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Random winner from a lotto with $N sim mathrmPois(1)$ other people

0

$begingroup$

What is the better answer to this first year probability test question?

You enter a lottery with $N+1$ people total

$N sim mathrmPoisson(1)$

A random person is selected to be the winner, what is your probability of winning?

Answer A: Your probability of winning is $frac1N+1$ which is a RV from $0$ to $1$

Answer B: Your probability of winning is about 63%

$$ P(W) = sum_n P(W mid N = n) P(N=n) = sum_n frac11+n frace^-1n! = 1 - e^-1 approx .63$$

My question: What is the better answer here, A or B? In the real world I would be a little hesitant to say 63% because that loses the fact that if there are many lottos then the probability of winning each one could be different. 63% seems more like an expectation which I suppose the total law of probability is an expectation.

It does make sense to me if someone said "$P(W)$ is the unconditional probability of winning" and $1/(n+1)$ is the probability of winning given $N = n$ but it also makes sense to me if someone said the probability of winning is a RV.

I know this is boring question but can you help clear my confusion. Thanks for your help and patience.

probability expected-value

share|cite|improve this question

edited 16 hours ago

Lee David Chung Lin

4,50841342

asked Mar 25 at 18:37

HJ_beginnerHJ_beginner

9601415

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The distinction is essentially the difference between the conditional probability (conditioned on $N$) and the [unconditional] probability.
  $endgroup$
  – angryavian
  Mar 25 at 18:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @angryavian Thanks for your help. Ahhh so if I say the probability of winning is the RV $1/(1+N)$, I am implicitly conditioning on the RV $N$? So given the RV $N$ then the probability of winning is a RV $1/(1+N)$... The conditioning is very clear to me for a statement like $P(W mid N=n) = 1/(1+n)$
  $endgroup$
  – HJ_beginner
  Mar 25 at 19:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @angryavian Hmm is this something that makes sense, $P(W mid N) = 1/(1+N)$? I know you can do things like $E[E[W|N]]$ but didn't think about doing that for a probability $P()$
  $endgroup$
  – HJ_beginner
  Mar 25 at 19:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MinusOne-Twelfth Excellent thanks!
  $endgroup$
  – HJ_beginner
  Mar 25 at 19:17
  

  
  
  
  

add a comment | 

0

$begingroup$

What is the better answer to this first year probability test question?

You enter a lottery with $N+1$ people total

$N sim mathrmPoisson(1)$

A random person is selected to be the winner, what is your probability of winning?

Answer A: Your probability of winning is $frac1N+1$ which is a RV from $0$ to $1$

Answer B: Your probability of winning is about 63%

$$ P(W) = sum_n P(W mid N = n) P(N=n) = sum_n frac11+n frace^-1n! = 1 - e^-1 approx .63$$

My question: What is the better answer here, A or B? In the real world I would be a little hesitant to say 63% because that loses the fact that if there are many lottos then the probability of winning each one could be different. 63% seems more like an expectation which I suppose the total law of probability is an expectation.

It does make sense to me if someone said "$P(W)$ is the unconditional probability of winning" and $1/(n+1)$ is the probability of winning given $N = n$ but it also makes sense to me if someone said the probability of winning is a RV.

I know this is boring question but can you help clear my confusion. Thanks for your help and patience.

probability expected-value

share|cite|improve this question

edited 16 hours ago

Lee David Chung Lin

4,50841342

asked Mar 25 at 18:37

HJ_beginnerHJ_beginner

9601415

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The distinction is essentially the difference between the conditional probability (conditioned on $N$) and the [unconditional] probability.
  $endgroup$
  – angryavian
  Mar 25 at 18:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @angryavian Thanks for your help. Ahhh so if I say the probability of winning is the RV $1/(1+N)$, I am implicitly conditioning on the RV $N$? So given the RV $N$ then the probability of winning is a RV $1/(1+N)$... The conditioning is very clear to me for a statement like $P(W mid N=n) = 1/(1+n)$
  $endgroup$
  – HJ_beginner
  Mar 25 at 19:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @angryavian Hmm is this something that makes sense, $P(W mid N) = 1/(1+N)$? I know you can do things like $E[E[W|N]]$ but didn't think about doing that for a probability $P()$
  $endgroup$
  – HJ_beginner
  Mar 25 at 19:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MinusOne-Twelfth Excellent thanks!
  $endgroup$
  – HJ_beginner
  Mar 25 at 19:17
  

  
  
  
  

add a comment | 

$begingroup$

What is the better answer to this first year probability test question?

You enter a lottery with $N+1$ people total

$N sim mathrmPoisson(1)$

A random person is selected to be the winner, what is your probability of winning?

Answer A: Your probability of winning is $frac1N+1$ which is a RV from $0$ to $1$

Answer B: Your probability of winning is about 63%

$$ P(W) = sum_n P(W mid N = n) P(N=n) = sum_n frac11+n frace^-1n! = 1 - e^-1 approx .63$$

My question: What is the better answer here, A or B? In the real world I would be a little hesitant to say 63% because that loses the fact that if there are many lottos then the probability of winning each one could be different. 63% seems more like an expectation which I suppose the total law of probability is an expectation.

It does make sense to me if someone said "$P(W)$ is the unconditional probability of winning" and $1/(n+1)$ is the probability of winning given $N = n$ but it also makes sense to me if someone said the probability of winning is a RV.

I know this is boring question but can you help clear my confusion. Thanks for your help and patience.

probability expected-value

share|cite|improve this question

edited 16 hours ago

Lee David Chung Lin

4,50841342

asked Mar 25 at 18:37

HJ_beginnerHJ_beginner

9601415

$endgroup$

share|cite|improve this question

edited 16 hours ago

Lee David Chung Lin

4,50841342

asked Mar 25 at 18:37

HJ_beginnerHJ_beginner

9601415

share|cite|improve this question

edited 16 hours ago

Lee David Chung Lin

4,50841342

asked Mar 25 at 18:37

HJ_beginnerHJ_beginner

9601415

edited 16 hours ago

Lee David Chung Lin

4,50841342

edited 16 hours ago

Lee David Chung Lin

4,50841342

asked Mar 25 at 18:37

HJ_beginnerHJ_beginner

9601415

asked Mar 25 at 18:37

HJ_beginnerHJ_beginner

9601415