For those of you familiar (and not familiar) with mahjong.
This game is played with 136 tiles and there are 4 identical copies of each tile in a wall. When a game starts, each player draws 4 tiles from the wall in clockwise order until they get a starting hand of 13 tiles (4+4+4+1) (14 tiles for a dealer (4+4+4+2)). I want to know what is the probability of getting exactly 2 identical tiles in a starting hand (let the tile be X). Assume we are the dealer.
To count this I used the binomial formula:
$C(4,2)*0.03^2*0.97^2=0.005$
where 4 is the number of draws and 2 is the number of successes (the average probability of drawing X with each draw is about 0.03, since first we draw 4 tiles out of 136 tiles, then we draw 4 tiles out of 120 tiles etc.). So, we have the probability 0.5% of getting 2 identical tiles X in a starting hand. My personal experience in this game tells me that something is wrong, that it must be higher.
One problem we face here is that with each draw we may draw from 0 up to 4 X's, or that by our second draw there are no X's left in the wall since other players might have drawn it - how could we account for that? Is there any way to account for that at all? I used $p\approx0.03$ but that makes sense only if after each draw we return all tiles into the wall.
I'm completely new to this part of maths, so please point out possible mistakes if there are any.
You want to chose one type of tile for the pair ($34$ possibilities) and ${4 \choose 2}=6$ ways of choosing two of that type. Although you pick in groups of four, you can imagine this as picking singly so they can have been picked in any position for your hand so $14\times 13=182$ pairs of positions
You then want to choose $12$ types from the remaining $33$ for the remaining $12$ positions in your hand so $\frac{33!}{(33-12)!}$ possibilities for the types and any one of the four of each type so $4^{12}$ for the individual tiles.
Overall you could have had any $14$ of the $136$ tiles in a particular order so $\frac{136!}{(136-14)!}$ possibilities.
That makes the probability $\dfrac{34 \times 6 \times 182 \times \frac{33!}{21!} \times 4^{12} }{\frac{136!}{122!}} \approx 0.2857$