Which sum shows up most often when you roll 10 dice?
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Which sum shows up most often when you roll 10 dice?
probability
edited Feb 10 '11 at 16:12
Rasmus
14.1k14378
asked Feb 10 '11 at 15:04
Ssegawa Victor
443155
add a comment |Â
up vote
1
down vote
favorite
Which sum shows up most often when you roll 10 dice?
probability
edited Feb 10 '11 at 16:12
Rasmus
14.1k14378
asked Feb 10 '11 at 15:04
Ssegawa Victor
443155
4
Please turn off caps lock when you write the title. What have you tried-do you know the answer for 2 dice?
– Ross Millikan
Feb 10 '11 at 15:06
9
please note that this is at least the fourth time in the recent days another member of the site has requested that you not to use all caps in the question title. Consider this your moderator warning. Any further abuse will result in graduated suspension from the site.
– Willie Wong
Feb 10 '11 at 15:19
2
How many sides do your dice have?
– user489
Feb 10 '11 at 18:47
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Which sum shows up most often when you roll 10 dice?
probability
edited Feb 10 '11 at 16:12
Rasmus
14.1k14378
asked Feb 10 '11 at 15:04
Ssegawa Victor
443155
Which sum shows up most often when you roll 10 dice?
probability
edited Feb 10 '11 at 16:12
Rasmus
14.1k14378
asked Feb 10 '11 at 15:04
Ssegawa Victor
443155
edited Feb 10 '11 at 16:12
Rasmus
14.1k14378
edited Feb 10 '11 at 16:12
Rasmus
14.1k14378
edited Feb 10 '11 at 16:12
Rasmus
14.1k14378
14.1k14378
asked Feb 10 '11 at 15:04
Ssegawa Victor
443155
asked Feb 10 '11 at 15:04
Ssegawa Victor
443155
asked Feb 10 '11 at 15:04
Ssegawa Victor
443155
443155
4
Please turn off caps lock when you write the title. What have you tried-do you know the answer for 2 dice?
– Ross Millikan
Feb 10 '11 at 15:06
9
please note that this is at least the fourth time in the recent days another member of the site has requested that you not to use all caps in the question title. Consider this your moderator warning. Any further abuse will result in graduated suspension from the site.
– Willie Wong
Feb 10 '11 at 15:19
2
How many sides do your dice have?
– user489
Feb 10 '11 at 18:47
add a comment |Â
4
Please turn off caps lock when you write the title. What have you tried-do you know the answer for 2 dice?
– Ross Millikan
Feb 10 '11 at 15:06
9
please note that this is at least the fourth time in the recent days another member of the site has requested that you not to use all caps in the question title. Consider this your moderator warning. Any further abuse will result in graduated suspension from the site.
– Willie Wong
Feb 10 '11 at 15:19
2
How many sides do your dice have?
– user489
Feb 10 '11 at 18:47
4
4
Please turn off caps lock when you write the title. What have you tried-do you know the answer for 2 dice?
– Ross Millikan
Feb 10 '11 at 15:06
Please turn off caps lock when you write the title. What have you tried-do you know the answer for 2 dice?
– Ross Millikan
Feb 10 '11 at 15:06
9
9
please note that this is at least the fourth time in the recent days another member of the site has requested that you not to use all caps in the question title. Consider this your moderator warning. Any further abuse will result in graduated suspension from the site.
– Willie Wong
Feb 10 '11 at 15:19
please note that this is at least the fourth time in the recent days another member of the site has requested that you not to use all caps in the question title. Consider this your moderator warning. Any further abuse will result in graduated suspension from the site.
– Willie Wong
Feb 10 '11 at 15:19
2
2
How many sides do your dice have?
– user489
Feb 10 '11 at 18:47
How many sides do your dice have?
– user489
Feb 10 '11 at 18:47
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
4
down vote
The sum that shows up most often when you roll 10 dice is 35.
Are you also interested to see a reason why this is so?
answered Feb 10 '11 at 16:10
Rasmus
14.1k14378
I am interested
– ajbeaven
Jun 22 '11 at 22:26
@ajbeaven: I think Yuval Filmus suggestion for a proof is very good. See also this question.
