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Expected value of dice rolls to get a non decreasing sequence of roll values.
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What is the expected number of rolls to get a non decreasing sequence of dice roll values ? Suppose one rolls 1-2-5-6-4, then after rolling 6 he will stop, since 4 > 6.
probability expected-value
asked Sep 2 at 4:13
Bibhas Ranjan Das
164
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up vote
3
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favorite
What is the expected number of rolls to get a non decreasing sequence of dice roll values ? Suppose one rolls 1-2-5-6-4, then after rolling 6 he will stop, since 4 > 6.
probability expected-value
asked Sep 2 at 4:13
Bibhas Ranjan Das
164
1
I think where you write "since $4lt6$" you mean "since $4le6$"? (Otherwise it should be "weakly increasing" instead of "strictly increasing".)
– joriki
Sep 2 at 5:36
Please correct the question. The question should be understandable on its own, without the comments.
– joriki
Sep 2 at 9:58
You could've just changed $4lt6$ to $4le6$. :-)
– joriki
Sep 2 at 11:16
Now you've changed "strictly increasing" to "non-decreasing". That turns this into a whole new question. That's a very bad idea, since there are already three answers to the original question. Please reinstate the original question; if you want to ask the same question about non-decreasing sequences, please ask a new, separate question.
– joriki
Sep 2 at 22:25
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up vote
3
down vote
favorite
up vote
3
down vote
favorite
What is the expected number of rolls to get a non decreasing sequence of dice roll values ? Suppose one rolls 1-2-5-6-4, then after rolling 6 he will stop, since 4 > 6.
probability expected-value
asked Sep 2 at 4:13
Bibhas Ranjan Das
164
What is the expected number of rolls to get a non decreasing sequence of dice roll values ? Suppose one rolls 1-2-5-6-4, then after rolling 6 he will stop, since 4 > 6.
probability expected-value
asked Sep 2 at 4:13
Bibhas Ranjan Das
164
edited Sep 2 at 13:34
asked Sep 2 at 4:13
Bibhas Ranjan Das
164
asked Sep 2 at 4:13
Bibhas Ranjan Das
164
asked Sep 2 at 4:13
Bibhas Ranjan Das
164
164
1
I think where you write "since $4lt6$" you mean "since $4le6$"? (Otherwise it should be "weakly increasing" instead of "strictly increasing".)
– joriki
Sep 2 at 5:36
Please correct the question. The question should be understandable on its own, without the comments.
– joriki
Sep 2 at 9:58
You could've just changed $4lt6$ to $4le6$. :-)
– joriki
Sep 2 at 11:16
Now you've changed "strictly increasing" to "non-decreasing". That turns this into a whole new question. That's a very bad idea, since there are already three answers to the original question. Please reinstate the original question; if you want to ask the same question about non-decreasing sequences, please ask a new, separate question.
– joriki
Sep 2 at 22:25
add a comment |Â
1
I think where you write "since $4lt6$" you mean "since $4le6$"? (Otherwise it should be "weakly increasing" instead of "strictly increasing".)
– joriki
Sep 2 at 5:36
Please correct the question. The question should be understandable on its own, without the comments.
– joriki
Sep 2 at 9:58
You could've just changed $4lt6$ to $4le6$. :-)
– joriki
Sep 2 at 11:16
Now you've changed "strictly increasing" to "non-decreasing". That turns this into a whole new question. That's a very bad idea, since there are already three answers to the original question. Please reinstate the original question; if you want to ask the same question about non-decreasing sequences, please ask a new, separate question.
– joriki
Sep 2 at 22:25
1
1
I think where you write "since $4lt6$" you mean "since $4le6$"? (Otherwise it should be "weakly increasing" instead of "strictly increasing".)
– joriki
Sep 2 at 5:36
I think where you write "since $4lt6$" you mean "since $4le6$"? (Otherwise it should be "weakly increasing" instead of "strictly increasing".)
– joriki
Sep 2 at 5:36
Please correct the question. The question should be understandable on its own, without the comments.
– joriki
Sep 2 at 9:58
Please correct the question. The question should be understandable on its own, without the comments.
– joriki
Sep 2 at 9:58
You could've just changed $4lt6$ to $4le6$. :-)
– joriki
Sep 2 at 11:16
You could've just changed $4lt6$ to $4le6$. :-)
– joriki
Sep 2 at 11:16
Now you've changed "strictly increasing" to "non-decreasing". That turns this into a whole new question. That's a very bad idea, since there are already three answers to the original question. Please reinstate the original question; if you want to ask the same question about non-decreasing sequences, please ask a new, separate question.
