Expected Time on dice roll
$begingroup$
Wanted to just confirm whether I was doing the right thing:
Question: Jack repeatedly rolls a fair die. If it comes up 1, he instantly wins (and stops playing with no time elapsed). If it comes up $n$, where $n ∈ {2, 3, 4}$, he waits $n$ minutes and then rolls again. If it comes up $n$, where $n ∈ {5, 6}$, he waits $n^{2}$ minutes and then rolls again. What is the expected elapsed time from when he starts rolling until he wins?
I've begun by conditioning on the outcome of the roll and the corresponding time elapsed.
Hence, $E[X] = frac{1}{6}(0)+frac{1}{6}(2)+frac{1}{6}(3)+frac{1}{6}(4)+frac{1}{6}(25)+frac{1}{6}(36)=frac{35}{3} approx :11.67$
Much thanks in advance.
probability combinatorics
edited Dec 9 '18 at 4:22
bob
1089
asked Dec 9 '18 at 4:10
RaviRavi
85
$endgroup$
add a comment |
$begingroup$
Wanted to just confirm whether I was doing the right thing:
Question: Jack repeatedly rolls a fair die. If it comes up 1, he instantly wins (and stops playing with no time elapsed). If it comes up $n$, where $n ∈ {2, 3, 4}$, he waits $n$ minutes and then rolls again. If it comes up $n$, where $n ∈ {5, 6}$, he waits $n^{2}$ minutes and then rolls again. What is the expected elapsed time from when he starts rolling until he wins?
I've begun by conditioning on the outcome of the roll and the corresponding time elapsed.
Hence, $E[X] = frac{1}{6}(0)+frac{1}{6}(2)+frac{1}{6}(3)+frac{1}{6}(4)+frac{1}{6}(25)+frac{1}{6}(36)=frac{35}{3} approx :11.67$
Much thanks in advance.
probability combinatorics
edited Dec 9 '18 at 4:22
bob
1089
asked Dec 9 '18 at 4:10
RaviRavi
85
$endgroup$
add a comment |
$begingroup$
Wanted to just confirm whether I was doing the right thing:
Question: Jack repeatedly rolls a fair die. If it comes up 1, he instantly wins (and stops playing with no time elapsed). If it comes up $n$, where $n ∈ {2, 3, 4}$, he waits $n$ minutes and then rolls again. If it comes up $n$, where $n ∈ {5, 6}$, he waits $n^{2}$ minutes and then rolls again. What is the expected elapsed time from when he starts rolling until he wins?
I've begun by conditioning on the outcome of the roll and the corresponding time elapsed.
Hence, $E[X] = frac{1}{6}(0)+frac{1}{6}(2)+frac{1}{6}(3)+frac{1}{6}(4)+frac{1}{6}(25)+frac{1}{6}(36)=frac{35}{3} approx :11.67$
Much thanks in advance.
probability combinatorics
edited Dec 9 '18 at 4:22
bob
1089
asked Dec 9 '18 at 4:10
RaviRavi
85
$endgroup$
Wanted to just confirm whether I was doing the right thing:
Question: Jack repeatedly rolls a fair die. If it comes up 1, he instantly wins (and stops playing with no time elapsed). If it comes up $n$, where $n ∈ {2, 3, 4}$, he waits $n$ minutes and then rolls again. If it comes up $n$, where $n ∈ {5, 6}$, he waits $n^{2}$ minutes and then rolls again. What is the expected elapsed time from when he starts rolling until he wins?
I've begun by conditioning on the outcome of the roll and the corresponding time elapsed.
Hence, $E[X] = frac{1}{6}(0)+frac{1}{6}(2)+frac{1}{6}(3)+frac{1}{6}(4)+frac{1}{6}(25)+frac{1}{6}(36)=frac{35}{3} approx :11.67$
Much thanks in advance.
probability combinatorics
probability combinatorics
edited Dec 9 '18 at 4:22
bob
1089
asked Dec 9 '18 at 4:10
RaviRavi
85
edited Dec 9 '18 at 4:22
bob
1089
asked Dec 9 '18 at 4:10
RaviRavi
85
edited Dec 9 '18 at 4:22
bob
1089
edited Dec 9 '18 at 4:22
bob
1089
edited Dec 9 '18 at 4:22
bob
1089
1089
asked Dec 9 '18 at 4:10
RaviRavi
85
asked Dec 9 '18 at 4:10
RaviRavi
85
asked Dec 9 '18 at 4:10
RaviRavi
85
85
add a comment |
add a comment |
1 Answer
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$begingroup$
This is a really good start. Note that if he does not roll a $1$, he has to wait and then roll again. In other words, the process hasn't finished when Jack rolls a non-1 number. What you have calculated is exactly that.
Instead, what happens if he rolls a non-1? The process essentially starts over, except some number of minutes has been added to his time. We can reflect that in your equation by adding the original value $E[X]$ to each time in your sum:
$$E[X] = frac{1}{6}(0) + frac{1}{6}(2 + E[X]) + frac{1}{6}(3+E[X]) + frac{1}{6}(4+E[X]) + frac{1}{6}(25+E[X]) + frac{1}{6}(36+E[X])$$.
After this, simply solve for $E[X]$ and you're done!