– Rasmus
Jun 23 '11 at 20:52
add a comment |Â
up vote
3
down vote
The key to this question is the concept of unimodality (q.v.).
Try to prove by induction that if you take the sum of $n geq 2$ cubes then the distribution has the following form:
- It is supported (has non-zero probability) on the numbers $n, ldots, 6n$.
- It is symmetric around $3.5n$.
- It is unimodal: the sequence of probabilities increases monotonically until it reaches its mode (the most probable value), then decreases monotonically.
- If $n$ is even then the mode is unique (and so $3.5n$).
- If $n$ is odd then there are two modes ($lfloor 3.5n rfloor$ and $lceil 3.5n rceil$).
If you find the proof difficult, try again with two-sided dice.
Of course, as $n rightarrow infty$ the distribution will approach (in some sense) a Gaussian, but I'm not sure this is enough to prove anything specific about the mode (other than its asymptotics).
answered Mar 26 '11 at 17:37
Yuval Filmus
47.2k369142
add a comment |Â
up vote
-3
down vote
For each individual die, its 'expected value' is 7/2. This is because it can be either 1, 2, 3, 4, 5 or 6, each with a probability of 1/6 (assuming it's a fair die). If you want to know what value will come up most often for 1 die, this is the sum of each of it's possible values times the probability of that value. Which means 1(1/6) + 2(1/6) + 3(1/6) + 4(1/6) + 5(1/6) + 6(1/6) = 1/6(1 + 2 + 3 + 4 + 5 + 6) = 7/2. Obviously, 3.5 will never show up on a die, and this makes sense because if each side has equal probability, you shouldn't expect to see one side more than the others.
To find the expected value of multiple die, you can simply add up each individual expected value. For all 10 dice, this value is 7/2, so you have 7/2 * 10 = 35
Thus, the expected value of rolling 10 dice is 35.
Wikipedia has more information on probability and expected values, if it interests you:
http://en.wikipedia.org/wiki/Expected_value
answered Mar 25 '11 at 16:46
mozi
312
2
Expected value != Most likely value.
– Aryabhata
Mar 25 '11 at 17:16
1
hmm, would you mind explaining how to go about solving this problem then? I'm curious now!
– mozi
Mar 25 '11 at 17:57
Perhaps this will help: math.stackexchange.com/questions/4632/… . Computer software can easily compute the the required coefficients. There might even be a proof that 35 has the highest coefficient...
– Aryabhata
Mar 25 '11 at 19:19
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
The sum that shows up most often when you roll 10 dice is 35.
Are you also interested to see a reason why this is so?
answered Feb 10 '11 at 16:10
Rasmus
14.1k14378
I am interested
– ajbeaven
Jun 22 '11 at 22:26
@ajbeaven: I think Yuval Filmus suggestion for a proof is very good. See also this question.
– Rasmus
Jun 23 '11 at 20:52
add a comment |Â
up vote
4
down vote
The sum that shows up most often when you roll 10 dice is 35.
Are you also interested to see a reason why this is so?
answered Feb 10 '11 at 16:10
Rasmus
14.1k14378
I am interested
– ajbeaven
Jun 22 '11 at 22:26
@ajbeaven: I think Yuval Filmus suggestion for a proof is very good. See also this question.
– Rasmus
Jun 23 '11 at 20:52
add a comment |Â
up vote
4
down vote
up vote
4
down vote
The sum that shows up most often when you roll 10 dice is 35.
Are you also interested to see a reason why this is so?
answered Feb 10 '11 at 16:10
Rasmus
14.1k14378
The sum that shows up most often when you roll 10 dice is 35.
Are you also interested to see a reason why this is so?
answered Feb 10 '11 at 16:10
Rasmus
14.1k14378
answered Feb 10 '11 at 16:10
Rasmus
14.1k14378
answered Feb 10 '11 at 16:10
Rasmus
14.1k14378
answered Feb 10 '11 at 16:10
Rasmus
14.1k14378
14.1k14378
I am interested
– ajbeaven
Jun 22 '11 at 22:26
@ajbeaven: I think Yuval Filmus suggestion for a proof is very good. See also this question.
