How do we play poker without those cards?

Briony takes a standard pack of 52 cards and throws some cards away. However, she makes sure she keeps all four aces among the remaining cards. She then selects four cards at random from these remaining cards. If the probability of her selecting the four aces is 1 1001 \dfrac {1}{1001} , how many cards did she throw away?


The answer is 38.

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1 solution

Sharky Kesa
Jul 2, 2015

Assume that Briony has k k cards left in the pack after she threw some cards away. Then, there are k ( k 1 ) ( k 2 ) ( k 3 ) 24 \dfrac {k(k-1)(k-2)(k-3)}{24} different ways of selecting four cards. Since there is only one way of picking up four aces, the probability of such an event is 24 k ( k 1 ) ( k 2 ) ( k 3 ) \dfrac {24}{k(k-1)(k-2)(k-3)} . Hence

24 k ( k 1 ) ( k 2 ) ( k 3 ) = 1 1001 \dfrac {24}{k(k-1)(k-2)(k-3)} = \dfrac {1}{1001}

Therefore k ( k 1 ) ( k 2 ) ( k 3 ) = 24 × 1001 k(k-1)(k-2)(k-3) = 24 \times 1001 . There is only one value of k k that satisfies this equation, and the fact that 1001 = 7 × 11 × 13 1001 = 7 \times 11 \times 13 helps to quickly find it: k = 14 k = 14 . Thus the answer is 52 14 = 38 52 - 14 = \boxed{38} .

Why can't we throw away 63 63 cards so that we have a negative probability of 4 11 -\dfrac { 4 }{ 11 } of drawing the first ace, 3 12 -\dfrac { 3 }{ 12 } of drawing the second ace, 2 13 -\dfrac { 2 }{ 13 } of drawing the third ace, and 1 14 -\dfrac { 1 }{ 14 } of drawing the fourth, so that the product is 1 1001 \dfrac { 1 }{ 1001 } ? Or am I just being nonsensical?

Michael Mendrin - 5 years, 11 months ago

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The question stated that all four aces remained in the pack.

Sharky Kesa - 5 years, 11 months ago

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...eh, a technicality

Michael Mendrin - 5 years, 11 months ago

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