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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:


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:)

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up vote 4 down vote accepted

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.

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.

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I can't understand a couple of things:first the cards are 13 of each seed,why you use 12 cards?Second thing is:why every card should kill me $\frac{2}{3}$ of the time?Thanks a lot for your answer anyway. – gyosko Jul 26 '12 at 13:24
@wild91: Four aces, four $2$s and four $3$s -- those are the ones that can kill you. – joriki Jul 26 '12 at 13:26
@joriki I was counting the wrong cards,I got it now. Thanks:) – gyosko Jul 26 '12 at 13:28

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).

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So it will be something like this right?:$P(L)=\frac{4}{52} + (\frac{4}{52} * \frac{3}{51} + \frac{48}{52} * \frac{4}{51})$ ... and so on.. – gyosko Jul 26 '12 at 13:49
Yes, but keeping in mind that on your third move, you have to take in account that up to certain probabilities you might have removed zero, one, or even two threes from the deck, this might get quite messy... – roman Jul 27 '12 at 7:29

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.

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