– Rasmus
Jun 23 '11 at 20:52
add a comment |Â
I am interested
– ajbeaven
Jun 22 '11 at 22:26
@ajbeaven: I think Yuval Filmus suggestion for a proof is very good. See also this question.
– Rasmus
Jun 23 '11 at 20:52
I am interested
– ajbeaven
Jun 22 '11 at 22:26
I am interested
– ajbeaven
Jun 22 '11 at 22:26
@ajbeaven: I think Yuval Filmus suggestion for a proof is very good. See also this question.
– Rasmus
Jun 23 '11 at 20:52
@ajbeaven: I think Yuval Filmus suggestion for a proof is very good. See also this question.
– Rasmus
Jun 23 '11 at 20:52
add a comment |Â
up vote
3
down vote
The key to this question is the concept of unimodality (q.v.).
Try to prove by induction that if you take the sum of $n geq 2$ cubes then the distribution has the following form:
- It is supported (has non-zero probability) on the numbers $n, ldots, 6n$.
- It is symmetric around $3.5n$.
- It is unimodal: the sequence of probabilities increases monotonically until it reaches its mode (the most probable value), then decreases monotonically.
- If $n$ is even then the mode is unique (and so $3.5n$).
- If $n$ is odd then there are two modes ($lfloor 3.5n rfloor$ and $lceil 3.5n rceil$).
If you find the proof difficult, try again with two-sided dice.
Of course, as $n rightarrow infty$ the distribution will approach (in some sense) a Gaussian, but I'm not sure this is enough to prove anything specific about the mode (other than its asymptotics).
answered Mar 26 '11 at 17:37
Yuval Filmus
47.2k369142
add a comment |Â
up vote
3
down vote
The key to this question is the concept of unimodality (q.v.).
Try to prove by induction that if you take the sum of $n geq 2$ cubes then the distribution has the following form:
- It is supported (has non-zero probability) on the numbers $n, ldots, 6n$.
- It is symmetric around $3.5n$.
- It is unimodal: the sequence of probabilities increases monotonically until it reaches its mode (the most probable value), then decreases monotonically.
- If $n$ is even then the mode is unique (and so $3.5n$).
- If $n$ is odd then there are two modes ($lfloor 3.5n rfloor$ and $lceil 3.5n rceil$).
If you find the proof difficult, try again with two-sided dice.
Of course, as $n rightarrow infty$ the distribution will approach (in some sense) a Gaussian, but I'm not sure this is enough to prove anything specific about the mode (other than its asymptotics).
answered Mar 26 '11 at 17:37
Yuval Filmus
47.2k369142
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The key to this question is the concept of unimodality (q.v.).
Try to prove by induction that if you take the sum of $n geq 2$ cubes then the distribution has the following form:
- It is supported (has non-zero probability) on the numbers $n, ldots, 6n$.
- It is symmetric around $3.5n$.
- It is unimodal: the sequence of probabilities increases monotonically until it reaches its mode (the most probable value), then decreases monotonically.
- If $n$ is even then the mode is unique (and so $3.5n$).
- If $n$ is odd then there are two modes ($lfloor 3.5n rfloor$ and $lceil 3.5n rceil$).
If you find the proof difficult, try again with two-sided dice.
Of course, as $n rightarrow infty$ the distribution will approach (in some sense) a Gaussian, but I'm not sure this is enough to prove anything specific about the mode (other than its asymptotics).
answered Mar 26 '11 at 17:37
Yuval Filmus
47.2k369142
The key to this question is the concept of unimodality (q.v.).
Try to prove by induction that if you take the sum of $n geq 2$ cubes then the distribution has the following form:
- It is supported (has non-zero probability) on the numbers $n, ldots, 6n$.
- It is symmetric around $3.5n$.
- It is unimodal: the sequence of probabilities increases monotonically until it reaches its mode (the most probable value), then decreases monotonically.
- If $n$ is even then the mode is unique (and so $3.5n$).
