Expected waiting time at tennis court [simplified version]
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I want to understand that why the answers for both these scenarios are the same.
Q1. An athletic facility has 1 tennis court. Pairs of players arrive at the courts and use a court for an exponentially distributed amount of time with mean 40 minutes. Suppose a pair of players arrives and finds all courts busy and also that k other teams are waiting in queue. What is the expected waiting time to get a court?.
Q2. An athletic facility has 1 tennis court. Pairs of players arrive at the courts and use a court for an exponentially distributed amount of time with mean 40 minutes. Suppose a pair of players arrives and finds that a team just went inside and after that k other teams are waiting in queue. What is the expected waiting time to get a court?.
I feel that the answer to Q1 should be 40k + 20 and answer to Q2 should also be 40k + 40. My TA says that answer to both the questions is 40k+40 . I am not getting the reason.
probability expected-value
asked Nov 21 at 13:38
Raj Rajvir
161
add a comment |
up vote
0
down vote
favorite
I want to understand that why the answers for both these scenarios are the same.
Q1. An athletic facility has 1 tennis court. Pairs of players arrive at the courts and use a court for an exponentially distributed amount of time with mean 40 minutes. Suppose a pair of players arrives and finds all courts busy and also that k other teams are waiting in queue. What is the expected waiting time to get a court?.
Q2. An athletic facility has 1 tennis court. Pairs of players arrive at the courts and use a court for an exponentially distributed amount of time with mean 40 minutes. Suppose a pair of players arrives and finds that a team just went inside and after that k other teams are waiting in queue. What is the expected waiting time to get a court?.
I feel that the answer to Q1 should be 40k + 20 and answer to Q2 should also be 40k + 40. My TA says that answer to both the questions is 40k+40 . I am not getting the reason.
probability expected-value
asked Nov 21 at 13:38
Raj Rajvir
161
Because the exponential distribution has no memory. So the probability of the players staying on the court another 40 min from when you first observe them is the same as the probability of them staying on the court for 40 min from when they first started.
– user121049
Nov 21 at 18:50
Expanding on my previous comment. Say they have already being playing for a time $t_1$ when you arrive, then the probability that they will play a further amount $Delta t$ is given by $P(T>t_1+Delta t|T>t_1)$ It does not take much to show that this is equal to $P(T>t_1)$. So the expected time they will continue to play is still the mean play time of 40 min.
– user121049
Nov 21 at 20:35
A mistake in my comment which I can no longer edit. Should read more like $P(T>t_1+Delta t|T>t_1)=P(T>Delta t)$
– user121049
Nov 22 at 14:12
I understood it. Thank you very much @user121049
– Raj Rajvir
Nov 22 at 16:38
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to understand that why the answers for both these scenarios are the same.
Q1. An athletic facility has 1 tennis court. Pairs of players arrive at the courts and use a court for an exponentially distributed amount of time with mean 40 minutes. Suppose a pair of players arrives and finds all courts busy and also that k other teams are waiting in queue. What is the expected waiting time to get a court?.
Q2. An athletic facility has 1 tennis court. Pairs of players arrive at the courts and use a court for an exponentially distributed amount of time with mean 40 minutes. Suppose a pair of players arrives and finds that a team just went inside and after that k other teams are waiting in queue. What is the expected waiting time to get a court?.
I feel that the answer to Q1 should be 40k + 20 and answer to Q2 should also be 40k + 40. My TA says that answer to both the questions is 40k+40 . I am not getting the reason.
probability expected-value
asked Nov 21 at 13:38
Raj Rajvir
161
I want to understand that why the answers for both these scenarios are the same.
Q1. An athletic facility has 1 tennis court. Pairs of players arrive at the courts and use a court for an exponentially distributed amount of time with mean 40 minutes. Suppose a pair of players arrives and finds all courts busy and also that k other teams are waiting in queue. What is the expected waiting time to get a court?.
Q2. An athletic facility has 1 tennis court. Pairs of players arrive at the courts and use a court for an exponentially distributed amount of time with mean 40 minutes. Suppose a pair of players arrives and finds that a team just went inside and after that k other teams are waiting in queue. What is the expected waiting time to get a court?.
