# 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

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.