Math - Dice Game Probability

The Dice Game

I played this game at a local festival to raise money for a good cause. The game is deceiving. I naively thought that the worst I could do is break even and the House would not make any money.    But the math shows I was wrong in my thought process. 

Here are the rules: 

1) The House has three dice
2) Players place a bet on a number from 1 to 6 (For the sake of argument lets say the player puts a $1 bet on a specific number)
3) For each round the House throws the three dice
4) If one of the dice comes up with your number you get your bet back and $1
5) If two of the dice come up with your number you get your bet back and $2
6) If three of the dice come up with your number you get your bet back and $3
7) If none of the dice come up with your number you lose your $1 bet

My naïve thinking goes like this:

The probability of one dice landing with my number is 1/6.
Now add the probability of another dice landing  with my number as 1/6.
Therefore: 1/6  +  1/6   =    2/6    =    1/3     =     33.33% probability.
Now add the probability of the third dice landing with my number as 1/6.
Therefore 1/6 + 1/6 + 1/6   =   3/6   =   1/2   =   50%   chance that one of the dice lands with my number.
But that is not true.
If it were true then the probability when using 6 dice of one of them landing with my number would be:
1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6  =  6/6  = 100%
Yes the probability when using 6 dice of one landing  with my number is higher than one dice, but it is *not* 100%.

Let's look closely at the case when using two dice:

The probability when using two dice of at least one landing on my number can be broken down into these cases:
1) First die lands with my number.  Second die does not.
2) First die does not land with my number.  Second die lands with my number.
3) Both dice land with my number.
4) Neither die lands with my number.

The probability of one or both dice landing with my number is the probability for cases 1, 2, and 3: 

Ok, using the same method let's break down the three dice game into individual cases:

1) None of the three dice match my number
2) One of the three dice match my number.
3) Two of the three dice match my number. 
4) All three of the dice match my number.

Below is the Probability for each Combination of Dice for a $1 bet put on a specific number.  

Below Shows that the sum of the Probabilities of every combination for a bet put on a single number is 100%

i.e. The probability that you either win or lose is 100% 

Below Shows your expected Payback for each $1 bet placed on a specific number. 

Suppose the Player starts with $10 in their pocket and the House has a $1 minimum.
How many rounds can the Player play until they have less than $1 left? 

1) The Player sometimes wins $2, sometimes wins $3 and sometimes wins $4.
2) But on average the Player loses $0.08 on each bet.  

How many rounds does it take the Player to use up $9?

        $9.00         =       112.5  rounds
        $0.08                                  

Therefore the Player can play on average about 112 rounds to use up $9.00
The Player wins enough to keep interest, but the House wins in the long term.