A die is tossed until the sum of the rolls is 15 or greater.

What is the most likely total? What is the the least likely total which can occur?

To solve this problem of dice rolling, begin by listing all the possible roles that sum to 15. For example, 3,6,6 or 5,5,5. Note that permutations of a role are not needed (otherwise you’ll be listing roles until tomorrow and doing unnecessary extra work). I uncovered a total of 103 different combinations of roles that lead to 15. I found that listing the dice roles in ascending order helped to make sure I didn’t duplicate or miss anything.

The next challenge is to find the total number of alternative roles. That means, you role the last die and the sum of your roles exceeds 15 without landing on it. For example, in the case of 5,5,5, you role a 6 instead of the last five. If you role a 4 instead of the last 5, you still have a chance to total to 15 by rolling a 1.

To count the total number of alternative combinations, we have to assume that the dice can role the numbers in any order. Thus, for 3,6,6, we assume that 6,3, 3,6 and 6,6 are possible first two role pairs. The fact that 6,6 will only occur 1/5 of the time compared to 6,3 and 3,6’s 2/5 probability is irrelevant to just counting the total number of alternative solutions.

For each combination that sums to 15, use the following formula:
If any number of 1’s are present, add 5 to the total number of alternative solutions
If any number of 2’s are present, add 4
If any number of 3’s are present, add 3
If any number of 4’s are present, add 2
If any number of 5’s are present, add 1
If any number of 6’s are present, add 0

Why is this? Let’s look at the combination of 15 1’s. Let’s say you have rolled 14 1’s so far. If you don’t role a 1 on your last role, you pass 15. 2, 3, 4, 5, and 6 are alternative roles, so there are 5 total alternative results. Some people may wonder why we don’t add 5 alternatives results for each 1. The reason is that we are only concerned with the very last role. Imagine you having 15 different dice, and you role them all one by one. The first 14 dice all show 1’s. You have only a 1/6 chance that the final role will be a 1, bringing your total to 15. Otherwise you have a 5/6 chance that the total sums to greater than 15 – 5 alternative solutions. If you repeat this same experiment indefinitely, you will always have only 5 alternative solutions at the end.

Now, this changes with 13 1’s and a 2. At the end of 13 roles, you can have 13 1’s, or you can have 12 1’s and a 2. In the former situation, you need a 2 to hit 15. If you role a 3, 4, 5, or 6, you pass 15. But if you role a 1, you are only at 14, so a 1 takes you to the previously described situation. So, if you need a 2, only 4 alternative solutions exist. If, on the other hand, you have rolled 12 1’s and a 2, you now need a 1. This goes back to the previously described scenario, where there are 5 alternative solutions. So, there are a total of 9 alternative scenarios. You can apply this logic to the remaining 101 dice role combinations. I discovered that in all 961 alternative results could be rolled.

Add the alternative results to the to the combinations that sum to 15 to get 1064 total possible combinations given that you must stop rolling once you hit 15 or greater. Divide the 103 sum-to-15 combinations by 1064 to discover that there is only a 9.7% chance of getting 15.

To calculate the most frequently occurring number, go back through the 103 combinations and count the number combinations that have any number of 1’s. Do the same for 2’s, 3’s, 4’s, and 5’s. The total number of alternative solutions that sum to 16 equals the results for 1’s + the results for 2’s + … + the results for 5’s. The total number of alternative solutions that sum to 17 equals the results for 1’s + 2’s + 3’s + 4’s. 18 is 1’s + 2’s + 3’s, 19 is 1’s + 2’s, and 20 is only 1’s. I found the following:

Sum of dice roles, total number of occurrences
15, 103
16, 278
17, 246
18, 204
19, 149
20, 84

Thus, 16 is the most common result.

Posted by draistal on 2004-01-20 11:58:08