Wednesday, 2 December 2020

On the Colour of the Third Card


Three cards are dealt from a normal deck.  You don't see them.  You are told that the first two are the same colour, but not what colour they are.  What is the probability that the next card is the same colour?

\[\frac{1}{2},\;\frac{1}{4},\;\frac{{12}}{{25}},\;or\;\frac{4}{{17}}\;?\]


This beautiful, onthefaceofit innocuous little problem comes from the great Martin Gardner's 'Modelling Mathematics with Playing Cards'.  It's a problem that never fails to invoke heated discussion and vehement argument when I share it with students to play with.  The four solutions and reasoning sketched out below are those typically proffered (and invariably staunchly defended) by students.  Which solution would you go with, and why?  Do you have a different solution?  And what argument would you give to those who firmly hold the solution to be one of the others to show them that they are wrong and you are right?  (Note that Gardner's solution was solution 2 [1])


Solution 1

The last card could either be 1) the same colour by being red, or 2) a different colour by being red, or 3) the same colour by being black, or 4) a different colour by being black.  So there are four possibilities and two of these result in the final card being the same colour as the previous two.  So:

\[\frac{2}{4} = \frac{1}{2}\]


Solution 2

There are eight ways that the arrangement of the colours of the three cards can occur, namely RRR, RRB, RBR, BRR, RBB, BRB, BBR, or BBB (where R = Red and B = Black), and two of these arrangements have the final card as the same colour as the previous two.  So:

\[\frac{2}{8} = \frac{1}{4}\]


Solution 3

After the first two cards, fifty cards remain in the deck.  If the first two cards are both red, there remain twenty-four cards that are red.  Similarly, if the first two cards are both black, there remain twenty-four cards that are black.  So: 

\[\frac{24}{50} = \frac{12}{25}\]


Solution 4

For the last card to be the same colour as the first two, the colours of all three cards are either RRR or BBB.  You have fifty-two cards to choose from for your first card, and twenty-six of these are red and twenty-six are black.  After taking the first card from the deck, you have fifty-one cards left to choose from for your second card, and assuming that the first card was red, twenty-five of the fifty-one cards left are also red.  After taking the second card from the deck, you have fifty cards left to choose from for your third and final card, and assuming that the first two cards were red, twenty-four of the fifty cards left are also red.  The same would be true if the cards pulled from the deck were black.  So: 

\[\left( {\frac{{26}}{{52}} \times \frac{{25}}{{51}} \times \frac{{24}}{{50}}} \right) \times 2 = \frac{{31200}}{{132600}} = \frac{4}{{17}}\]




Notes, References & Links:

[1] This is Gardner's set up and solution: 'Three face-down cards [are] dealt from a shuffled deck.  A friend looks at their faces and turns over two that are the same color.  What's the probability that the remaining face-down card is the same color as the two face-up cards?  You might think it is 1/2.  Actually it is 1/4.  Here's the proof.  The probability that three randomly selected cards are the same color is two out of eight equal possibilities, or 1/4. Subtract 1/4 from 1 (the card must be red or black) and you get 3/4 for the probability that the face-down card differs in color from the two face-up cards. This is the basis for an ancient sucker bet. If you are the operator, you can offer even odds that the card is of opposite color from the two face-up cards, and win the bet three out of four times.'

3 comments:

  1. Wording is crucial here, no? In the original, Gardner states "A friend looks at their faces and turns over two that are the same color. What's the probability that the remaining face-down card is the same color as the two face-up cards?" That's not what you wrote and I think the probabilities are, then, different.

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