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:
\[\left( {\frac{{26}}{{52}} \times \frac{{25}}{{51}} \times \frac{{24}}{{50}}} \right) \times 2 = \frac{{31200}}{{132600}} = \frac{4}{{17}}\]
Notes, References & Links:
[1] This is Gardner's set up and suggested 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.'
Is Gardner correct? Do each of the eight possibilities have equal chance of occurring? Do we need to consider the probabilities of the first two cards being the same colour, given that we know they are?
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.
ReplyDeleteI find that when logic and probability puzzles have heated proponents of several answers, the problem is usually in the wording.
DeleteHow does the wording make a difference here? How does the friend looking in their faces change the answer?
(I do think Gardiner is wrong though, and I’d like to read his reasoning.)
Yes, getting your head around the wording of such problems is the thing. The 'friend' is the person looking at the cards. S/he therefore *knows* that the first two cards are the same colour. So the probability that the third is the same colour shouldn't take into account the probability that the first two were, hence my tending toward the third solution. Gardner's solution also doesn't take into account that the possibility of say RRR occurring is not the same as say RRB. (His reasoning is quoted in the notes, by the way.)
DeleteWe are talking about the 3th card. The 3th card does not care how the first 2 became the same, it could have been random or put there on purpose. The only thing that counts is that when you pick the 3th card you have 26 of one colour and 24 of the other colour.
ReplyDelete