- If $n$ is odd then there are two modes ($lfloor 3.5n rfloor$ and $lceil 3.5n rceil$).
If you find the proof difficult, try again with two-sided dice.
Of course, as $n rightarrow infty$ the distribution will approach (in some sense) a Gaussian, but I'm not sure this is enough to prove anything specific about the mode (other than its asymptotics).
answered Mar 26 '11 at 17:37
Yuval Filmus
47.2k369142
answered Mar 26 '11 at 17:37
Yuval Filmus
47.2k369142
answered Mar 26 '11 at 17:37
Yuval Filmus
47.2k369142
answered Mar 26 '11 at 17:37
Yuval Filmus
47.2k369142
47.2k369142
add a comment |Â
add a comment |Â
up vote
-3
down vote
For each individual die, its 'expected value' is 7/2. This is because it can be either 1, 2, 3, 4, 5 or 6, each with a probability of 1/6 (assuming it's a fair die). If you want to know what value will come up most often for 1 die, this is the sum of each of it's possible values times the probability of that value. Which means 1(1/6) + 2(1/6) + 3(1/6) + 4(1/6) + 5(1/6) + 6(1/6) = 1/6(1 + 2 + 3 + 4 + 5 + 6) = 7/2. Obviously, 3.5 will never show up on a die, and this makes sense because if each side has equal probability, you shouldn't expect to see one side more than the others.
To find the expected value of multiple die, you can simply add up each individual expected value. For all 10 dice, this value is 7/2, so you have 7/2 * 10 = 35
Thus, the expected value of rolling 10 dice is 35.
Wikipedia has more information on probability and expected values, if it interests you:
http://en.wikipedia.org/wiki/Expected_value
answered Mar 25 '11 at 16:46
mozi
312
2
Expected value != Most likely value.
– Aryabhata
Mar 25 '11 at 17:16
1
hmm, would you mind explaining how to go about solving this problem then? I'm curious now!
– mozi
Mar 25 '11 at 17:57
Perhaps this will help: math.stackexchange.com/questions/4632/… . Computer software can easily compute the the required coefficients. There might even be a proof that 35 has the highest coefficient...
– Aryabhata
Mar 25 '11 at 19:19
add a comment |Â
up vote
-3
down vote
For each individual die, its 'expected value' is 7/2. This is because it can be either 1, 2, 3, 4, 5 or 6, each with a probability of 1/6 (assuming it's a fair die). If you want to know what value will come up most often for 1 die, this is the sum of each of it's possible values times the probability of that value. Which means 1(1/6) + 2(1/6) + 3(1/6) + 4(1/6) + 5(1/6) + 6(1/6) = 1/6(1 + 2 + 3 + 4 + 5 + 6) = 7/2. Obviously, 3.5 will never show up on a die, and this makes sense because if each side has equal probability, you shouldn't expect to see one side more than the others.
To find the expected value of multiple die, you can simply add up each individual expected value. For all 10 dice, this value is 7/2, so you have 7/2 * 10 = 35
Thus, the expected value of rolling 10 dice is 35.
Wikipedia has more information on probability and expected values, if it interests you:
http://en.wikipedia.org/wiki/Expected_value
answered Mar 25 '11 at 16:46
mozi
312
2
Expected value != Most likely value.
– Aryabhata
Mar 25 '11 at 17:16
1
hmm, would you mind explaining how to go about solving this problem then? I'm curious now!
– mozi
Mar 25 '11 at 17:57
Perhaps this will help: math.stackexchange.com/questions/4632/… . Computer software can easily compute the the required coefficients. There might even be a proof that 35 has the highest coefficient...
– Aryabhata
Mar 25 '11 at 19:19
add a comment |Â
up vote
-3
down vote
up vote
-3
down vote
For each individual die, its 'expected value' is 7/2. This is because it can be either 1, 2, 3, 4, 5 or 6, each with a probability of 1/6 (assuming it's a fair die). If you want to know what value will come up most often for 1 die, this is the sum of each of it's possible values times the probability of that value. Which means 1(1/6) + 2(1/6) + 3(1/6) + 4(1/6) + 5(1/6) + 6(1/6) = 1/6(1 + 2 + 3 + 4 + 5 + 6) = 7/2. Obviously, 3.5 will never show up on a die, and this makes sense because if each side has equal probability, you shouldn't expect to see one side more than the others.
