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# Rolling Total of a Die

## Question

An ordinary die is rolled until the running total of the rolls first exceeds 12. What is the most likely final total that will be obtained?

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If the die stops being rolled once the cummulative total is larger than twelve, then the highest possible total is 18, and there is only one previous total (12) that could have made that happen. If the previous total had been 11, 18 could not have been reached. If it had been 13, the die would not have rolled again.

With that in mind, the the second highest total (17) could have two different previous totals (11 and 12), 16 could possibly have any of three previous totals, and on down until we reach 13 which could have been made from any of six previous totals. Given that it has the most previous totals leading up to it, 13 should be the final total most likely obtained.

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If the die stops being rolled once the cummulative total is larger than twelve, then the highest possible total is 18, and there is only one previous total (12) that could have made that happen. If the previous total had been 11, 18 could not have been reached. If it had been 13, the die would not have rolled again.

With that in mind, the the second highest total (17) could have two different previous totals (11 and 12), 16 could possibly have any of three previous totals, and on down until we reach 13 which could have been made from any of six previous totals. Given that it has the most previous totals leading up to it, 13 should be the final total most likely obtained.

you may want to consider the likelihood of rolling the numbers below 13 as well

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you may want to consider the likelihood of rolling the numbers below 13 as well

For numbers between 1-12,

1. The least frequent number encountered is 1.
2. The most frequent number encountered is 6.
3. The ocurrances of 7-12 are roughly the same

(this is an implicit assumption in BobbyGo's analysis)

between 5.2 and 6 times the frequency of the number 18.

For numbers greater than 12, and because the six preceeding numbers occur with similar

frequencies, the ocurrences are roughly in the proportion of 6 5 4 3 2 1. For a million rolls,

ending when the total exceeds 12, and normalized to the ocurrances of the number 18,

the following frequency distribution obtains:

(The average number of rolls to exceed 12 was 4.197)

1 3.415

2 4.011

3 4.656

4 5.442

5 6.328

6 7.404 <= most frequently encountered total.

7 5.206

8 5.524

9 5.751

10 5.937

11 6.000

12 5.988

13 5.742 (roughly 6)

14 4.879 (roughly 5)

15 3.937 (roughly 4)

16 2.995 (roughly 3)

17 1.984 (roughly 2)

18 1

If the rolls are stopped when the total exceeds a high number, say 1000,

the ocurrances of the six numbers preceeding 1000 is virtually identical,

and the relative ocurrances of 1001 1002 1003 1004 1005 1006 become

arbitrarily close to 6 5 4 3 2 1.

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you may want to consider the likelihood of rolling the numbers below 13 as well

For numbers between 1-12,

1. The least frequent number encountered is 1.
2. The most frequent number encountered is 6.
3. The numbers 7-12 are remarkably flat (this condition

is an implicit assumption in BobbyGo's analysis)

at about six times the frequency of the number 18.

For numbers greater than 12, the ocurrences are roughly in the proportion of 6 5 4 3 2 1.

For a million rolls, ending when the total exceeds 12 normalized to the ocurrances of the number 18,

this frequency distribution obtains: (average of 4.197 rolls needed to exceed 12)

1 3.415

2 4.011

3 4.656

4 5.442

5 6.328

6 7.404 <= most frequently encountered number.

7 5.206

8 5.524

9 5.751

10 5.937

11 6.000

12 5.988

13 5.742 (roughly 6)

14 4.879 (roughly 5)

15 3.937 (roughly 4)

16 2.995 (roughly 3)

17 1.984 (roughly 2)

18 1

If the rolls are stopped when the total exceeds a high number, say 1000,

the ocurrances of the six numbers preceeding 1000 is virtually constant,

and the relative ocurrances of 1001 1002 1003 1004 1005 1006 become

arbitrarily close to 6 5 4 3 2 1.

So this is based off simulation? My method led to a different conclusion but it wouldn't be the first time I made a mistake.

