Question

My favorite game, Maplestory M, features a mini-game that allows you to roll dice in an attempt to get a particular number as the sum of the dice rolled. For this case, that number is 30.

The way the game works, one rolls dice one at a time, and one obtains the prizes associated with each running sum. For instance, if I rolled a 1 , 3, 4, 5, 6, respectively, I'd get the prize associated with 1, 4, 8, 13, and 18.

The 30 square prize is by far the best prize, and as such I'd like to examine it further.

Answer 1

a. Normally, there are exactly 5 dice given, and you may choose to roll to try to get 30. What is the probability of doing so?

Answer 2

b. The game has many events, and one of these allows you an additional die roll for free. Note that the game counts a running sum, so if you get 30 with 5 dice, you win the prize, but if not, this is where the extra die would come in and you could roll it to see if you could get 30 after the first 5 dice failed.

Answer 3

c. For real money, you may elect to purchase up to three additional die rolls. Given that the dies are rolled one by one as mentioned before (and not all at once), what is the chance of winning the game with 7 dice? What about 8 dice? And finally, what about 9 dice? Show all work.

Answer 4

@Vocaloid

Answer 5

Oh boy, there's a lot of combinations (or, more accurately, permutations) to work with here.
Let's start with 5 dice. Obviously, the only way to get 30 with only 5 dice is to roll the high 6 on all 5. That's 5 events with a 1/6 chance which happen independently of each other, so the probability of getting 30 this way is 1/6^5,
which is also equal to...
1/7776 chance. Wow, look at all those lucky 7's!

Answer 6

Now, since this probability is extremely low, you're probably interested in how the odds change when you add more dice.
Well, when you have 6 dice, there are 3 ways to get exactly 30. First, you can get five 6's, and I guess the game ends, so you don't need the sixth die.
Secondly, you could get six 5's, but there's actually less chance of this happening than just getting five 6's. In fact, it's six times less likely, since instead of only five exact rolls, you're depending on six exact rolls.
Thirdly, you could roll any combination of four 6's, plus two more numbers which add up to six. This third option actually improves your chances by quite a bit.
To see why, let's take a look at all the ways two rolls can add up to six: 3+3//2+4//1+5
Then, consider that the six numbers can be rolled in any order, and you still win.
For instance,
6,6,6,6,1,5
is a different outcome than
1,5,6,6,6,6
But they both represent a win, and they both count as independent possibilities of the dice rolls.
Now, just how many such permutations do we get from these options?
To answer that, I'll be making use of something called factorials. I'll explain below.

Answer 7

Now, whenever we have different objects arranged in a particular order, we can calculate the different arrangement possibilities (called permutations) by simply taking the factorial of the amount of objects. The factorial of a number is simply the product of that number and every whole positive integer below it.
For instance, the factorial of 6 would be 6 * 5 * 4 * 3* 2 * 1
This relates to our dice roll scenario because we can treat every dice value as a unique "object", then use factorials to figure out how many arrangements give us the outcome we're looking for.
The method won't be as simple as taking the factorial of 6, however. That method works if all the objects being arranged are unique. But in this case, four of the objects--the dice that come up as '6'--are the same, while the only unique ones are the other two, which add up to '6'.

Answer 8

For simplicity's sake, let's forget about factorials for now. They may be useful for us later.
Also, in this scenario, I'll refer to the two digits which add up to 6 as 'a' and 'b'.
Now, let's consider the cases where 'a' is the first dice roll. In this case, 'b' can be the second, third, fourth, fifth, or sixth die, and all the others must have a value of '6'.
Likewise, if 'a' comes second, 'b' can be first, third, fourth, fifth, or sixth, while all the others have a value of '6'. You may see where I'm going with this.
For every position that 'a' can occupy, there are five positions that 'b' can occupy. And since 'a' can occupy six positions, that gives us 30 possibilities so far.
The only pair of 'a' and 'b' for which this does not hold true is '3' and '3'. Normally, an order like (a,b,6,6,6,6,6) and (b,a,6,6,6,6) would be two separate outcomes, but if both 'a' and 'b' are equal, the two outcomes become one and the same. For this reason, '3' and '3' gives us only 15 possibilities.
So, we take the 30 possibilities from '1' and '5', 30 from '2' and '4' and 15 from '3' and '3', and we get 75 possibilities. But out of how many?
Well, when we roll 6 dice, there are 6^6 total possibilities, or 46,656 total.
75/46656 is 0.0016, or about one in 625.
Don't forget that the original chance with five dice (1/7776) still exists, and there is that very tiny chance of rolling all 5's with six dice (1/46,656). Adding all these together, your chances of getting 30 with six dice is... 83/46656, or about one in 588.

