Mixing
A dice game that allows infinite rolls
Clash Royale CLAN TAG#URR8PPP
up vote
9
down vote
favorite
You will play a game with a fair 6-sided die.
You will throw the die and as long as the result of the throw is greater than or equal to the previous throw, you will continue throwing. If the throw is lower than the previous one, you will stop and get as many points as the sum of all throws, including the last one.
For example, if you get 2, 5, 5, and 3 as a result of 4 throws, the game will end with 15 points.
What is the expected value of the points you will get at the end of the game?
probability sequences-and-series
asked 3 hours ago
Taner
1055
add a comment |Â
up vote
9
down vote
favorite
You will play a game with a fair 6-sided die.
You will throw the die and as long as the result of the throw is greater than or equal to the previous throw, you will continue throwing. If the throw is lower than the previous one, you will stop and get as many points as the sum of all throws, including the last one.
For example, if you get 2, 5, 5, and 3 as a result of 4 throws, the game will end with 15 points.
What is the expected value of the points you will get at the end of the game?
probability sequences-and-series
asked 3 hours ago
Taner
1055
I tried a number of things, but the repetition of the same roll is where i am stuck. My approach is this: The question suggests that if you roll 1-2-1, you'll get 4 points. And the possibility of that happening is 1/216. When we multiply these we get 4/216. If we do this for all sequences and sum them all, we may have the answer. But i dont know how to calculate the 1-1-1-...-1-2-1 situation.
– Taner
3 hours ago
Also the fact that more than one number can repeat is confusing.
– Taner
3 hours ago
add a comment |Â
up vote
9
down vote
favorite
up vote
9
down vote
favorite
You will play a game with a fair 6-sided die.
You will throw the die and as long as the result of the throw is greater than or equal to the previous throw, you will continue throwing. If the throw is lower than the previous one, you will stop and get as many points as the sum of all throws, including the last one.
For example, if you get 2, 5, 5, and 3 as a result of 4 throws, the game will end with 15 points.
What is the expected value of the points you will get at the end of the game?
probability sequences-and-series
asked 3 hours ago
Taner
1055
You will play a game with a fair 6-sided die.
You will throw the die and as long as the result of the throw is greater than or equal to the previous throw, you will continue throwing. If the throw is lower than the previous one, you will stop and get as many points as the sum of all throws, including the last one.
For example, if you get 2, 5, 5, and 3 as a result of 4 throws, the game will end with 15 points.
What is the expected value of the points you will get at the end of the game?
probability sequences-and-series
probability sequences-and-series
asked 3 hours ago
Taner
1055
asked 3 hours ago
Taner
1055
asked 3 hours ago
Taner
1055
asked 3 hours ago
Taner
1055
asked 3 hours ago
Taner
1055
1055
I tried a number of things, but the repetition of the same roll is where i am stuck. My approach is this: The question suggests that if you roll 1-2-1, you'll get 4 points. And the possibility of that happening is 1/216. When we multiply these we get 4/216. If we do this for all sequences and sum them all, we may have the answer. But i dont know how to calculate the 1-1-1-...-1-2-1 situation.
– Taner
3 hours ago
Also the fact that more than one number can repeat is confusing.
– Taner
3 hours ago
add a comment |Â
I tried a number of things, but the repetition of the same roll is where i am stuck. My approach is this: The question suggests that if you roll 1-2-1, you'll get 4 points. And the possibility of that happening is 1/216. When we multiply these we get 4/216. If we do this for all sequences and sum them all, we may have the answer. But i dont know how to calculate the 1-1-1-...-1-2-1 situation.
– Taner
3 hours ago
Also the fact that more than one number can repeat is confusing.
– Taner
3 hours ago
I tried a number of things, but the repetition of the same roll is where i am stuck. My approach is this: The question suggests that if you roll 1-2-1, you'll get 4 points. And the possibility of that happening is 1/216. When we multiply these we get 4/216. If we do this for all sequences and sum them all, we may have the answer. But i dont know how to calculate the 1-1-1-...-1-2-1 situation.
