Two Aces?

Probability Level 4

Three playing cards, two Aces and a King, are on the table upside down.

One of the three is then returned to the deck, leaving only two cards on the table.

  • Ryan picks one of the two at random. It's an Ace.
  • Maya picks one of the two at random. It's an Ace.
  • Hana picks one of the two at random. It's an Ace.
  • Jaycie picks one of the two at random. It's an Ace.

If the probability that both cards are Aces is a b \dfrac{a}{b} , where a a and b b are coprime positive integers, what is a + b a+b ?

Image credit: http://www.goodpokergames.com


The answer is 17.

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Geoff Pilling by Geoff Pilling , 53, USA

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1 solution

Geoff Pilling
May 9, 2017

Let 2 A T 2A_T = Probability that two Aces were left on the table.

Let 4 A P 4A_P = Probability that an Ace was picked 4 times.

Then, by Bayes' theorem ,

P ( 2 A L ∣ 4 A P ) = P ( 4 A P ∣ 2 A L ) P ( 2 A L ) P ( 4 A P ) P(2A_L|4A_P) = \dfrac{P(4A_P|2A_L)P(2A_L)}{P(4A_P)}

= 1 ⋅ 1 3 1 3 ⋅ 1 + 2 3 ⋅ 1 16 = \dfrac{1 \cdot \frac{1}{3}}{\frac{1}{3}\cdot 1+\frac{2}{3}\cdot \frac{1}{16}}

Note : The denominator is derived as follows: There is a 1 3 \frac{1}{3} probability there are two Aces on the table, in which case an ace was picked four times with certainty. There is a 2 3 \frac{2}{3} probability there was an Ace and a King on the table, in which case there was a 1 16 \frac{1}{16} chance that an Ace was picked four times.

P ( 2 A L ∣ 4 A P ) = 1 1 + 1 8 = 8 9 P(2A_L|4A_P) = \dfrac{1}{1+\frac{1}{8}} = \dfrac{8}{9}

8 + 9 = 17 8+9= \boxed{17}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

It's interesting that if we stopped at Ryan's pick, then the probability would be 1 / 2 1/2 ; I would have anticipated higher. So in general, if n n people in a row pick an Ace then the probability that both cards are Aces is 2 n − 1 2 n − 1 + 1 \dfrac{2^{n-1}}{2^{n-1} + 1} .

I'm experimenting with various initial conditions now to see if any surprising scenarios emerge ....

Brian Charlesworth - 4 years, 1 month ago

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Surprise #1: If we start with m m Aces and 1 1 King then after n n Aces picked in a row the probability that all m m remaining cards are Aces is m n − 1 m n − 1 + 1 \dfrac{m^{n-1}}{m^{n-1} + 1} , the surprise being that the probability after n = 1 n = 1 is always 1 / 2 1/2 , regardless of what m m is.

Brian Charlesworth - 4 years, 1 month ago

Good problem related to Bayes theorem...

Indraneel Mukhopadhyaya - 4 years, 1 month ago

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