# Two Card Probability Problem

I came across this puzzle somewhere on the interwebs. Has stumped me for an answer for quite sometime. Helpplease?

Someone picks, at their will, two cards from a deck of cards. The cards have different numbers, i.e one card is higher than the other. (In other words, the person picks two distinct numbers in the inclusive range 1 through 13. [11-13 : J-K]

The cards are placed face down on a table in front of you. You get to choose one of the cards and turn it face up.

Now, you will select one of the two cards (one of whose face you can see, the other one you can’t). If you select the highest card, you win.

Is there a card-selection strategy for which your chance of winning is strictly greater than 50% ?

• Does "at their will" mean that they choose the value of both cards? i.e. if they want to pick Q & K, they can? – Joel Rondeau Sep 18 '14 at 18:01
• I guess it just means that all 13*12 possible pairs are possible. – Yaitzme Sep 18 '14 at 18:08
• Could you specify what you tried to do and what problem do you experience answering this question? For me it looks very simple. – klm123 Sep 18 '14 at 18:21
• I thought I had solved the question until I realized that my naive strategy (if openedcard > 7, pick same card. Else pick other card) had winning probability of exactly 50%. – Yaitzme Sep 18 '14 at 18:35
• Without the at will part it would be trivial - one rule: always keep the card if it is not a 1 you always win 50%, on a 1 you win 100%, so overall > 50% – Falco Sep 26 '14 at 13:14

## 2 Answers

Use a random source to pick a number from 2 to 13. If it is equal or less than the value of the card you picked, keep the card. If it is greater, switch. Thus, if you turned over a 1, you always switch; and a 13, you never switch.

If the card numbers are $a$ and $b$, with $a > b$, then you win if you flip $a$ ($50\%$ chance) and don't switch (probability $(a-1)/12$), or if you flip $b$ and do switch (probability $50\% \cdot (13-b)/12$). Your total chance of winning is: $$P = 0.5 \cdot{a-1\over 12} + 0.5 \cdot{13-b\over 12} = 0.5 \cdot {a - 1 + 13 - b \over 12} = 0.5 +{a-b\over 24}$$

Since $a > b$, your odds of winning are greater than $50\%$.

Simply choosing to keep a card that is higher than $7$ will not work, as they are not picking cards at random. They could deliberately choose $9$ and $10$, so that strategy would have only a $50\%$ chance of winning.

• Fair point, I did not notice the "at will" distinction. – kaine Sep 18 '14 at 18:23
• I think your math is a bit wrong in the last step. Should be about 0.5 + (a-b)/24. – Fabian Sep 18 '14 at 21:16
• Where a=13 and b=1, (0.5+(a-b)/12)=1.5. Yep, should be (0.5+(a-b)/24). – Brilliand Sep 18 '14 at 23:01
• This is the general solution, which surprisingly even works for any two numbers the other one can choose (even without limit!) – Falco Sep 26 '14 at 13:15
• Here is the general problem with any two real numbers, beautifully explained: puzzling.stackexchange.com/questions/2523/… – Bogdan Alexandru Jan 9 '15 at 8:32

The card you have facing up has a value of $X$.

The probability of randomly picking a card greater than $X$ is $\frac{4*(13-X)}{13*4-4}=\frac{13-X}{12}$. The chance that it is smaller than $X$ is $\frac{X-1}{12}$

The second card is, essentially, randomly chosen from the cards in the deck not equal to $X$. You can therefore pick the face up card if $X$ is greater than $7$. If $X$ is less than $7$ then pick the other card. In each of these cases, you have a probability of winning greater than 50%. It is actually equal to $\frac{|X-7|}{12}+\frac{1}{2}$. If, however, $X=7$ then you have a 50% probability for either one being higher.

• This works assuming the cards are chosen randomly. But from the problem statement it looks like the first person chooses the card however he likes. He could restrict the choice to the range 1..6. In this case your strategy would win with probability 50% only. – Florian F Sep 20 '14 at 12:16