The computer model: You needn´t really model, the drawing of 4 cards out of 27 rares. You make 3 arrays (vectors) with 4 entries each - the 3 slots - and randomly choose natural numbers between 0 and 3. To regard the fact, that some of the chosen 4 cards can be pairs, you calculate chances of that, and regard that in your model, using another random number.
Here is a table, using combinatorics for the 4 chosen rares.
First (weakest) Estimation (Computer based, Java Program) Spoiler for Hidden:
First, we simulate X games, and count, how many rares are in all 3 slots. These are the rares, which are theoretical winnable. Here´s a table for 10.000.000 games. A direct win counts as one available rare.
This shows an interesting fact: The player can often not choose, which rare he´ll win, because in 64% of all games, he can has max one choice. However he can try to choose a strategy to maximize the chance to get one. This table also gives us some (weak) estimations of win chances.
Upper border: The player can win in
max 87,3 % of all games simply because in 12,7 % of all games, there´s no rare card to win. This estimation is independent of the players strategy.
Lower border: Imagine, the player gets shown all 4 rares, and he randomly picks one of them. Then he wins ~ 12,7%*0+ 52,0%*0,25+ 30,8%*0,5+ 3,3%*0,75+ (8,36%+0,06%)*1 ~ 33,8% of all games.
33,8% < win chance < 87,3 % But we can do better:
Second better Estimation (Mathematical way) Spoiler for Hidden:
Now, we include a spin strategy of players, but we also simplify a bit. We say that always 4 different rares are chosen. Then, the calculated probabilty is a bit lower than the real one.
We use the following strategy of the user:
A pair: Spin the remaining card as much as possible and if it doesn´t match with the others,spin the other 2 slots again.
Three different cards: Different strategies are possible. I used the following one: Spin slot 1, until you get a pair, then spin the other one, till you get the card or til you can´t spin the slot anymore. If you don´t win, spin the other 2 again as often as possible/needed.
We get:
A chance estimation of the winchance is the scalarproduct of the two columns, which is ~68,24%. When we also consider the fact, that we can have pairs in our 4 chosen cards, then we get another border (using Table 1):
74,8%.
The chance to win a rare is between 68,2% and 74,8%, when you choose the described strategy for 3 different cards in the initial spin.It´s interesting, that OldTrees got a similar result using a different strategy, when you find no pairs in your initial spin (assuming 4 different cards are chosen). His estimation is quite much hidden, not many players have ever seen it, or understood, what it means.
OldTrees gets 69,84% and would get 76,1% for an upper border. Means OldTrees´ strategy seems to be a bit better. However, we don´t know what happens, when there are pairs in the 4 chosen cards. Here´s his estimation:
Third (best) Estimation (Computer based, Java Program) The last logical step is to use the chances of Table 1 to really simulate games and also consider the fact that some of the 4 chosen cards may be the same card.
I also did 10.000.000 tests
I got 72,5 %, which is within the borders of part 2.