To find the expected value of multiple die, you can simply add up each individual expected value. For all 10 dice, this value is 7/2, so you have 7/2 * 10 = 35
Thus, the expected value of rolling 10 dice is 35.
Wikipedia has more information on probability and expected values, if it interests you:
http://en.wikipedia.org/wiki/Expected_value
answered Mar 25 '11 at 16:46
mozi
312
For each individual die, its 'expected value' is 7/2. This is because it can be either 1, 2, 3, 4, 5 or 6, each with a probability of 1/6 (assuming it's a fair die). If you want to know what value will come up most often for 1 die, this is the sum of each of it's possible values times the probability of that value. Which means 1(1/6) + 2(1/6) + 3(1/6) + 4(1/6) + 5(1/6) + 6(1/6) = 1/6(1 + 2 + 3 + 4 + 5 + 6) = 7/2. Obviously, 3.5 will never show up on a die, and this makes sense because if each side has equal probability, you shouldn't expect to see one side more than the others.
To find the expected value of multiple die, you can simply add up each individual expected value. For all 10 dice, this value is 7/2, so you have 7/2 * 10 = 35
Thus, the expected value of rolling 10 dice is 35.
Wikipedia has more information on probability and expected values, if it interests you:
http://en.wikipedia.org/wiki/Expected_value
answered Mar 25 '11 at 16:46
mozi
312
answered Mar 25 '11 at 16:46
mozi
312
answered Mar 25 '11 at 16:46
mozi
312
answered Mar 25 '11 at 16:46
mozi
312
312
2
Expected value != Most likely value.
– Aryabhata
Mar 25 '11 at 17:16
1
hmm, would you mind explaining how to go about solving this problem then? I'm curious now!
– mozi
Mar 25 '11 at 17:57
Perhaps this will help: math.stackexchange.com/questions/4632/… . Computer software can easily compute the the required coefficients. There might even be a proof that 35 has the highest coefficient...
– Aryabhata
Mar 25 '11 at 19:19
add a comment |Â
2
Expected value != Most likely value.
– Aryabhata
Mar 25 '11 at 17:16
1
hmm, would you mind explaining how to go about solving this problem then? I'm curious now!
– mozi
Mar 25 '11 at 17:57
Perhaps this will help: math.stackexchange.com/questions/4632/… . Computer software can easily compute the the required coefficients. There might even be a proof that 35 has the highest coefficient...
– Aryabhata
Mar 25 '11 at 19:19
2
2
Expected value != Most likely value.
– Aryabhata
Mar 25 '11 at 17:16
Expected value != Most likely value.
– Aryabhata
Mar 25 '11 at 17:16
1
1
hmm, would you mind explaining how to go about solving this problem then? I'm curious now!
– mozi
Mar 25 '11 at 17:57
hmm, would you mind explaining how to go about solving this problem then? I'm curious now!
– mozi
Mar 25 '11 at 17:57
Perhaps this will help: math.stackexchange.com/questions/4632/… . Computer software can easily compute the the required coefficients. There might even be a proof that 35 has the highest coefficient...
– Aryabhata
Mar 25 '11 at 19:19
Perhaps this will help: math.stackexchange.com/questions/4632/… . Computer software can easily compute the the required coefficients. There might even be a proof that 35 has the highest coefficient...
– Aryabhata
Mar 25 '11 at 19:19
add a comment |Â
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4
Please turn off caps lock when you write the title. What have you tried-do you know the answer for 2 dice?
– Ross Millikan
Feb 10 '11 at 15:06
9
please note that this is at least the fourth time in the recent days another member of the site has requested that you not to use all caps in the question title. Consider this your moderator warning. Any further abuse will result in graduated suspension from the site.
– Willie Wong
Feb 10 '11 at 15:19
2
How many sides do your dice have?
– user489
Feb 10 '11 at 18:47