– Taner
3 hours ago
I tried a number of things, but the repetition of the same roll is where i am stuck. My approach is this: The question suggests that if you roll 1-2-1, you'll get 4 points. And the possibility of that happening is 1/216. When we multiply these we get 4/216. If we do this for all sequences and sum them all, we may have the answer. But i dont know how to calculate the 1-1-1-...-1-2-1 situation.
– Taner
3 hours ago
Also the fact that more than one number can repeat is confusing.
– Taner
3 hours ago
Also the fact that more than one number can repeat is confusing.
– Taner
3 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
Consider the expected number of points you will obtain after rolling a particular number.
e.g. Suppose we have just rolled a 6. We need to roll another 6 to keep playing, otherwise we stop. Using conditional expectation, we can compute the number of future points we expect to obtain, $E_6$: $$E_6 = frac16(E_6 + 6) + frac56times 3.$$
The first term corresponds to rolling a 6, and we are back in the same position as before, just with 6 extra points. The second term gives expected number of points obtained, given that we roll a lower number and stop playing. We solve this to obtain $$E_6 = frac215.$$
Now suppose that, in a new game, we have just rolled a 5. What is the expected number of points from this point, $E_5$? Using the same conditional expectation rules as before: $$E_5 = frac16(E_5+5) + frac16(E_6+6) + frac46 times frac52.$$ Since we know $E_6$ from above, we can now solve for $E_5$.
Repeat this for $E_4, dots, E_1$, and we know the expected number of future points given the most recent roll number.
Finally, the expected number of points, $E$, will be the conditional sum of these values, i.e.
beginequation
E = sum_n=1^6 frac16(E_n + n). tag1
endequation
Edit: following an observation by Taner, I thought I'd add a few more lines.
After some generalizing and rearranging, we obtain $$E_n = frac15left(sum_m=n+1^6 E_m + 21right),$$ and we can set up a recurrence relation for $E_n$ so we don't have to do this sum every time to compute it. We have
beginalign
E_n+1 &= frac15left(sum_m=n+2^6 E_m + 21right) \ &= frac15left(sum_m=n+1^6 E_m + 21 - E_n+1right) \ &= E_n - frac15E_n+1,
endalign
which we rearrange to obtain $$E_n+1 = frac56E_n,$$ which has solution $$E_n = C left(frac56right)^n,$$ where $C$ is some constant, for which we can solve by setting $n=6$ and using our value of $E_6$. We obtain (allowing for possible arithmetic errors made by me) $$E_n = frac215left(frac56right)^n-6.$$
We can substitute this into Equation (1) and use geometric and arithmetic sum formulae to obtain the final answer.
Edit 2: as Alex Zorn pointed out, we don't even need to do this geometric and arithmetic sum stuff in Equation (1). Note that the game is the same before and after rolling a 1, so this tells us straight away that $$E = E_1 = frac215left(frac56right)^-5 = frac16329615625 approx 10.45.$$
answered 3 hours ago
combinatoricky
1014
New contributor
combinatoricky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Very well and clearly explained. Thanks a lot!
– Taner
3 hours ago
There are a few simplifications that can be made, and one can derive a slightly more general formula for $E_n$ (given the "higher" ones) which would save you from having to manually compute each one, and of course you can use the formula for summing integers if you want (although summing 1 to 6 manually won't consume too much time!). Anyway, go ahead and mark the answer as accepted if you feel it adequately answered your question, and good luck!
– combinatoricky
1 hour ago
I noticed that the E values can be obtained by simply multiplicating the previous one by 6/5. But im calculating each one anyway lol. Thanks again.
– Taner
1 hour ago
Good observation, you're right! I've added more lines on to my answer to derive and build on this observation.
– combinatoricky
56 mins ago
3
You don't even need a geometric sum formula: The game after you've rolled a 1 isn't different at all from the game at the start, so the expected value of this game is simply $E_1$.
