Probability of winning the game "1-2-3" Ok, game is as follow, with spanish cards (you can do it with poker cards using the As as a 1)
You shuffle, put the deck face bottom, and start turning the cards one by one, saying a number each time you turn a card around ---> 1, 2, 3; 1, 2, 3; etc. If when you say 1 a 1 comes out, you lose, same with 2 and 3. If you finish the deck without losing, you win.
I know some basics of probabilities, but is there a way to calculate the probability of winning the game, given a random shuffled deck?
 A: Another update:
As explained in the paper below, you can use rook polynomials to solve such problems. 
Playing with a full deck of 52 cards we will call "one" 18 times, 
we will call "two" 17 times, and we will call "three" 17 times.
The forbidden positions in the 52 by 52 board consist of three "independent" 
complete rectangles; one $18\times 4$ and the other two $17\times 4$. 
The rook polynomial for a full $m\times n$ rectangle with $m\geq n$ is 
$$\sum_{k=0}^n{m\choose k}\, {n!\over (n-k)!}\, x^k. $$
Multiply the polynomials for these three rectangles to give us the rook polynomial for our problem
$$R(x)=(1+73440x^4+19584x^3+1836x^2+72x)(1+57120x^4+16320x^3+1632x^2+68x)^2.$$
The number of winning deck orders is 
$$\int_0^\infty x^N R(-1/x) \exp(-x)\,dx $$
so the probability is this divided by $N!$, i.e.
$$\mathbb{P}(\text{win})=  24532967512/3004641364725= 0.008165023553.$$ 
Reference: F. F. Knudsen and I. Skau, 
On the Asymptotic Solution of a Card-Matching Problem,
 Mathematics Magazine 69 (1996), 190-197.   

Update: The solution below is for a simplified version of the problem where you work 
with a deck of size 12: four each of ace, deuce, and trey.

This is a problem in generalized derangements and joriki's answer here 
tells you what to do. In general, the number of deck orders that lead to a win is
$$\int_0^\infty L_{n_1}(x)\cdots L_{n_r}(x)\,\mathrm e^{-x}\mathrm dx.$$ 
In this problem, we have $r=3$ and $n_1=n_2=n_3=4$. The fourth Laguerre 
polynomial is $L_4(x)=(x^4-16x^3+72x^2-96x+24)/24$. Raising this to the third 
power and integrating against $\exp(-x)$ gives $346$.
That is, there are $346$ ways to order the deck that give a win. 
Divide this by the total number of orders $12!/(4!)^3$, to give 
 $$\mathbb{P}(\text{win})=173/17325=0.00998.$$ 
A: For $i,j\in\{1,2,3\}$, let $a_{i,j}$ denote the number of $i$ cards being dealt with number $j$ spoken.
We have $\sum_j a_{i,j}=4$ and for a winning game $a_{i,i}=0$.
The number of winning positions for a given $(a_{i,j})$ is
$$\frac{18!}{a_{2,1}!a_{3,1}!(18-a_{2,1}-a_{3,1})!}\cdot\frac{17!}{a_{1,2}!a_{3,2}!(17-a_{1,2}-a_{3,2})!}\cdot\frac{17!}{a_{1,3}!a_{2,3}!(17-a_{1,3}-a_{2,3})!}. $$
We need to sum this over all $(a_{i,j})$ and divide by the total count $$ \frac{52!}{4!4!4!40!}.$$
(Actually, we need just let $a_{1,2}, a_{2,3}, a_{3,1}$ run from $0$ to $4$ and this determines $a_{1,3}=4-a_{1,2}$ etc.)
The final result is
$$p=\frac{58388462678560}{7151046448045500}=\frac{24532967512}{3004641364725}\approx 0.008165 $$
(I just noted that Harold has performed a Monte Carlo simulation with matching result)
A: This is a hard question, if the player is using optimal strategy rather than just cycling through numbers.  
For example, once the deck is down to 2 cards the player is guaranteed a win, because those two cards are known and (even if they're different) the player can name the third number.  If the deck is down to 3 cards the player is guaranteed a win unless those last three cards are all different, in which case the player can't do better than guessing at random (2/3 chance) of winning.
Full analysis for a deck of size 6: 2 each of $1,2,3$.
Card 1: random, 1/3 chance of loss
Card 2: guess whatever card 1 was, 1/5 chance of loss (if top two cards are the same)
Card 3: guess either of the first two cards, 1/4 chance of loss
We now have two situations.  If the first three cards are all different, there is a further 1/3 chance of loss, based on the analysis in the paragraph above, otherwise a win is assured.  Each case happens half the time; two each of the four cards lead to the two cases.
Altogether, the probability of loss is: $$\frac{1}{3}+\frac{2}{3}\frac{1}{5} + \frac{2}{3}\frac{4}{5}\frac{1}{4} + \frac{2}{3}\frac{4}{5}\frac{3}{4}\frac{1}{2}\frac{1}{3}=\frac{2}{3}$$
This seems too good to be true, but there it is.