I feel that the answer to Q1 should be 40k + 20 and answer to Q2 should also be 40k + 40. My TA says that answer to both the questions is 40k+40 . I am not getting the reason.
probability expected-value
probability expected-value
asked Nov 21 at 13:38
Raj Rajvir
161
asked Nov 21 at 13:38
Raj Rajvir
161
edited Nov 21 at 18:36
asked Nov 21 at 13:38
Raj Rajvir
161
asked Nov 21 at 13:38
Raj Rajvir
161
asked Nov 21 at 13:38
Raj Rajvir
161
161
Because the exponential distribution has no memory. So the probability of the players staying on the court another 40 min from when you first observe them is the same as the probability of them staying on the court for 40 min from when they first started.
– user121049
Nov 21 at 18:50
Expanding on my previous comment. Say they have already being playing for a time $t_1$ when you arrive, then the probability that they will play a further amount $Delta t$ is given by $P(T>t_1+Delta t|T>t_1)$ It does not take much to show that this is equal to $P(T>t_1)$. So the expected time they will continue to play is still the mean play time of 40 min.
– user121049
Nov 21 at 20:35
A mistake in my comment which I can no longer edit. Should read more like $P(T>t_1+Delta t|T>t_1)=P(T>Delta t)$
– user121049
Nov 22 at 14:12
I understood it. Thank you very much @user121049
– Raj Rajvir
Nov 22 at 16:38
add a comment |
Because the exponential distribution has no memory. So the probability of the players staying on the court another 40 min from when you first observe them is the same as the probability of them staying on the court for 40 min from when they first started.
– user121049
Nov 21 at 18:50
Expanding on my previous comment. Say they have already being playing for a time $t_1$ when you arrive, then the probability that they will play a further amount $Delta t$ is given by $P(T>t_1+Delta t|T>t_1)$ It does not take much to show that this is equal to $P(T>t_1)$. So the expected time they will continue to play is still the mean play time of 40 min.
– user121049
Nov 21 at 20:35
A mistake in my comment which I can no longer edit. Should read more like $P(T>t_1+Delta t|T>t_1)=P(T>Delta t)$
– user121049
Nov 22 at 14:12
I understood it. Thank you very much @user121049
– Raj Rajvir
Nov 22 at 16:38
Because the exponential distribution has no memory. So the probability of the players staying on the court another 40 min from when you first observe them is the same as the probability of them staying on the court for 40 min from when they first started.
– user121049
Nov 21 at 18:50
Because the exponential distribution has no memory. So the probability of the players staying on the court another 40 min from when you first observe them is the same as the probability of them staying on the court for 40 min from when they first started.
– user121049
Nov 21 at 18:50
Expanding on my previous comment. Say they have already being playing for a time $t_1$ when you arrive, then the probability that they will play a further amount $Delta t$ is given by $P(T>t_1+Delta t|T>t_1)$ It does not take much to show that this is equal to $P(T>t_1)$. So the expected time they will continue to play is still the mean play time of 40 min.
– user121049
Nov 21 at 20:35
Expanding on my previous comment. Say they have already being playing for a time $t_1$ when you arrive, then the probability that they will play a further amount $Delta t$ is given by $P(T>t_1+Delta t|T>t_1)$ It does not take much to show that this is equal to $P(T>t_1)$. So the expected time they will continue to play is still the mean play time of 40 min.
– user121049
Nov 21 at 20:35
A mistake in my comment which I can no longer edit. Should read more like $P(T>t_1+Delta t|T>t_1)=P(T>Delta t)$
– user121049
Nov 22 at 14:12
A mistake in my comment which I can no longer edit. Should read more like $P(T>t_1+Delta t|T>t_1)=P(T>Delta t)$
– user121049
Nov 22 at 14:12
I understood it. Thank you very much @user121049
– Raj Rajvir
Nov 22 at 16:38
I understood it. Thank you very much @user121049
– Raj Rajvir
Nov 22 at 16:38
add a comment |
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Because the exponential distribution has no memory. So the probability of the players staying on the court another 40 min from when you first observe them is the same as the probability of them staying on the court for 40 min from when they first started.
– user121049
Nov 21 at 18:50
Expanding on my previous comment. Say they have already being playing for a time $t_1$ when you arrive, then the probability that they will play a further amount $Delta t$ is given by $P(T>t_1+Delta t|T>t_1)$ It does not take much to show that this is equal to $P(T>t_1)$. So the expected time they will continue to play is still the mean play time of 40 min.
– user121049
Nov 21 at 20:35
A mistake in my comment which I can no longer edit. Should read more like $P(T>t_1+Delta t|T>t_1)=P(T>Delta t)$
– user121049
Nov 22 at 14:12
I understood it. Thank you very much @user121049
– Raj Rajvir
Nov 22 at 16:38