# Homework Help: Formula for a game of combinations of numbers

1. Oct 23, 2008

### cubanopro

1. The problem statement, all variables and given/known data

First of all, let's imagine that there are 3 types of balls in a bag. Balls a, b and c. There are infinite balls of each sort and we have to pick an x number of times. Not long ago I asked someone to help me find a formula wich states all the possibles combinations of numbers if we have tree variables (a,b,c) and an x number of trials. For example, if we have a, b and c and we do 5 picks, the number of possibles combinations would be 3 to the power of 5. The answer is 243. But now here is where the game gets harder.

Depending on what comes out of the bag we obtain different results in the game. For example, we have to pick 11 balls....So the possibles combinations are 177147. Out of those ones, we know that certain combinations make us lose the game, certain make us win and the rest are neutral. Here are the rules to follow:

The only way to lose is to have drawn a maximum of only two kinds of balls for a consecutive 11 picks. For example, aaaaaaaaaaa (we lose), acacacacaca (we lose), bbbbbbbbbbb (we lose)...you get the point.

In order to win:
First, we have to get the first 7 picks to be a combination of a MAXIMUM of two letters (except for a and b togeter). For example, aaaaaaa, bbbbbbb, aaaccac, bbbcbcc, and so on.

Second, once we have the first 7 balls as we stated earlier, we need to get at least once the missing letter before getting to 12. So for example, if the first seven balls are acaccca then in the last four picks left WE NEED to pick a B ball and vice-versa. Notice that in the conditions applied when loosing, there is never a combinations of the three letters.

In order to have a neutral result:
We need to have in the first 7 picks at least one A ball and B ball togeter (not in the last four picks...only in the first seven). For example, cbcacaababc makes it neutral since there is a at least one B ball and one A ball in the first 7 picks.

I was wondering if there was a way of calculating the winnings odds. At this moment, I have just been able to find the number of losing combinations which is (correct me if I'm wrong) (2to the power of 11) times 2.

The answer is 4096 losing combinations out of a total of 177147. Now we still need to find a way of calculating the possibles winning combinations and the rest which are the neutral ones. I found it very hard to calculate it by myself alone...maybe you guys could help me.