Conditional Probability: There are three six-chambered revolvers....

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Jeff1801

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There are three six-chambered revolvers. The first has no bullets; the second, one bullet; and
the third, two bullets in consecutive chambers. The cylinder advances automatically as the
trigger is pulled. A madman grabs a revolver at random, aims it at his own head, 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 that you are shot?
 
Jeff1801 said:
There are three six-chambered revolvers. The first has no bullets; the second, one bullet; and
the third, two bullets in consecutive chambers. The cylinder advances automatically as the
trigger is pulled. A madman grabs a revolver at random, aims it at his own head, 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 that you are shot?
Click to expand...
I have managed to get the instance when t = 0,
P(gets shot | t = 0) = 0 + (1/6)(1/3) + (2/6)(1/3) = 0.167
for the others I am going wrong somewhere and wanted to know the error in my logic.
P(gets shot | t=1) = 0 + (5/6)(1/5)(1/3) + (4/6)(1/4)(1/3)
 
\(\displaystyle \Pr(\text{you get shot} | \text{t times}) = 1- \Pr(\text{madman gets shot} | \text{t times} )\)

[imath] \Pr(\text{madman gets shot} | \text{t times} ) = \Pr(\text{madman gets shot within t times}| \text{Revolver 1} )\Pr(\text{Revolver 1}) + \\ \Pr(\text{madman gets shot within t times}| \text{Revolver 2} )\Pr(\text{Revolver 2}) +\\ \Pr(\text{madman gets shot within t times}| \text{Revolver 3} )\Pr(\text{Revolver 3}) [/imath]

Where:
[imath] \Pr(\text{Revolver 1}) = \Pr(\text{Revolver 2}) = \Pr(\text{Revolver 3}) = \dfrac{1}{3} [/imath]

[imath]\Pr(\text{madman gets shot within t times}| \text{Revolver 1} ) = (1)^{t-1}0 = 0[/imath]

[imath]\Pr(\text{madman gets shot within t times}| \text{Revolver 2} ) = \left(\dfrac{5}{6}\right)^{t-1}\left(\dfrac{1}{6}\right)[/imath]

[imath]\Pr(\text{madman gets shot within t times}| \text{Revolver 3} ) = \left(\dfrac{4}{6}\right)^{t-1}\left(\dfrac{2}{6}\right)[/imath]

It follows:
[imath]\Pr(\text{madman get shot} | \text{t times} ) = \dfrac{1}{3}\left[ 0 + \left(\dfrac{5}{6}\right)^{t-1}\left(\dfrac{1}{6}\right) +\left(\dfrac{4}{6}\right)^{t-1}\left(\dfrac{2}{6}\right)\right] [/imath]

Therefore:
[imath]\Pr(\text{you get shot} | \text{t times} )= 1 - \dfrac{1}{3}\left[\left(\dfrac{5}{6}\right)^{t-1}\left(\dfrac{1}{6}\right) +\left(\dfrac{4}{6}\right)^{t-1}\left(\dfrac{2}{6}\right)\right] [/imath]
 
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Let’s try to stay close to your logic and calculate the probability that you end up dead.

[math] P(\text {you dead}) = P(\text {empty revolver and you dead}) + \\ P((\text {one bullet and you dead}) +\\ P(\text {two bullets and you dead}) = \\ P(\text {you dead | empty}) * P(\text {empty}) +\\ P(\text {you dead | 1 bullet}) * P(\text {one bullet}) + \\ P(\text {you dead | 2 bullets}) * P(\text {2 bullets}) =\\ 0 * \dfrac{1}{3} + P(\text {u dead | 1 bullet}) * \dfrac{1}{3} + P(\text {u dead | 2 bullets}) =\\ \dfrac{1}{3} * \{ P(\text {u dead | 1 bullet}) + P(\text {u dead | 2 bullets}) \}. [/math]
If that was your logic to start, it is fine. We can now concentrate on the two remaining probabilities. It may be easier to label the chambers that are loaded. Let’s say from chambers A, B, C, D, E, and F that only chamber F is loaded (the one bullet case).
If t = 0, you end up dead only if chamber F will fire first: probability = 1/6. If t = 1, you end up dead only if chamber E fires first: probability 1/6. If t = 2, you end up dead only if chamber D fires first: probability 1/6. If t = 3, the dangerous chamber is C: 1/6. And so on. You can follow the same type of reasoning for two bullets in chambers E and F. If t > 5, the problem is trivial.

