Help with a math problem?

Check out the binomial distribution (http://en.wikipedia.org/wiki/Binomial_distribution). There's a catch which I'll discuss at the end.

In your case,

p = 1/6 (the probability of rolling a 6),
n = the number of rolls for which you want to compute the chances of getting the third 6,
k = 3 (the number of suksesses, meaning "6" results, which, for you, is always 3 times). Sorry about the "successes" mnemonic -- I know it hurts.

The general formula is

n! / (k! * (n - k)!) all times p ^ k * (1 - p) ^ (n - k) = Probability of success

where ^ means "to the power of".

So, to find the odds of rolling 3 sixes in 4 rolls, you would have:

p = 1/6
n = 4
k = 3

Probability of success = 4! / (3! * (4 - 3)!) * (1/6) ^ 3 * (5/6) ^ (4 - 3)
or
P(success) = 4! / (3! * 1!) * (1/6)^3 * (5/6)^1
P(success) = 1 * 2 * 3 * 4 / (1 * 2 *3 * 1) * (1/6) * (1/6) * (1/6) * (5/6)
P(success) = 4 * (1/6) * (1/6) * (1/6) * (5/6)
P(success) = 4 * 5 / (6 * 6 * 6 * 6)
P(success) = 5 / (3 * 3 * 6 * 6)
P(success) = 5 / (9 * 36) which is 5 / (270 + 54) or 5 / 324 or approximately 1 / 65.

So, what's the catch?

As far as I know (and I could be wrong), if you're computing the chance of success in 4 rolls, that includes the chance of succeeding in the first three rolls. In other words, your 4 rolls could look like any of these cases (where '*' means "wasn't a 6"):

* 6 6 6
6 * 6 6
6 6 * 6
6 6 6 *

But in the last case, you would never have reached the fourth roll -- the game ends after the 3rd six.

So, in order to use the binomial equation, you need to keep track of all previous success and subtract them out of later results.

For example, if P(3) is the probability of succeeding in exactly 3 rolls, then
P(4) = Binomial result with (n=4, k=3, and p=1/6) - P(3).

Similarly, P(5) -- the chance of succeeding in exactly 5 rolls -- is:

P(5) = Binomial result with (n=5, k=3, and p=1/6) - [P(4) + P(3)],

or, if you like:

P(5) = Binomial result with (n=5, k=3, and p=1/6) - Binomial result with (n=4, k=3, and p=1/6).

So, really, to get the results for any number of rolls N, just compute the Binomial result for N and subtract off the Binomial result for N - 1.

Now a caveat on the caveat -- I'm not good at statistics, and I probably messed that up, so for goodness' sake, *double check this* with someone who can actually do math!

MarkKreiter's analysis is correct, as far as I can see. The caveat he mentions is also true, and unfortunately this makes the analysis significantly more complicated. That is, "What is the probability of the game length being 10 turns or less" is easy to calculate using the formula Mark provided, but "What is the probablility of the game length being exactly 10 turns" is much harder to calculate.

Adding the rolls together would certainly be an option as an alternative end-game mechanism. Note that the greater the variation in your min and max values on the die the more widely your game length is going to vary -- in particular the max game length will go rapidly to the right.

Example 1: Roll 1d6 per turn, end game when total is 20 or more.
- Min game length: 4
- Ave game length: 6
- Max game length: 20

Example 2: Roll 1d20 per turn, end game when total is 60 or more.
- Min game length: 3
- Ave game length: 6
- Max game length: 60

In examples 1 and 2 the average game length is the same (6 turns, near enough), and the min game length is almost the same, but the max game length is vastly different.

Personally, I would go with something that has much less variability than either of those examples, because (in general) I want more predictability in game length than "4 - 20 turns". Also, I'd want something that was easy to keep track of with a small number of counters. Say, roll 1d6: 1-3 place 1 counter, 4-6 place 2 counters. This translates into:

Example 3: Roll 1d2 per turn, end game when total is 8 or more.
- Min game length: 4
- Ave game length: 6
- Max game length: 8

(Note to all the mathemagicians out there: Yes, all numbers are rounded up to the nearest integer.)

Regards,
kos

Square Bashing (miniatures game) uses an end-of-game counter that starts at 30 and decreases by 1d6 per turn, but not if the attacker made two successful attacks in that turn. That means the attacker will not run out of time as long as things are going well, but once the attack bogs down for whatever reason the game will end eventually.

KAndrw, I assume you mean A+B tiles with identical backs, and not B+C? Since A+C + B-C = A+B+C-C = A+B.

As far as the 3 sixes go, though, my math works out differently.

The chance that the game starts ending at round n (P) is the chance that you have thrown exactly 2 sixes (Q), times the chance that you throw the last six in this round (R), i.e. P(n) = R(n) * Q(n).

The chance that you have thrown exactly 2 sixes in the last n-1 rounds, is as follows:
Q(n) = (1/6)^2 * (5/6)^(n-3) * 1/2*(n-1)*(n-2).
The (1/6)^2 part means that you have thrown exactly 2 sixes, whereas (5/6)^(n-3) means that you have thrown n-1-2 = n-3 non-sixes. The last bit, 1/2*(n-1)*(n-2), is a bit harder to explain, but account to the fact that there are many different ways to through 2 sixes in n-1 throws; you have n-1 turns to throw one six, and n-2 turns to throw another.* (Mathematicians: (n-1)*(n-2)*1/2 = (n-1)! / 2! / (n-3)!, the binomial coefficient of n-1 and 2.)

(* I'm pretty sure it doesn't matter, but maybe I made a mistake where you cannot throw the first six at round 12 and the second at round 4.)

Ofcourse, the chance that you throw the last six, R, is equal to the chance you throw any six, i.e. 1/6.

That would leave us with the chance of ending the game at exactly n turns of P(n) = 1/6 * (1/6)^2 * (5/6)^(n-3) * 1/2*(n-1)*(n-2) = (1/6)^3 * (5/6)^(n-3) * 1/2*(n-1)*(n-2).

This all boils down to 0.46% chance to end the game after 3 rounds, and just as much chance to end it after 39 rounds. To me, saying your game takes about 3-39 rounds isn't really appetiting, to say the least. Even more so because out of 200 games you play, 1 will take longer than 60 turns.

So my suggestion would be to go with either kos' or pelle's or KAndrw's idea; 4-8 seams like a very reasonable roundcount (and you could easily scale this to 12-24 turns, for instance).

I'm going to have to use a system like this in a game of my own. I love how simple it is, both in implementation and in understanding how and why it works. I have to ask, was this system borrowed from/inspired by any other games in particular? I've seen a few games that use a fill-the-track system to determine when a game might end, but the idea of using a fixed number of tiles with blanks to force both minimum and maximum length is new to me.

I believe that increased values of C would favor a shorter game over a longer one, however, since the extra tiles all have stars on them. More stars means a greater chance that a tile is added to the track each turn.

I'm pretty sure I made that system up, though obviously it owes a lot to systems like the Catan Event Cards and other deck-of-dice concepts.

Gah!