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A deck of cards is shuffled and the cards are then turned over one at a time until the first ace appears. Given that the first ace is the 20th card to appear, what is the conditional prob that the card following it is a) the ace of spades and b) the two of clubs?

I can do both parts easily - but I tried it another way and can't get the correct answer. So my first way I used the definition $P(E|F) = P(EF)/P(F)$ Let $E$ be the event that the 21st card is the ace of spades, $H$ the event that the 20th card is the first ace and $F$ the event that the 20th card is the ace of spades.

So for the second way let$$P(E|H) = P(E|FH)P(F|H) + P(E|F^cH)P(F^c|H)$$ The first term here is $0$ and so we need only consider the second term.

I said $P(E|F^cH) = (48C19)(3C1)(1C1)(31!)/(52!),$ i.e the first 19 cards have to not be any aces $(48C19)$, the 20th card has to be an ace but not the ace of spades,$(3C1)$, the 21st card is ace of spades (1C1) and the rest can be any of the remaining cards.

Similarly, $P(F^c|H) = (48C19)(3C1)(32!)/(52!)$ Or would this simply be 3/4? My reasoning being you know the 20th card is an ace. So the probability of it being the ace of spades is 1/4 => not ace of spades with prob 3/4? Thanks

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  • $\begingroup$ $P(F^C|H) = \frac34$ for the reason you said. Was that your only question? $\endgroup$ – Jonathan Christensen Nov 30 '12 at 21:32
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I said $P(E|F^cH) = (48C19)(3C1)(1C1)(31!)/(52!),$ i.e the first 19 cards have to not be any aces $(48C19)$, the 20th card has to be an ace but not the ace of spades,$(3C1)$, the 21st card is ace of spades (1C1) and the rest can be any of the remaining cards.

As your denominator of $52!$ makes clear, you’re counting permutations. There are $\binom{48}{19}$ ways to choose the first $19$ cards, but they can then be arranged in $19!$ different orders, so the first factor should be $19!\cdot\binom{48}{19}=\frac{48!}{29!}$: there are

$$\frac{48!}{29!}\cdot3\cdot 31!=3\cdot30\cdot31\cdot48!$$

permutations that result in the desired outcome $EH$, and the probability is therefore

$$\frac{3\cdot30\cdot31}{49\cdot50\cdot51\cdot52}=\frac{93}{5\cdot17\cdot49\cdot52}=\frac{93}{216580}\approx0.0004294\;.$$

Similarly, $P(F^c|H) = (48C19)(3C1)(32!)/(52!)$ Or would this simply be 3/4? My reasoning being you know the 20th card is an ace. So the probability of it being the ace of spades is 1/4 => not ace of spades with prob 3/4?

Yes, it’s simply $\frac34$, for the reason that you give.

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  • $\begingroup$ The answer you gave multiplied by $\frac{3}{4} $ gives the incorrect result. In my first attempt, using the form $P(E|H)= P(EH)/P(H)$, I said this was equal to $ (48C19)(3C1)(1C1)(31!)/(52!) $all divided by $(48C19)(4C1)(32!)/(52!)$. The former term being $P(EH) $ and the latter $P(H)$. This gives the correct answer. But now that I think about it, your idea of multiplying by an extra $19!$ makes sense. Why do I have the right answer? $\endgroup$ – CAF Dec 1 '12 at 9:42
  • $\begingroup$ I see why your answer multiplied by 3/4 gives the incorrect answer- that was event EH you calculated and not what I thought you computed. So I now have the correct answer both ways, but without considering that 19!. Why? Thanks! $\endgroup$ – CAF Dec 1 '12 at 9:58
  • $\begingroup$ @CAF: Your $$\frac{\binom{48}{19}\binom31\binom11 31!}{\binom{48}{19}\binom41 32!}$$ gives the correct result because you’re missing a factor of $19!$ in both the numerator and the denominator, so the two omissions cancel out. $\endgroup$ – Brian M. Scott Dec 1 '12 at 10:04

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