Now the problem is not well stated. We do not know what t is. Is it a certain number specified in advance or is it random itself. The rationality-challenged person (“madman” is clearly a non-permitted word) may for example decide on the number of times to pull the trigger based on the roll of a fair die. I do not know how to proceed in the absence of more information on t.
 
BigBeachBanana said:
\(\displaystyle \Pr(\text{you get shot} | \text{t times}) = 1- \Pr(\text{madman gets shot} | \text{t times} )\)

[imath] \Pr(\text{madman gets shot} | \text{t times} ) = \Pr(\text{madman gets shot within t times}| \text{Revolver 1} )\Pr(\text{Revolver 1}) + \\ \Pr(\text{madman gets shot within t times}| \text{Revolver 2} )\Pr(\text{Revolver 2}) +\\ \Pr(\text{madman gets shot within t times}| \text{Revolver 3} )\Pr(\text{Revolver 3}) [/imath]

Where:
[imath] \Pr(\text{Revolver 1}) = \Pr(\text{Revolver 2}) = \Pr(\text{Revolver 3}) = \dfrac{1}{3} [/imath]

[imath]\Pr(\text{madman gets shot within t times}| \text{Revolver 1} ) = (1)^{t-1}0 = 0[/imath]

[imath]\Pr(\text{madman gets shot within t times}| \text{Revolver 2} ) = \left(\dfrac{5}{6}\right)^{t-1}\left(\dfrac{1}{6}\right)[/imath]

[imath]\Pr(\text{madman gets shot within t times}| \text{Revolver 3} ) = \left(\dfrac{4}{6}\right)^{t-1}\left(\dfrac{2}{6}\right)[/imath]

It follows:
[imath]\Pr(\text{madman get shot} | \text{t times} ) = \dfrac{1}{3}\left[ 0 + \left(\dfrac{5}{6}\right)^{t-1}\left(\dfrac{1}{6}\right) +\left(\dfrac{4}{6}\right)^{t-1}\left(\dfrac{2}{6}\right)\right] [/imath]

Therefore:
[imath]\Pr(\text{you get shot} | \text{t times} )= 1 - \dfrac{1}{3}\left[\left(\dfrac{5}{6}\right)^{t-1}\left(\dfrac{1}{6}\right) +\left(\dfrac{4}{6}\right)^{t-1}\left(\dfrac{2}{6}\right)\right] [/imath]
Click to expand...
I don't think the 2 events make up the whole sample space. For example you can have the instance where no one gets shot.
 
NotJeffM said:
Let’s try to stay close to your logic and calculate the probability that you end up dead.

[math] P(\text {you dead}) = P(\text {empty revolver and you dead}) + \\ P((\text {one bullet and you dead}) +\\ P(\text {two bullets and you dead}) = \\ P(\text {you dead | empty}) * P(\text {empty}) +\\ P(\text {you dead | 1 bullet}) * P(\text {one bullet}) + \\ P(\text {you dead | 2 bullets}) * P(\text {2 bullets}) =\\ 0 * \dfrac{1}{3} + P(\text {u dead | 1 bullet}) * \dfrac{1}{3} + P(\text {u dead | 2 bullets}) =\\ \dfrac{1}{3} * \{ P(\text {u dead | 1 bullet}) + P(\text {u dead | 2 bullets}) \}. [/math]
If that was your logic to start, it is fine. We can now concentrate on the two remaining probabilities. It may be easier to label the chambers that are loaded. Let’s say from chambers A, B, C, D, E, and F that only chamber F is loaded (the one bullet case).
If t = 0, you end up dead only if chamber F will fire first: probability = 1/6. If t = 1, you end up dead only if chamber E fires first: probability 1/6. If t = 2, you end up dead only if chamber D fires first: probability 1/6. If t = 3, the dangerous chamber is C: 1/6. And so on. You can follow the same type of reasoning for two bullets in chambers E and F. If t > 5, the problem is trivial.

