# Conditional probability of guessing the correct rank of a card drawn from a standard deck

Alice draws a card from a standard deck, doesn't look at it, and gives it to Bob and he looks at it and says "it's not a King" (a spoiler of this type will, henceforth, be referred to as a Bob-Spoiler). Alice wants the probability that she'll draw a Queen next:

There are 51 cards left and 4 cards left for Alice to draw (3 if it has been drawn already). On average, it has not been drawn 11 out of 12 times (13 ranks, but we know that we can disregard the King from Bob's spoiler). Therefore I claim that the exact probability is:

$$\frac{4*\frac{11}{12}}{51}$$

And if Alice continues after n rounds (she lost $n-1$ guesses so far), we have:

$$\frac{4*\frac{13-n}{13-n+1}}{52-n+1}$$

(Please let me know if you disagree that this is the exact probability.)

My real intention of asking this question is me wanting an approximate and simple expression for the probability (which doesn't lose much accuracy even if Alice guesses after, say, a dozen cards have already been drawn (Bob-Spoilers have been dispensed every time). He varies this rank in his spoilers every time (distinct) so the maximum this little game can be played is with 13 drawn cards.)

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I have calculated the first probability: it is not quite the same as yours. I do not understand how the rest of the game goes well enough to calculate other probabilities. – André Nicolas Sep 4 '12 at 2:28
@AndréNicolas well,. the original game actually goes like this: a players guesses that the first card will be a 2, then if he does not win, he claims that the next card will be a 3, if he guesses incorrectly again he repeats the whole thing until he guesses that an ace comes on the 13th hand (so worst case he draws 13 cards without sucsess. he wins the game if he can guess a card right and the game will stop then and there. im trying to find the probability of winning this game and i think i can generalize it now. thanks for your clear answer. – Wuschelbeutel Kartoffelhuhn Sep 4 '12 at 2:32
That looks like a different game. I think exact probabilities can be calculated. – André Nicolas Sep 4 '12 at 2:43

Let us calculate the probability that Alice will draw a Queen next, given that the first card is not a King.

Given that the first card is not a King, the probability it is a Queen is $\frac{4}{48}$, and the probability it is not a Queen is $\frac{44}{48}$.

If the first card is a Queen, the probability that the second is a Queen is $\frac{3}{51}$. If the first card is not a Queen, the probability the second is a Queen is $\frac{4}{51}$. Thus given that the first card is not a King, the probability the second is a Queen is $$\frac{4}{48}\cdot\frac{3}{51}+\frac{44}{48}\cdot \frac{4}{51}.$$ This simplifies to $\frac{1}{12}\cdot\frac{47}{51}$. That is somewhat larger than the answer that you gave, so even at the start ($n=2$) your expression does not give the exact probability.

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Is there a nice approximation for, say, the fourth card? It appears as you go further the logical tree gets more and more complex (AND OR combinations)... – Wuschelbeutel Kartoffelhuhn Sep 4 '12 at 7:40
Could think about it if I knew what the game was. But for me the description of how the game proceeds after the second card is not clear. Or are you talking about what you called the "original game" in an earlier comment? In any case, cannot do anything until morning, it is very late. – André Nicolas Sep 4 '12 at 7:47
Yes, the original game. After the one guesses wrong that the i'th card drawn is not of rank i, one draws another card up to i=13 (i=1 means ace, i=13 means king). if by the 13th card one does not guess right, the game is lost. i'm trying to get the probability of winning this game. i got a sim up and running but i cant solve it by hand. – Wuschelbeutel Kartoffelhuhn Sep 5 '12 at 2:37