Answer 9

Next up, seven dice. Keep in mind that the possibilities from five and six dice still exist.
The new possibilities are as follows:
1. Four 6's and (a + b + c), where a+b+c=6
2. Four 5's and (a + b + c), where a+b+c=10
3. Five 5's and (a + b), where a+b=5
This is where factorials may be useful for us.
In scenario 1. and 2. we are arranging 3 unique objects among 7 positions. So, the formula for the possibilities is (7!)/(7-3)!. Written out, the calculation would look like (7*6*5*4*3*2*1)/(4*3*2*1), which we can simplify to (7*6*5). That gives us 210 possibilities for each combination of a+b+c=6 as well as for a+b+c=10.
Now, what combinations fit this formula? We can get 6 by summing up 1, 2, and 3, as well as 2, 2, and 2, and finally, 4, 1, and 1. Things become a bit more complicated here because only 1, 2, and 3 are unique, while 2,2,2 are all the same and 4,1,1 has two identical objects. It isn't too difficult to calculate for these differences. For any combination of three objects a,b,c, there are six possible orderings: a,b,c; a,c,b; b,a,c; b,c,a; c,b,a; c,a,b.
When the three objects become the same, there is only one possible ordering: a,a,a.
Thus, 2,2,2 should have 1/6 as many possibilities as 1,2,3.
210/6 is 35
In the same way, a,b,b has only 3 orderings: a,b,b; b,a,b; b,b,a.
So, 4,1,1 should have 1/2 as many possibilities as 1,2,3.
210/2 is 105
The same principles can be applied for possibilities 2. and 3. mentioned above.

Answer 10

I'm sure I could continue with the calculations, but I think that I have given you all the tools you would need to solve this problem yourself at this point.
It is getting late, and, while I appreciate your dedication to both math and gaming, I ought to be heading off. Perhaps we'll pick this up again at another time?

Answer 11

Aw heck, I'm not quite ready to sleep yet, so I'll get you started with possibility 2.
The combinations for a+b+c=10 are as follows:
2,2,6
1,3,6
2,3,5
3,3,4
1,4,5
2,4,4
Note that 2,2,6; 3,3,4; and 2,4,4 all have two identical objects. The rest of the combinations are all unique.
Since we've already established that unique combinations of this kind yield 210 possibilities, and that combinations with two identical objects yield 105, we should be able to add them all up to get... 945.
OK now I'm really done though. Good night.

Answer 12

Valiant effort. I've calculated the probability mass functions of the multinomial coefficients in mathematica, yielding the chances of getting exactly a sum of 30. Unfortunately, there is one major complication in this problem
Let's say you roll 6 dice. The first five dice might give you a 30, and they might not. The chances of them giving you a 30 are 1/7776. The chances of them not giving you a 30 are 7775/7776. The sixth roll must exclude this event, so you would have to calculate as follows:

P[6 individual roll success]=P[5 roll success]+(1-P[5 roll success])*P[6 roll success]
wherein the 6 roll success event is merely the probability of rolling six dice all at once and getting a 30.

Similarily, you must consider when rolling 7 dice that both the five dice event and the six dice event must be excluded. For simplicity's sake, let's call a5 the probability of getting a 30 when rolling 5 dice, a6 the probability of getting a 30 when rolling 6 dice, a7 7 dice, etc.

Then for the case of 7 dice:
P[7 individual roll success]=a5+(1-a5)(a6+(1-a6)*a7)

Following this line of reasoning, I've calculated the results as follows:

Answer 13

5: 0.0128601%
6: 0.990101%
7: 5.27573%
8: 12.3584%
9: 18.7993%

Answer 14

Very impressive! Technology really is amazing

Answer 15

there are closed form expressions for the multinomial theorem but they are quite messy indeed, and since we have 6 terms it's far easier to use a CAS system