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you may want to consider the likelihood of rolling the numbers below 13 as well

For numbers between 1-12,

1. The least frequent number encountered is 1.
2. The most frequent number encountered is 6.
3. The numbers 7-12 are remarkably flat (this condition

is an implicit assumption in BobbyGo's analysis)

at about six times the frequency of the number 18.

For numbers greater than 12, the ocurrences are roughly in the proportion of 6 5 4 3 2 1.

For a million rolls, ending when the total exceeds 12 normalized to the ocurrances of the number 18,

this frequency distribution obtains: (average of 4.197 rolls needed to exceed 12)

1 3.415

2 4.011

3 4.656

4 5.442

5 6.328

6 7.404 <= most frequently encountered number.

7 5.206

8 5.524

9 5.751

10 5.937

11 6.000

12 5.988

13 5.742 (roughly 6)

14 4.879 (roughly 5)

15 3.937 (roughly 4)

16 2.995 (roughly 3)

17 1.984 (roughly 2)

18 1

If the rolls are stopped when the total exceeds a high number, say 1000,

the ocurrances of the six numbers preceeding 1000 is virtually constant,

and the relative ocurrances of 1001 1002 1003 1004 1005 1006 become

arbitrarily close to 6 5 4 3 2 1.

So this is based off simulation? My method led to a different conclusion but it wouldn't be the first time I made a mistake.

The numbers given were from a simulation.

I tried calculating it exactly but with average rolls exceeding 4 the numbers got too big to count.

The only surprise to me was the frequency of 6 being markedly higher than 5 or 7.

BG needed a somewhat flat distribution preceding 12, and it turned out flatter than I anticipated.

But his 6 5 4 3 2 1 argument would dominate small fluctuations.

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yeah i find it truly interesting.

I calculated more ways to make 7 after two rolls meaning that it would have a higher theoretical probability than 6 as opposed to your experimental probability.

The count definitely gets big as we are dealing with a massive tree diagram of possible outcomes of up to 13 possible rolls.

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yeah i find it truly interesting.

I calculated more ways to make 7 after two rolls meaning that it would have a higher theoretical probability than 6 as opposed to your experimental probability.

The count definitely gets big as we are dealing with a massive tree diagram of possible outcomes of up to 13 possible rolls.

While the individual chances of totaling to 7 is greater than totaling to 6 after the first roll, the cumulative totals always favor 6. For example, the chances of landing a 7 or 6 on the second throw is 6/36 and 5/36 respectively, but 6 already had a 1/6 chance added from the first roll. Bonanova's simulations produced results very close to the actual figures. (Although, I'm not sure how well Excel handles rounding, so he very well might be spot on.)

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yeah i find it truly interesting.

I calculated more ways to make 7 after two rolls meaning that it would have a higher theoretical probability than 6 as opposed to your experimental probability.

The count definitely gets big as we are dealing with a massive tree diagram of possible outcomes of up to 13 possible rolls.

While the individual chances of totaling to 7 is greater than totaling to 6 after the first roll, the cumulative totals always favor 6. For example, the chances of landing a 7 or 6 on the second throw is 6/36 and 5/36 respectively, but 6 already had a 1/6 chance added from the first roll. Bonanova's simulations produced results very close to the actual figures. (Although, I'm not sure how well Excel handles rounding, so he very well might be spot on.)

This is a good point! I will reexamine my numbers.

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Giving a more generic proof where you are finding the most probable sum X or higher from dice rolls where X is greater than 6 the number will always be X.

The reason for this is that the probability of rolling a value bringing the sum to X on dice roll Y will be 1/6 times the sum probabilities of having obtained X-1 to X-6 after roll Y-1. In laymans terms the odds of rolling a 1 if the sum is X-1 plus rolling a 2 if it is X-2 and so on. The probability of obtaining X+1 is 1/6 of the sum probabilities of X-1 to X-5 on the Y-1 roll which will always be lower. This pattern holds true for X+2 and beyond.

Sorry for any typos I'm doing this on my phone

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