– Alex Zorn
30 mins ago
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Consider the expected number of points you will obtain after rolling a particular number.
e.g. Suppose we have just rolled a 6. We need to roll another 6 to keep playing, otherwise we stop. Using conditional expectation, we can compute the number of future points we expect to obtain, $E_6$: $$E_6 = frac16(E_6 + 6) + frac56times 3.$$
The first term corresponds to rolling a 6, and we are back in the same position as before, just with 6 extra points. The second term gives expected number of points obtained, given that we roll a lower number and stop playing. We solve this to obtain $$E_6 = frac215.$$
Now suppose that, in a new game, we have just rolled a 5. What is the expected number of points from this point, $E_5$? Using the same conditional expectation rules as before: $$E_5 = frac16(E_5+5) + frac16(E_6+6) + frac46 times frac52.$$ Since we know $E_6$ from above, we can now solve for $E_5$.
Repeat this for $E_4, dots, E_1$, and we know the expected number of future points given the most recent roll number.
Finally, the expected number of points, $E$, will be the conditional sum of these values, i.e.
beginequation
E = sum_n=1^6 frac16(E_n + n). tag1
endequation
Edit: following an observation by Taner, I thought I'd add a few more lines.
After some generalizing and rearranging, we obtain $$E_n = frac15left(sum_m=n+1^6 E_m + 21right),$$ and we can set up a recurrence relation for $E_n$ so we don't have to do this sum every time to compute it. We have
beginalign
E_n+1 &= frac15left(sum_m=n+2^6 E_m + 21right) \ &= frac15left(sum_m=n+1^6 E_m + 21 - E_n+1right) \ &= E_n - frac15E_n+1,
endalign
which we rearrange to obtain $$E_n+1 = frac56E_n,$$ which has solution $$E_n = C left(frac56right)^n,$$ where $C$ is some constant, for which we can solve by setting $n=6$ and using our value of $E_6$. We obtain (allowing for possible arithmetic errors made by me) $$E_n = frac215left(frac56right)^n-6.$$
We can substitute this into Equation (1) and use geometric and arithmetic sum formulae to obtain the final answer.
Edit 2: as Alex Zorn pointed out, we don't even need to do this geometric and arithmetic sum stuff in Equation (1). Note that the game is the same before and after rolling a 1, so this tells us straight away that $$E = E_1 = frac215left(frac56right)^-5 = frac16329615625 approx 10.45.$$
answered 3 hours ago
combinatoricky
1014
New contributor
combinatoricky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Very well and clearly explained. Thanks a lot!
– Taner
3 hours ago
There are a few simplifications that can be made, and one can derive a slightly more general formula for $E_n$ (given the "higher" ones) which would save you from having to manually compute each one, and of course you can use the formula for summing integers if you want (although summing 1 to 6 manually won't consume too much time!). Anyway, go ahead and mark the answer as accepted if you feel it adequately answered your question, and good luck!
– combinatoricky
1 hour ago
I noticed that the E values can be obtained by simply multiplicating the previous one by 6/5. But im calculating each one anyway lol. Thanks again.
– Taner
1 hour ago
Good observation, you're right! I've added more lines on to my answer to derive and build on this observation.
– combinatoricky
56 mins ago
3
You don't even need a geometric sum formula: The game after you've rolled a 1 isn't different at all from the game at the start, so the expected value of this game is simply $E_1$.
– Alex Zorn
30 mins ago
 |Â
show 2 more comments
up vote
4
down vote
accepted
Consider the expected number of points you will obtain after rolling a particular number.
e.g. Suppose we have just rolled a 6. We need to roll another 6 to keep playing, otherwise we stop. Using conditional expectation, we can compute the number of future points we expect to obtain, $E_6$: $$E_6 = frac16(E_6 + 6) + frac56times 3.$$
The first term corresponds to rolling a 6, and we are back in the same position as before, just with 6 extra points. The second term gives expected number of points obtained, given that we roll a lower number and stop playing. We solve this to obtain $$E_6 = frac215.$$
Now suppose that, in a new game, we have just rolled a 5. What is the expected number of points from this point, $E_5$? Using the same conditional expectation rules as before: $$E_5 = frac16(E_5+5) + frac16(E_6+6) + frac46 times frac52.$$ Since we know $E_6$ from above, we can now solve for $E_5$.