– joriki
Sep 2 at 22:25
Now you've changed "strictly increasing" to "non-decreasing". That turns this into a whole new question. That's a very bad idea, since there are already three answers to the original question. Please reinstate the original question; if you want to ask the same question about non-decreasing sequences, please ask a new, separate question.
– joriki
Sep 2 at 22:25
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
Ross has already shown that the expected sum is simply $n$.
For the expected number of rolls, note that $binom nk$ out of the $n^k$ possible results of $k$ rolls are strictly increasing sequences. Thus the expected value of the number $K$ of rolls including the last, non-increasing roll is
$$
mathsf E[K]=sum_k=0^nmathsf P(Kgt k)=sum_k=0^nbinom nkleft(frac1nright)^k=left(1+frac1nright)^ntomathrm equadtextas $ntoinfty$;.
$$
I'm not sure whether you wanted to count the last roll; if not, you'd obviously have to subtract $1$ from that.
See also Probability of winning dice game and Interview Question on Probability: A and B toss a dice with 1 to n faces in an alternative way.
answered Sep 2 at 5:52
joriki
167k10180333
add a comment |Â
up vote
2
down vote
Start from the top. If you roll a $6$ the expected sum is $6$ because you have to stop. If you roll a $5$ the expected sum is $5+frac 16cdot 6$ because you have $frac 16$ chance to roll a $6$. If you roll a $4$ the expected sum is $4 + frac 16cdot 6 + frac 16cdot 6$ because you have $frac 16$ chance to roll each of $5$ or $6$. You should be able to see the pattern-the expected value is $6$
For $n$ sided dice, the pseudocode for the sum would be
return n
The same approach works for number of rolls. If you roll a $6$ there will be just $1$. If you roll a $5$ the expected number is $frac 76$. Keep going down the chain, then average them all for the first roll.
answered Sep 2 at 5:10
Ross Millikan
279k22189355
If you roll a 4, why is the expected sum 4+(1/6)⋅6 [why is this 6?] +(1/6) ⋅6?
– Bibhas Ranjan Das
Sep 4 at 12:31
@BibhasRanjanDas: Because the expected addition for rolling a $5$ is $6$ as I just calculated.
– Ross Millikan
Sep 4 at 13:58
add a comment |Â
up vote
2
down vote
Let $X_1,X_2,....X_m,...$ be the dice rolls for an $n$-sided dice.
P(Rolls > $m$) = $sum_i_1 < i_2 < i_3 ... < i_m prod_j=1^m P(X_j = i_j)$
= $n choose m frac1n^m$
answered Sep 2 at 11:11
Balaji sb
37315
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Ross has already shown that the expected sum is simply $n$.
For the expected number of rolls, note that $binom nk$ out of the $n^k$ possible results of $k$ rolls are strictly increasing sequences. Thus the expected value of the number $K$ of rolls including the last, non-increasing roll is
$$
mathsf E[K]=sum_k=0^nmathsf P(Kgt k)=sum_k=0^nbinom nkleft(frac1nright)^k=left(1+frac1nright)^ntomathrm equadtextas $ntoinfty$;.
$$
I'm not sure whether you wanted to count the last roll; if not, you'd obviously have to subtract $1$ from that.
See also Probability of winning dice game and Interview Question on Probability: A and B toss a dice with 1 to n faces in an alternative way.
answered Sep 2 at 5:52
joriki
167k10180333
add a comment |Â
up vote
3
down vote
Ross has already shown that the expected sum is simply $n$.
For the expected number of rolls, note that $binom nk$ out of the $n^k$ possible results of $k$ rolls are strictly increasing sequences. Thus the expected value of the number $K$ of rolls including the last, non-increasing roll is
$$
mathsf E[K]=sum_k=0^nmathsf P(Kgt k)=sum_k=0^nbinom nkleft(frac1nright)^k=left(1+frac1nright)^ntomathrm equadtextas $ntoinfty$;.
$$
I'm not sure whether you wanted to count the last roll; if not, you'd obviously have to subtract $1$ from that.
See also Probability of winning dice game and Interview Question on Probability: A and B toss a dice with 1 to n faces in an alternative way.
answered Sep 2 at 5:52
joriki
167k10180333
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Ross has already shown that the expected sum is simply $n$.
