The chances that the 3 Quintessences are on position x, y and z is 1 / Binomial[30,3]
How many positive cases do we have, with 2 Quintessences in a row?
I count 28+27+26...+1+1+2+...+27=1/2*28*(28+1)+1/2*27*(27+1)=784
Binomial[30,3]=4060
784/4060~0,193=19,3%, you had it twice in 10 games ,that´s 20%
I don't follow what you're doing here:
1. I don't understand how you're getting 784 for your "positive cases" (I don't see what method you're using).
2. In your binomial, you're finding the number of ways to choose 3 cards from 30, but you only have positive cases for drawing 2 Quints. How can you compare these if they are for a different number of draws?
The chances that the Quintessences are on position x,y,z are equal to the chances that they are on position a,b,c
Then I made an intuitive counting, how many ways do I have to get a row of 2
1= Quintessence
0= the Rest
111000....
11010000...
1100100000...
...
01110000...
01101000.....
Then, I forgot the pairs with the 3rd Quintessence above the pair e.g.
1011000000
That lets me come to that number
The chances are the same for all cases, exactly one of this cases must be, if you draw cards, so I come to 19,3%.
I came up with my numbers like this:
Drawing 3 Shards in a row:
Probability of drawing 1 Shard (# of Shards/Total number of cards) * Probability of drawing a second Shard (-1 potential Shard, -1 cards total) *
Probability of drawing a third Shard (-1 potential Shard, -1 cards total)
This works out to:
(4/30) * (3/29) * (2/28) = 9.85x10^-4 = 0.0985% ~ 0.1%
In my understanding, this are the chances, that 3 of the 4 shards are on position x, y and z. e.g. 1,2 and 3.
But thats not the same as the chances that you can find a row of 3 in your deck of 30 cards.
A case in which your way of calculation would produce results which are obviously wrong:
Have more than 15 Quantum Towers in your deck and calculate the chances of 2 in a row.
If you have more than 15 Towers, the chances for this are 100% but you calculate a value <100%