Now the problem is not well stated. We do not know what t is. Is it a certain number specified in advance or is it random itself. The rationality-challenged person (“madman” is clearly a non-permitted word) may for example decide on the number of times to pull the trigger based on the roll of a fair die. I do not know how to proceed in the absence of more information on t.
Click to expand...
I do have the final answers [0.167 (t = 0), 0.133 (t = 1), 0.154 (t = 2), 0.182 (t = 3), 0.222 (t = 4), 0.143 (t = 5), 0 (t ≥ 6)]. So it seems we need to calculate the probabilities for all possible values of t specified in advance. I have gotten the case of t = 0, but failed on the others.
 
Oh. It looks as though you are to calculate the answers given different values of t.
 
Jeff1801 said:
I don't think the 2 events make up the whole sample space. For example you can have the instance where no one gets shot.
Click to expand...
1) Of course, need to incorporate that.

2) I don't agree with the answer, in particular for t=0, because if you think about it logically if the madman fires 0 times, there's no chance you can get shot

3) Based on the given answer, you're looking for the conditional probability and not the unconditional

4) I assumed the revolver was being shot with the replacement above but in reality, the revolver rotates one barrel to the next so it should be without replacement.

Revised answer...
\(\displaystyle \Pr(\text{you get shot} | \text{t times}) = 1- \Pr(\text{madman gets shot} | \text{t times}) - \Pr(\text{no one get shot | \text{t times}})\)

Where:

[imath]\Pr(\text{Revolver 1}) = \Pr(\text{Revolver 2}) = \Pr(\text{Revolver 3}) = \dfrac{1}{3}[/imath]

[imath]\Pr(\text{madman gets shot within t times}| \text{Revolver 1} ) = 0[/imath]

[imath]\Pr(\text{madman gets shot within t times}| \text{Revolver 2} ) = \dfrac{t}{6}[/imath]

[imath]\Pr(\text{madman gets shot within t times}| \text{Revolver 3} ) =\dfrac{1}{30}(11t-t^2)[/imath]

It follows:
[imath]\Pr(\text{madman get shot} | \text{t times} ) = \dfrac{1}{3}\left[ \dfrac{t}{6} + \dfrac{1}{30}(11t-t^2)\right] [/imath]

[imath]\Pr(\text{both live}| \text{Revolver 1} ) = 1[/imath]

[imath]\Pr(\text{both live}| \text{Revolver 2} ) = \Pr(\text{madman live within t times}| \text{Revolver 2} ) \times \Pr(\text{you live within t times}| \text{Revolver 2})\\ = \left[1-\dfrac{1}{3}\left( \dfrac{t}{6} + \dfrac{1}{30}(11t-t^2)\right)\right] \times \left(\dfrac{5-t}{6-t}\right) \text{ for t < 5, else 1}[/imath]

[imath]\Pr(\text{both live}| \text{Revolver 3} ) = \Pr(\text{madman live within t times}| \text{Revolver 3} ) \times \Pr(\text{you live within t times}| \text{Revolver 3} ) \\ = \left[1-\dfrac{1}{3}\left( \dfrac{t}{6} + \dfrac{1}{30}(11t-t^2)\right)\right] \times \left(\dfrac{4-t}{6-t}\right) \text{ for t < 4, else 1} [/imath]

Results:

t timesPr(you get shot | t times)
10.1444...
20.1222...
30.1000...
40.0555...
50
60
 
@BigBeachBanana The variable t appears to be the number of times the rationality-challenged person pulls the trigger, but the trigger may not be pulled when aiming at himself. All we know is that no shot was fired before the revolver is aimed at you. The problem is not well worded. I am going to think about it more.
 
