number of shots needed for appearing a bullet if at random a revolver is picked and is shooted several times.

Then $X$ takes values in $left1,2,3,4,5,6,inftyright$

Note that there are $18$ chambers in total having equal probability
to be chosen for the first shot, and it is not really difficult to find that:

$Pleft(X=1right)=frac318$

$Pleft(X=2right)=Pleft(X=3right)=Pleft(X=4right)=Pleft(X=5right)=frac218$

$Pleft(X=6right)=frac118$

$Pleft(X=inftyright)=frac618$

Based on this you can find $P(X=t+1mid X>t)$.

Observe that this probability takes value $0$ if $t>5$ so the task can be completed by finding expressions for $tin0,1,2,3,4,5$.

I leave the rest to you.

share|cite|improve this answer

edited yesterday

Rodrigo de Azevedo

13k41960

answered yesterday

drhabdrhab

103k545136

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i'm struggling to understand your answer, could you possible show how I would obtain the answer for t=1 using your way please.
  $endgroup$
  – KombatWombat
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm more confused as to how you set up the problem, i.e how you calculate P(X=2) and probability (X=infinity). I understand for x = 1 is number of bullets(3) divided by number of chambers.
  $endgroup$
  – KombatWombat
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also, how do you know that you should combine them making it out of 18 instead of multiplying each one out of 6 by 1/3?
  $endgroup$
  – KombatWombat
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For $t=1$ we find $P(X=2mid X>1)=P(X=2)/P(X>1)=2/(18-3)=2/15$. Note that there are $15$ empty chambers and exactly $2$ of them are followed by a chamber that contains a bullet. Concerning your last comment: it is not so much a matter of "knowing" but more a (nice) way to approach the problem. Instead of looking at $3$ guns we are looking at $18$ equiprobable chambers.
  $endgroup$
  – drhab
  yesterday
  
  

  
  
  
  

add a comment | 

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2

$begingroup$

Setup:

Let $X$ denote the number of shots needed for appearing a bullet if at random a revolver is picked and is shooted several times.

Then $X$ takes values in $left1,2,3,4,5,6,inftyright$

Note that there are $18$ chambers in total having equal probability
to be chosen for the first shot, and it is not really difficult to find that:

$Pleft(X=1right)=frac318$

$Pleft(X=2right)=Pleft(X=3right)=Pleft(X=4right)=Pleft(X=5right)=frac218$

$Pleft(X=6right)=frac118$

$Pleft(X=inftyright)=frac618$

Based on this you can find $P(X=t+1mid X>t)$.

Observe that this probability takes value $0$ if $t>5$ so the task can be completed by finding expressions for $tin0,1,2,3,4,5$.

I leave the rest to you.

share|cite|improve this answer

edited yesterday

Rodrigo de Azevedo

13k41960

answered yesterday

drhabdrhab

103k545136

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i'm struggling to understand your answer, could you possible show how I would obtain the answer for t=1 using your way please.
  $endgroup$
  – KombatWombat
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm more confused as to how you set up the problem, i.e how you calculate P(X=2) and probability (X=infinity). I understand for x = 1 is number of bullets(3) divided by number of chambers.
  $endgroup$
  – KombatWombat
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also, how do you know that you should combine them making it out of 18 instead of multiplying each one out of 6 by 1/3?
  $endgroup$
  – KombatWombat
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For $t=1$ we find $P(X=2mid X>1)=P(X=2)/P(X>1)=2/(18-3)=2/15$. Note that there are $15$ empty chambers and exactly $2$ of them are followed by a chamber that contains a bullet. Concerning your last comment: it is not so much a matter of "knowing" but more a (nice) way to approach the problem. Instead of looking at $3$ guns we are looking at $18$ equiprobable chambers.
  $endgroup$
  – drhab
  yesterday
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Setup:

Let $X$ denote the number of shots needed for appearing a bullet if at random a revolver is picked and is shooted several times.

Then $X$ takes values in $left1,2,3,4,5,6,inftyright$

Note that there are $18$ chambers in total having equal probability
to be chosen for the first shot, and it is not really difficult to find that:

$Pleft(X=1right)=frac318$

$Pleft(X=2right)=Pleft(X=3right)=Pleft(X=4right)=Pleft(X=5right)=frac218$

$Pleft(X=6right)=frac118$

$Pleft(X=inftyright)=frac618$

Based on this you can find $P(X=t+1mid X>t)$.

Observe that this probability takes value $0$ if $t>5$ so the task can be completed by finding expressions for $tin0,1,2,3,4,5$.

I leave the rest to you.

share|cite|improve this answer

edited yesterday

Rodrigo de Azevedo

13k41960

answered yesterday

drhabdrhab

103k545136

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i'm struggling to understand your answer, could you possible show how I would obtain the answer for t=1 using your way please.
  $endgroup$
  – KombatWombat
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm more confused as to how you set up the problem, i.e how you calculate P(X=2) and probability (X=infinity). I understand for x = 1 is number of bullets(3) divided by number of chambers.
  $endgroup$
  – KombatWombat
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also, how do you know that you should combine them making it out of 18 instead of multiplying each one out of 6 by 1/3?
  $endgroup$
  – KombatWombat
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For $t=1$ we find $P(X=2mid X>1)=P(X=2)/P(X>1)=2/(18-3)=2/15$. Note that there are $15$ empty chambers and exactly $2$ of them are followed by a chamber that contains a bullet. Concerning your last comment: it is not so much a matter of "knowing" but more a (nice) way to approach the problem. Instead of looking at $3$ guns we are looking at $18$ equiprobable chambers.
  $endgroup$
  – drhab
  yesterday
  
  

  
  
  
  

add a comment | 

$begingroup$

Setup:

Let $X$ denote the number of shots needed for appearing a bullet if at random a revolver is picked and is shooted several times.

Then $X$ takes values in $left1,2,3,4,5,6,inftyright$

Note that there are $18$ chambers in total having equal probability
to be chosen for the first shot, and it is not really difficult to find that:

$Pleft(X=1right)=frac318$

$Pleft(X=2right)=Pleft(X=3right)=Pleft(X=4right)=Pleft(X=5right)=frac218$

$Pleft(X=6right)=frac118$

$Pleft(X=inftyright)=frac618$

Based on this you can find $P(X=t+1mid X>t)$.

Observe that this probability takes value $0$ if $t>5$ so the task can be completed by finding expressions for $tin0,1,2,3,4,5$.

I leave the rest to you.

share|cite|improve this answer

edited yesterday

Rodrigo de Azevedo

13k41960

answered yesterday

drhabdrhab

103k545136

$endgroup$

share|cite|improve this answer

edited yesterday

Rodrigo de Azevedo

13k41960

answered yesterday

drhabdrhab

103k545136

edited yesterday

Rodrigo de Azevedo

13k41960

edited yesterday

Rodrigo de Azevedo

13k41960

answered yesterday

drhabdrhab

103k545136

answered yesterday

drhabdrhab

103k545136