Note: The reason we don't add $E[X]$ to the first $0$ is because the process actually does finish in that case.
answered Dec 9 '18 at 4:40
Adam CartisanoAdam Cartisano
1764
$endgroup$
$begingroup$
Ah, this makes sense, this is similar to the coal miner problem. Now I get it, thanks
$endgroup$
– Ravi
Dec 9 '18 at 4:44
add a comment |
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1 Answer
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1 Answer
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$begingroup$
This is a really good start. Note that if he does not roll a $1$, he has to wait and then roll again. In other words, the process hasn't finished when Jack rolls a non-1 number. What you have calculated is exactly that.
Instead, what happens if he rolls a non-1? The process essentially starts over, except some number of minutes has been added to his time. We can reflect that in your equation by adding the original value $E[X]$ to each time in your sum:
$$E[X] = frac{1}{6}(0) + frac{1}{6}(2 + E[X]) + frac{1}{6}(3+E[X]) + frac{1}{6}(4+E[X]) + frac{1}{6}(25+E[X]) + frac{1}{6}(36+E[X])$$.
After this, simply solve for $E[X]$ and you're done!
Note: The reason we don't add $E[X]$ to the first $0$ is because the process actually does finish in that case.
answered Dec 9 '18 at 4:40
Adam CartisanoAdam Cartisano
1764
$endgroup$
$begingroup$
Ah, this makes sense, this is similar to the coal miner problem. Now I get it, thanks
$endgroup$
– Ravi
Dec 9 '18 at 4:44
add a comment |
$begingroup$
This is a really good start. Note that if he does not roll a $1$, he has to wait and then roll again. In other words, the process hasn't finished when Jack rolls a non-1 number. What you have calculated is exactly that.
Instead, what happens if he rolls a non-1? The process essentially starts over, except some number of minutes has been added to his time. We can reflect that in your equation by adding the original value $E[X]$ to each time in your sum:
$$E[X] = frac{1}{6}(0) + frac{1}{6}(2 + E[X]) + frac{1}{6}(3+E[X]) + frac{1}{6}(4+E[X]) + frac{1}{6}(25+E[X]) + frac{1}{6}(36+E[X])$$.
After this, simply solve for $E[X]$ and you're done!
Note: The reason we don't add $E[X]$ to the first $0$ is because the process actually does finish in that case.
answered Dec 9 '18 at 4:40
Adam CartisanoAdam Cartisano
1764
$endgroup$
$begingroup$
Ah, this makes sense, this is similar to the coal miner problem. Now I get it, thanks
$endgroup$
– Ravi
Dec 9 '18 at 4:44
add a comment |
$begingroup$
This is a really good start. Note that if he does not roll a $1$, he has to wait and then roll again. In other words, the process hasn't finished when Jack rolls a non-1 number. What you have calculated is exactly that.
Instead, what happens if he rolls a non-1? The process essentially starts over, except some number of minutes has been added to his time. We can reflect that in your equation by adding the original value $E[X]$ to each time in your sum:
$$E[X] = frac{1}{6}(0) + frac{1}{6}(2 + E[X]) + frac{1}{6}(3+E[X]) + frac{1}{6}(4+E[X]) + frac{1}{6}(25+E[X]) + frac{1}{6}(36+E[X])$$.
After this, simply solve for $E[X]$ and you're done!
Note: The reason we don't add $E[X]$ to the first $0$ is because the process actually does finish in that case.
answered Dec 9 '18 at 4:40
Adam CartisanoAdam Cartisano
1764
$endgroup$
This is a really good start. Note that if he does not roll a $1$, he has to wait and then roll again. In other words, the process hasn't finished when Jack rolls a non-1 number. What you have calculated is exactly that.
Instead, what happens if he rolls a non-1? The process essentially starts over, except some number of minutes has been added to his time. We can reflect that in your equation by adding the original value $E[X]$ to each time in your sum:
$$E[X] = frac{1}{6}(0) + frac{1}{6}(2 + E[X]) + frac{1}{6}(3+E[X]) + frac{1}{6}(4+E[X]) + frac{1}{6}(25+E[X]) + frac{1}{6}(36+E[X])$$.
After this, simply solve for $E[X]$ and you're done!
Note: The reason we don't add $E[X]$ to the first $0$ is because the process actually does finish in that case.
answered Dec 9 '18 at 4:40
Adam CartisanoAdam Cartisano
1764
answered Dec 9 '18 at 4:40
Adam CartisanoAdam Cartisano
1764
answered Dec 9 '18 at 4:40
Adam CartisanoAdam Cartisano
1764
answered Dec 9 '18 at 4:40
Adam CartisanoAdam Cartisano
1764
1764
$begingroup$
Ah, this makes sense, this is similar to the coal miner problem. Now I get it, thanks
$endgroup$
– Ravi
Dec 9 '18 at 4:44
add a comment |
$begingroup$
Ah, this makes sense, this is similar to the coal miner problem. Now I get it, thanks
$endgroup$
– Ravi
Dec 9 '18 at 4:44
$begingroup$
Ah, this makes sense, this is similar to the coal miner problem. Now I get it, thanks
$endgroup$
– Ravi
Dec 9 '18 at 4:44
$begingroup$
Ah, this makes sense, this is similar to the coal miner problem. Now I get it, thanks
$endgroup$
– Ravi
Dec 9 '18 at 4:44
add a comment |
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