BigBeachBanana said:
1) Of course, need to incorporate that.

2) I don't agree with the answer, in particular for t=0, because if you think about it logically if the madman fires 0 times, there's no chance you can get shot

3) Based on the given answer, you're looking for the conditional probability and not the unconditional

4) I assumed the revolver was being shot with the replacement above but in reality, the revolver rotates one barrel to the next so it should be without replacement.

Revised answer...
\(\displaystyle \Pr(\text{you get shot} | \text{t times}) = 1- \Pr(\text{madman gets shot} | \text{t times}) - \Pr(\text{no one get shot | \text{t times}})\)

Where:

[imath]\Pr(\text{Revolver 1}) = \Pr(\text{Revolver 2}) = \Pr(\text{Revolver 3}) = \dfrac{1}{3}[/imath]

[imath]\Pr(\text{madman gets shot within t times}| \text{Revolver 1} ) = 0[/imath]

[imath]\Pr(\text{madman gets shot within t times}| \text{Revolver 2} ) = \dfrac{t}{6}[/imath]

[imath]\Pr(\text{madman gets shot within t times}| \text{Revolver 3} ) =\dfrac{1}{30}(11t-t^2)[/imath]

It follows:
[imath]\Pr(\text{madman get shot} | \text{t times} ) = \dfrac{1}{3}\left[ \dfrac{t}{6} + \dfrac{1}{30}(11t-t^2)\right] [/imath]

[imath]\Pr(\text{both live}| \text{Revolver 1} ) = 1[/imath]

[imath]\Pr(\text{both live}| \text{Revolver 2} ) = \Pr(\text{madman live within t times}| \text{Revolver 2} ) \times \Pr(\text{you live within t times}| \text{Revolver 2})\\ = \left[1-\dfrac{1}{3}\left( \dfrac{t}{6} + \dfrac{1}{30}(11t-t^2)\right)\right] \times \left(\dfrac{5-t}{6-t}\right) \text{ for t < 5, else 1}[/imath]

[imath]\Pr(\text{both live}| \text{Revolver 3} ) = \Pr(\text{madman live within t times}| \text{Revolver 3} ) \times \Pr(\text{you live within t times}| \text{Revolver 3} ) \\ = \left[1-\dfrac{1}{3}\left( \dfrac{t}{6} + \dfrac{1}{30}(11t-t^2)\right)\right] \times \left(\dfrac{4-t}{6-t}\right) \text{ for t < 4, else 1} [/imath]

Results:

t timesPr(you get shot | t times)
10.1444...
20.1222...
30.1000...
40.0555...
50
60
Click to expand...
t is the number of times the madman pulls the trigger while pointing it at his own head. After that he points it at you and pulls the trigger once. So you can have the case where t = 0 and you get shot.
 
NotJeffM said:
@BigBeachBanana The variable t appears to be the number of times the rationality-challenged person pulls the trigger, but the trigger may not be pulled when aiming at himself. All we know is that no shot was fired before the revolver is aimed at you. The problem is not well worded. I am going to think about it more.
Click to expand...
The question does say that he pulls the trigger while aiming it at himself. He pulls the trigger t times at himself before aiming it at you and pulling the trigger once.
 
Jeff1801 said:
The question does say that he pulls the trigger while aiming it at himself. He pulls the trigger t times at himself before aiming it at you and pulling the trigger once.
Click to expand...
So can t be zero or not?

What are the exact words of the problem?
 
NotJeffM said:
So can t be zero or not?

What are the exact words of the problem?
Click to expand...
t can be zero, intuitively we only need to consider values from 0 - 5.

I copied the question verbatim and I provided the final solutions where you can see the values of t and the probabilities.
 
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