Repeat this for $E_4, dots, E_1$, and we know the expected number of future points given the most recent roll number.
Finally, the expected number of points, $E$, will be the conditional sum of these values, i.e.
beginequation
E = sum_n=1^6 frac16(E_n + n). tag1
endequation
Edit: following an observation by Taner, I thought I'd add a few more lines.
After some generalizing and rearranging, we obtain $$E_n = frac15left(sum_m=n+1^6 E_m + 21right),$$ and we can set up a recurrence relation for $E_n$ so we don't have to do this sum every time to compute it. We have
beginalign
E_n+1 &= frac15left(sum_m=n+2^6 E_m + 21right) \ &= frac15left(sum_m=n+1^6 E_m + 21 - E_n+1right) \ &= E_n - frac15E_n+1,
endalign
which we rearrange to obtain $$E_n+1 = frac56E_n,$$ which has solution $$E_n = C left(frac56right)^n,$$ where $C$ is some constant, for which we can solve by setting $n=6$ and using our value of $E_6$. We obtain (allowing for possible arithmetic errors made by me) $$E_n = frac215left(frac56right)^n-6.$$
We can substitute this into Equation (1) and use geometric and arithmetic sum formulae to obtain the final answer.
Edit 2: as Alex Zorn pointed out, we don't even need to do this geometric and arithmetic sum stuff in Equation (1). Note that the game is the same before and after rolling a 1, so this tells us straight away that $$E = E_1 = frac215left(frac56right)^-5 = frac16329615625 approx 10.45.$$
answered 3 hours ago
combinatoricky
1014
New contributor
combinatoricky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Very well and clearly explained. Thanks a lot!
– Taner
3 hours ago
There are a few simplifications that can be made, and one can derive a slightly more general formula for $E_n$ (given the "higher" ones) which would save you from having to manually compute each one, and of course you can use the formula for summing integers if you want (although summing 1 to 6 manually won't consume too much time!). Anyway, go ahead and mark the answer as accepted if you feel it adequately answered your question, and good luck!
– combinatoricky
1 hour ago
I noticed that the E values can be obtained by simply multiplicating the previous one by 6/5. But im calculating each one anyway lol. Thanks again.
– Taner
1 hour ago
Good observation, you're right! I've added more lines on to my answer to derive and build on this observation.
– combinatoricky
56 mins ago
3
You don't even need a geometric sum formula: The game after you've rolled a 1 isn't different at all from the game at the start, so the expected value of this game is simply $E_1$.
– Alex Zorn
30 mins ago
 |Â
show 2 more comments
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Consider the expected number of points you will obtain after rolling a particular number.
e.g. Suppose we have just rolled a 6. We need to roll another 6 to keep playing, otherwise we stop. Using conditional expectation, we can compute the number of future points we expect to obtain, $E_6$: $$E_6 = frac16(E_6 + 6) + frac56times 3.$$
The first term corresponds to rolling a 6, and we are back in the same position as before, just with 6 extra points. The second term gives expected number of points obtained, given that we roll a lower number and stop playing. We solve this to obtain $$E_6 = frac215.$$
Now suppose that, in a new game, we have just rolled a 5. What is the expected number of points from this point, $E_5$? Using the same conditional expectation rules as before: $$E_5 = frac16(E_5+5) + frac16(E_6+6) + frac46 times frac52.$$ Since we know $E_6$ from above, we can now solve for $E_5$.
Repeat this for $E_4, dots, E_1$, and we know the expected number of future points given the most recent roll number.
Finally, the expected number of points, $E$, will be the conditional sum of these values, i.e.
beginequation
E = sum_n=1^6 frac16(E_n + n). tag1
endequation
Edit: following an observation by Taner, I thought I'd add a few more lines.