For the expected number of rolls, note that $binom nk$ out of the $n^k$ possible results of $k$ rolls are strictly increasing sequences. Thus the expected value of the number $K$ of rolls including the last, non-increasing roll is
$$
mathsf E[K]=sum_k=0^nmathsf P(Kgt k)=sum_k=0^nbinom nkleft(frac1nright)^k=left(1+frac1nright)^ntomathrm equadtextas $ntoinfty$;.
$$
I'm not sure whether you wanted to count the last roll; if not, you'd obviously have to subtract $1$ from that.
See also Probability of winning dice game and Interview Question on Probability: A and B toss a dice with 1 to n faces in an alternative way.
answered Sep 2 at 5:52
joriki
167k10180333
Ross has already shown that the expected sum is simply $n$.
For the expected number of rolls, note that $binom nk$ out of the $n^k$ possible results of $k$ rolls are strictly increasing sequences. Thus the expected value of the number $K$ of rolls including the last, non-increasing roll is
$$
mathsf E[K]=sum_k=0^nmathsf P(Kgt k)=sum_k=0^nbinom nkleft(frac1nright)^k=left(1+frac1nright)^ntomathrm equadtextas $ntoinfty$;.
$$
I'm not sure whether you wanted to count the last roll; if not, you'd obviously have to subtract $1$ from that.
See also Probability of winning dice game and Interview Question on Probability: A and B toss a dice with 1 to n faces in an alternative way.
answered Sep 2 at 5:52
joriki
167k10180333
answered Sep 2 at 5:52
joriki
167k10180333
answered Sep 2 at 5:52
joriki
167k10180333
answered Sep 2 at 5:52
joriki
167k10180333
167k10180333
add a comment |Â
add a comment |Â
up vote
2
down vote
Start from the top. If you roll a $6$ the expected sum is $6$ because you have to stop. If you roll a $5$ the expected sum is $5+frac 16cdot 6$ because you have $frac 16$ chance to roll a $6$. If you roll a $4$ the expected sum is $4 + frac 16cdot 6 + frac 16cdot 6$ because you have $frac 16$ chance to roll each of $5$ or $6$. You should be able to see the pattern-the expected value is $6$
For $n$ sided dice, the pseudocode for the sum would be
return n
The same approach works for number of rolls. If you roll a $6$ there will be just $1$. If you roll a $5$ the expected number is $frac 76$. Keep going down the chain, then average them all for the first roll.
answered Sep 2 at 5:10
Ross Millikan
279k22189355
If you roll a 4, why is the expected sum 4+(1/6)⋅6 [why is this 6?] +(1/6) ⋅6?
– Bibhas Ranjan Das
Sep 4 at 12:31
@BibhasRanjanDas: Because the expected addition for rolling a $5$ is $6$ as I just calculated.
– Ross Millikan
Sep 4 at 13:58
add a comment |Â
up vote
2
down vote
Start from the top. If you roll a $6$ the expected sum is $6$ because you have to stop. If you roll a $5$ the expected sum is $5+frac 16cdot 6$ because you have $frac 16$ chance to roll a $6$. If you roll a $4$ the expected sum is $4 + frac 16cdot 6 + frac 16cdot 6$ because you have $frac 16$ chance to roll each of $5$ or $6$. You should be able to see the pattern-the expected value is $6$
For $n$ sided dice, the pseudocode for the sum would be
return n
The same approach works for number of rolls. If you roll a $6$ there will be just $1$. If you roll a $5$ the expected number is $frac 76$. Keep going down the chain, then average them all for the first roll.
answered Sep 2 at 5:10
Ross Millikan
279k22189355
If you roll a 4, why is the expected sum 4+(1/6)⋅6 [why is this 6?] +(1/6) ⋅6?
– Bibhas Ranjan Das
Sep 4 at 12:31
@BibhasRanjanDas: Because the expected addition for rolling a $5$ is $6$ as I just calculated.
– Ross Millikan
Sep 4 at 13:58
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Start from the top. If you roll a $6$ the expected sum is $6$ because you have to stop. If you roll a $5$ the expected sum is $5+frac 16cdot 6$ because you have $frac 16$ chance to roll a $6$. If you roll a $4$ the expected sum is $4 + frac 16cdot 6 + frac 16cdot 6$ because you have $frac 16$ chance to roll each of $5$ or $6$. You should be able to see the pattern-the expected value is $6$
For $n$ sided dice, the pseudocode for the sum would be
return n
The same approach works for number of rolls. If you roll a $6$ there will be just $1$. If you roll a $5$ the expected number is $frac 76$. Keep going down the chain, then average them all for the first roll.
