Probability with card game

Question

I need help calculating the chances of winning this strange game that I'm going to explain right now:

 You have a deck of 52 cards(4 suits,Ace to King). All you have to do is turning
 the cards one by one counting one,two,three while you turn them. If you get an 
 Ace when you count one or a two when you count two or a three when you count
 three you lose.

For example if you get:

 2(one),K(two),6(three),3(one),Q(two),3(three)

You lose,because you get a 3 when you counted three.

The only way I could think to resolve this problem is to calculate the chances of losing and then: \begin{equation} P(W)=1-P(L) \end{equation} where $ P(W) $ is chances of winning and $ P(L) $ is chances of losing. But how do I calculate $ P(L) $ ?

I've tried this,but I'm almost sure that's wrong: $P(L)=$chances of getting an ace in first position or chances of getting a 2 in second position or chances of gettin a 3 in third position or chances of getting an ace in fourth position and so on...

So: \begin{equation} P(L)=\frac{4}{52}+\frac{4}{51}+\frac{4}{50}+\frac{3}{49}+\frac{3}{48}+\frac{3}{47}+\frac{2}{46}+\frac{2}{45}+\frac{2}{44}+\frac{1}{43}+\frac{1}{42}+\frac{1}{41} \end{equation} Thanks everybody:)

Answer 1

Hint: Think of the initial configuration of the deck. You will count one $18$ times and the aces need to be among the other $34$ spaces. You will count two $17$ times and the twos need to be among the other $35$ spaces. You will count three $17$ times and the threes need to be among the other $35$ spaces. So $\frac {34}{52} \frac {33}{51} \ldots $The odds are not good. As a rough approximation, you would expect each of $12$ cards to kill you $\frac 13$ of the time each so you will win $\left( \frac 23 \right)^{12}\approx 0.77 \%$. It is actually better than this, as the aces are taking spaces that would let the twos and threes kill you.

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The claimed exact solution above is not correct. There are correlations between where the aces go and the places for twos and threes that are not considered. To see the problem, consider a deck having just one ace, one two, and one three. There are two winning decks: 23A and 3A2 for a probability of $\frac 13$. The above would say the ace can go two places, chance $\frac 23$ (correct) and the two can go two places, but the ace may have taken one. I think the proper answer would have to enumerate how many two slots are taken by aces, then how many three slots are taken by aces and twos. It is a lot of work. The approximate value is probably pretty close.

Answer 2

Aside from Ross' answer: You can take an iterative approach on the problem:

On you first move, the chance to pick an ace is $\frac{4}{52}$.

On you second move, you have to take in account that you might have already picked a two. So with a probability of $\frac{4}{52}$, your chance to pick a two is $\frac{3}{51}$, and with a probability of $\frac{48}{52}$, your chance to pick a two is $\frac{4}{51}$.

On you third move, you've already had two chances to pick a three, and so on...

But take in account that e.g. on your fourth move, you've only had 2 chances to pick an ace, and not three (since you're not allowed to pick an ace on your first move).

Answer 3

This question was later reasked as Probability of winning the game "1-2-3" and not identified as a duplicate, and Hagen and Byron both gave exact answers, in good agreement with Ross' estimate.