After some generalizing and rearranging, we obtain $$E_n = frac15left(sum_m=n+1^6 E_m + 21right),$$ and we can set up a recurrence relation for $E_n$ so we don't have to do this sum every time to compute it. We have
beginalign
E_n+1 &= frac15left(sum_m=n+2^6 E_m + 21right) \ &= frac15left(sum_m=n+1^6 E_m + 21 - E_n+1right) \ &= E_n - frac15E_n+1,
endalign
which we rearrange to obtain $$E_n+1 = frac56E_n,$$ which has solution $$E_n = C left(frac56right)^n,$$ where $C$ is some constant, for which we can solve by setting $n=6$ and using our value of $E_6$. We obtain (allowing for possible arithmetic errors made by me) $$E_n = frac215left(frac56right)^n-6.$$
We can substitute this into Equation (1) and use geometric and arithmetic sum formulae to obtain the final answer.
Edit 2: as Alex Zorn pointed out, we don't even need to do this geometric and arithmetic sum stuff in Equation (1). Note that the game is the same before and after rolling a 1, so this tells us straight away that $$E = E_1 = frac215left(frac56right)^-5 = frac16329615625 approx 10.45.$$
answered 3 hours ago
combinatoricky
1014
New contributor
combinatoricky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Consider the expected number of points you will obtain after rolling a particular number.
e.g. Suppose we have just rolled a 6. We need to roll another 6 to keep playing, otherwise we stop. Using conditional expectation, we can compute the number of future points we expect to obtain, $E_6$: $$E_6 = frac16(E_6 + 6) + frac56times 3.$$
The first term corresponds to rolling a 6, and we are back in the same position as before, just with 6 extra points. The second term gives expected number of points obtained, given that we roll a lower number and stop playing. We solve this to obtain $$E_6 = frac215.$$
Now suppose that, in a new game, we have just rolled a 5. What is the expected number of points from this point, $E_5$? Using the same conditional expectation rules as before: $$E_5 = frac16(E_5+5) + frac16(E_6+6) + frac46 times frac52.$$ Since we know $E_6$ from above, we can now solve for $E_5$.
Repeat this for $E_4, dots, E_1$, and we know the expected number of future points given the most recent roll number.
Finally, the expected number of points, $E$, will be the conditional sum of these values, i.e.
beginequation
E = sum_n=1^6 frac16(E_n + n). tag1
endequation
Edit: following an observation by Taner, I thought I'd add a few more lines.
After some generalizing and rearranging, we obtain $$E_n = frac15left(sum_m=n+1^6 E_m + 21right),$$ and we can set up a recurrence relation for $E_n$ so we don't have to do this sum every time to compute it. We have
beginalign
E_n+1 &= frac15left(sum_m=n+2^6 E_m + 21right) \ &= frac15left(sum_m=n+1^6 E_m + 21 - E_n+1right) \ &= E_n - frac15E_n+1,
endalign
which we rearrange to obtain $$E_n+1 = frac56E_n,$$ which has solution $$E_n = C left(frac56right)^n,$$ where $C$ is some constant, for which we can solve by setting $n=6$ and using our value of $E_6$. We obtain (allowing for possible arithmetic errors made by me) $$E_n = frac215left(frac56right)^n-6.$$
We can substitute this into Equation (1) and use geometric and arithmetic sum formulae to obtain the final answer.
Edit 2: as Alex Zorn pointed out, we don't even need to do this geometric and arithmetic sum stuff in Equation (1). Note that the game is the same before and after rolling a 1, so this tells us straight away that $$E = E_1 = frac215left(frac56right)^-5 = frac16329615625 approx 10.45.$$
answered 3 hours ago
combinatoricky
1014
New contributor
combinatoricky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 18 mins ago
answered 3 hours ago
combinatoricky
1014
New contributor
combinatoricky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 3 hours ago
combinatoricky
1014
answered 3 hours ago
combinatoricky
1014
1014
New contributor
combinatoricky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
combinatoricky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
combinatoricky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Very well and clearly explained. Thanks a lot!