answered Sep 2 at 5:10
Ross Millikan
279k22189355
Start from the top. If you roll a $6$ the expected sum is $6$ because you have to stop. If you roll a $5$ the expected sum is $5+frac 16cdot 6$ because you have $frac 16$ chance to roll a $6$. If you roll a $4$ the expected sum is $4 + frac 16cdot 6 + frac 16cdot 6$ because you have $frac 16$ chance to roll each of $5$ or $6$. You should be able to see the pattern-the expected value is $6$
For $n$ sided dice, the pseudocode for the sum would be
return n
The same approach works for number of rolls. If you roll a $6$ there will be just $1$. If you roll a $5$ the expected number is $frac 76$. Keep going down the chain, then average them all for the first roll.
answered Sep 2 at 5:10
Ross Millikan
279k22189355
answered Sep 2 at 5:10
Ross Millikan
279k22189355
answered Sep 2 at 5:10
Ross Millikan
279k22189355
answered Sep 2 at 5:10
Ross Millikan
279k22189355
279k22189355
If you roll a 4, why is the expected sum 4+(1/6)⋅6 [why is this 6?] +(1/6) ⋅6?
– Bibhas Ranjan Das
Sep 4 at 12:31
@BibhasRanjanDas: Because the expected addition for rolling a $5$ is $6$ as I just calculated.
– Ross Millikan
Sep 4 at 13:58
add a comment |Â
If you roll a 4, why is the expected sum 4+(1/6)⋅6 [why is this 6?] +(1/6) ⋅6?
– Bibhas Ranjan Das
Sep 4 at 12:31
@BibhasRanjanDas: Because the expected addition for rolling a $5$ is $6$ as I just calculated.
– Ross Millikan
Sep 4 at 13:58
If you roll a 4, why is the expected sum 4+(1/6)⋅6 [why is this 6?] +(1/6) ⋅6?
– Bibhas Ranjan Das
Sep 4 at 12:31
If you roll a 4, why is the expected sum 4+(1/6)⋅6 [why is this 6?] +(1/6) ⋅6?
– Bibhas Ranjan Das
Sep 4 at 12:31
@BibhasRanjanDas: Because the expected addition for rolling a $5$ is $6$ as I just calculated.
– Ross Millikan
Sep 4 at 13:58
@BibhasRanjanDas: Because the expected addition for rolling a $5$ is $6$ as I just calculated.
– Ross Millikan
Sep 4 at 13:58
add a comment |Â
up vote
2
down vote
Let $X_1,X_2,....X_m,...$ be the dice rolls for an $n$-sided dice.
P(Rolls > $m$) = $sum_i_1 < i_2 < i_3 ... < i_m prod_j=1^m P(X_j = i_j)$
= $n choose m frac1n^m$
answered Sep 2 at 11:11
Balaji sb
37315
add a comment |Â
up vote
2
down vote
Let $X_1,X_2,....X_m,...$ be the dice rolls for an $n$-sided dice.
P(Rolls > $m$) = $sum_i_1 < i_2 < i_3 ... < i_m prod_j=1^m P(X_j = i_j)$
= $n choose m frac1n^m$
answered Sep 2 at 11:11
Balaji sb
37315
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Let $X_1,X_2,....X_m,...$ be the dice rolls for an $n$-sided dice.
P(Rolls > $m$) = $sum_i_1 < i_2 < i_3 ... < i_m prod_j=1^m P(X_j = i_j)$
= $n choose m frac1n^m$
answered Sep 2 at 11:11
Balaji sb
37315
Let $X_1,X_2,....X_m,...$ be the dice rolls for an $n$-sided dice.
P(Rolls > $m$) = $sum_i_1 < i_2 < i_3 ... < i_m prod_j=1^m P(X_j = i_j)$
= $n choose m frac1n^m$
answered Sep 2 at 11:11
Balaji sb
37315
answered Sep 2 at 11:11
Balaji sb
37315
answered Sep 2 at 11:11
Balaji sb
37315
answered Sep 2 at 11:11
Balaji sb
37315
37315
add a comment |Â
add a comment |Â
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1
I think where you write "since $4lt6$" you mean "since $4le6$"? (Otherwise it should be "weakly increasing" instead of "strictly increasing".)
– joriki
Sep 2 at 5:36
Please correct the question. The question should be understandable on its own, without the comments.
– joriki
Sep 2 at 9:58
You could've just changed $4lt6$ to $4le6$. :-)
– joriki
Sep 2 at 11:16
Now you've changed "strictly increasing" to "non-decreasing". That turns this into a whole new question. That's a very bad idea, since there are already three answers to the original question. Please reinstate the original question; if you want to ask the same question about non-decreasing sequences, please ask a new, separate question.
– joriki
Sep 2 at 22:25