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What is the probability that the ace of spades is at the bottom of a standard deck of 52 cards given that the ace of hearts is not at the top?

I asked my older brother, and he said it should be $\frac{50}{51} \cdot \frac{1}{51}$ because that's $$\mathbb{P}(A\heartsuit \text{ not at top}) \times \mathbb{P}(A\spadesuit \text{ at bottom}),$$ but I'm not sure if I agree. Shouldn't the $\frac{50}{51}$ be $\frac{50}{52}$?

Thanks you!

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The number is correct but the reason isn't (or is unclear). It may have been a misprint. I would say it is $(50/51)(1/51)$=(probability that the ace of hearts is not at the bottom)$\times$(probability that the ace is at the bottom, out of the 51 non-ace-of-hearts positions). –  Jonas Meyer Jan 9 '12 at 4:35

3 Answers 3

The conditional probability is a quotient,

P(AS at bottom and AH not at top)/P(AH not at top).

(Here AS and AH are the ace of spades and ace of hearts respectively.)

The numerator here is equal to

P(AS at bottom) - P(AS at bottom and AH at top)

and you can rewrite the second term as

P(AS at bottom) - P(AS at bottom) P(AH at top | AS at bottom)

and now put in numbers to get

$1/52 - (1/52)(1/51) = (50/51)(1/52)$

The denominator of the original quotient is P(AH not at top) or $51/52$, so the answer is

$$ {(50/51)(1/52) \over (51/52)} = {50 \over 51^2} $$

which agrees with your brother's answer.

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You can also get the correct result simply by counting. There are $52!$ possible orders for the deck; $1/52$ of these have the ace of hearts at the top, so there are $$\frac{51}{52}\cdot 52!=51\cdot 51!$$ orders that do not have the ace of hearts at the top.

Now let’s count how many of these have the ace of spades at the bottom. The cards other than the aces of hearts and spades can appear in any of $50!$ different orders. The ace of spades must follow all of them, and the ace of hearts can then be inserted immediately after any of the first $50$ cards of the resulting $51$-card deck. Thus, there are $50\cdot 50!$ orders that have the ace of spades at the bottom and do not have the ace of hearts at the top.

The desired probability is therefore $$\frac{50\cdot 50!}{51\cdot 51!}=\frac{50\cdot 50!}{51\cdot 51\cdot 50!}=\frac{50}{51^2}\;,$$ your brother’s answer.

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The ace of hearts has 51 positions available (since it's not at the top).

Having placed it somewhere, there are 51 positions available for ace of spades, so

Pr = P(ace of spades not at bottom)*P(ace of diamonds at bottom)

= 50/51 *1/51 = 50/51²

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