– Taner
3 hours ago
There are a few simplifications that can be made, and one can derive a slightly more general formula for $E_n$ (given the "higher" ones) which would save you from having to manually compute each one, and of course you can use the formula for summing integers if you want (although summing 1 to 6 manually won't consume too much time!). Anyway, go ahead and mark the answer as accepted if you feel it adequately answered your question, and good luck!
– combinatoricky
1 hour ago
I noticed that the E values can be obtained by simply multiplicating the previous one by 6/5. But im calculating each one anyway lol. Thanks again.
– Taner
1 hour ago
Good observation, you're right! I've added more lines on to my answer to derive and build on this observation.
– combinatoricky
56 mins ago
3
You don't even need a geometric sum formula: The game after you've rolled a 1 isn't different at all from the game at the start, so the expected value of this game is simply $E_1$.
– Alex Zorn
30 mins ago
 |Â
show 2 more comments
Very well and clearly explained. Thanks a lot!
– Taner
3 hours ago
There are a few simplifications that can be made, and one can derive a slightly more general formula for $E_n$ (given the "higher" ones) which would save you from having to manually compute each one, and of course you can use the formula for summing integers if you want (although summing 1 to 6 manually won't consume too much time!). Anyway, go ahead and mark the answer as accepted if you feel it adequately answered your question, and good luck!
– combinatoricky
1 hour ago
I noticed that the E values can be obtained by simply multiplicating the previous one by 6/5. But im calculating each one anyway lol. Thanks again.
– Taner
1 hour ago
Good observation, you're right! I've added more lines on to my answer to derive and build on this observation.
– combinatoricky
56 mins ago
3
You don't even need a geometric sum formula: The game after you've rolled a 1 isn't different at all from the game at the start, so the expected value of this game is simply $E_1$.
– Alex Zorn
30 mins ago
Very well and clearly explained. Thanks a lot!
– Taner
3 hours ago
Very well and clearly explained. Thanks a lot!
– Taner
3 hours ago
There are a few simplifications that can be made, and one can derive a slightly more general formula for $E_n$ (given the "higher" ones) which would save you from having to manually compute each one, and of course you can use the formula for summing integers if you want (although summing 1 to 6 manually won't consume too much time!). Anyway, go ahead and mark the answer as accepted if you feel it adequately answered your question, and good luck!
– combinatoricky
1 hour ago
There are a few simplifications that can be made, and one can derive a slightly more general formula for $E_n$ (given the "higher" ones) which would save you from having to manually compute each one, and of course you can use the formula for summing integers if you want (although summing 1 to 6 manually won't consume too much time!). Anyway, go ahead and mark the answer as accepted if you feel it adequately answered your question, and good luck!
– combinatoricky
1 hour ago
I noticed that the E values can be obtained by simply multiplicating the previous one by 6/5. But im calculating each one anyway lol. Thanks again.
– Taner
1 hour ago
I noticed that the E values can be obtained by simply multiplicating the previous one by 6/5. But im calculating each one anyway lol. Thanks again.
– Taner
1 hour ago
Good observation, you're right! I've added more lines on to my answer to derive and build on this observation.
– combinatoricky
56 mins ago
Good observation, you're right! I've added more lines on to my answer to derive and build on this observation.
– combinatoricky
56 mins ago
3
3
You don't even need a geometric sum formula: The game after you've rolled a 1 isn't different at all from the game at the start, so the expected value of this game is simply $E_1$.
– Alex Zorn
30 mins ago
You don't even need a geometric sum formula: The game after you've rolled a 1 isn't different at all from the game at the start, so the expected value of this game is simply $E_1$.
– Alex Zorn
30 mins ago
 |Â
show 2 more comments
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I tried a number of things, but the repetition of the same roll is where i am stuck. My approach is this: The question suggests that if you roll 1-2-1, you'll get 4 points. And the possibility of that happening is 1/216. When we multiply these we get 4/216. If we do this for all sequences and sum them all, we may have the answer. But i dont know how to calculate the 1-1-1-...-1-2-1 situation.
– Taner
3 hours ago
Also the fact that more than one number can repeat is confusing.
– Taner
3 hours ago