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# Rolling Total of a Die

### #1

Posted 03 June 2013 - 09:28 PM

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?

### #2

Posted 04 June 2013 - 12:38 AM Best Answer

### #3

Posted 04 June 2013 - 01:15 AM

Spoiler for

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

### #4

Posted 04 June 2013 - 05:56 AM

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

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #5

Posted 04 June 2013 - 06:08 AM

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

Spoiler for It is about what you would expect

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.

### #6

Posted 04 June 2013 - 06:47 AM

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

Spoiler for It is about what you would expect

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.

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #7

Posted 04 June 2013 - 06:57 AM

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.

**Edited by BMAD, 04 June 2013 - 06:58 AM.**

### #8

Posted 04 June 2013 - 11:10 PM

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.)

### #9

Posted 05 June 2013 - 04:20 AM

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.

### #10

Posted 06 June 2013 - 02:03 AM

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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