Three cards are dealt face down from a normal deck (52 cards; 26 red suits and 26 black). Two are turned over and are the same colour. What is the probability that the third card is the same colour as the first two?
\[\frac{1}{2},\;\frac{1}{4},\;\frac{{12}}{{25}},\;or\;\frac{4}{{17}}\;?\]
This beautiful, onthefaceofit innocuous little problem comes from the wonderful 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 suggested solution was solution 2 [1].)
Solution 1
The third 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
\[\frac{2}{8} = \frac{1}{4}\]
Solution 3
As explained in 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. After the first two cards are turned over, fifty cards remain unturned, and twenty-four of them will be the same colour as the first two. It is given that the first two cards turned over are the same colour as the third, so the probability that the first two cards are the same colour as the third is 1. If the first two cards turned over 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:
\[{1} \times\frac{24}{50} = \frac{12}{25}\]
Solution 4
For the third card to be the same colour as the first two turned over, 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 turning the first card over, 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 turning the second card over, 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 turned over were black. So:
\[\left( {\frac{{26}}{{52}} \times \frac{{25}}{{51}} \times \frac{{24}}{{50}}} \right) \times 2 = \frac{{31200}}{{132600}} = \frac{4}{{17}}\]
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