Question: $4$ cards are shuffled and placed face down in front of you. Their hidden faces display 4 elements: water, earth, wind, fire. You turn over cards until win or lose. You win if you turn over water and earth. You lose if you turn over fire. What is the probability that you win?

I understand that wind is effectively absent from the sample space. Does not affect your chances of winning or losing. I also know that $\frac13$ (because we removed wind), you can pick fire where you lose the game.

  • 1
    $\begingroup$ What is the probability that fire is the last of the three cards that matter? That’s the probability that you lose. $\endgroup$ Apr 17, 2015 at 18:37
  • $\begingroup$ VMO, in your question, you state "you win if you turn over water AND earth". Did you mean instead to say that "you win if you turn over water OR earth"? $\endgroup$
    – user137481
    Apr 17, 2015 at 20:00
  • $\begingroup$ @user137481 I believe it is OR. If one picks the card that is Water or Earth, game ends and you win. $\endgroup$
    – AG10
    Apr 17, 2015 at 20:36
  • $\begingroup$ VMO, OK. Just note that Brian's answer is based on turning over water AND earth. $\endgroup$
    – user137481
    Apr 18, 2015 at 22:54
  • $\begingroup$ I read the problem the same as Brian did. I recommend changing the text to, "You win if you turn over either water or earth." Then add a line after the second paragraph explaining that the question originally read "water and earth." That way the correct answer will make sense, and so will the comments. $\endgroup$
    – David K
    Apr 18, 2015 at 23:22

3 Answers 3


There are two intepretations here. For both you can throw out the Wind card and only focus on the three other cards. Restating the question:

You shuffle four cards. There is a bijection between your four cards and Water, Earth, Wind, and Fire; i.e., unique assignments.

  1. You win if you flip over $W$ and $E$ (both) before an $F$.

    There are only $3\cdot 2 \cdot 1 = 6$ permutations: $\{WEF, EWF, WFE, EFW, FWE, FEW\}$. Only $WEF, EWF$ result in wins so $P(win) = \frac{1}{3}$. This is the same as the probability of getting $F$ as the last card.

  2. You win if you flip over either $W$ or $E$ (inclusive) before an $F$.

    Looking at the permutations in 2a, you lose if $F$ is the first card. Since only $FWE, FEW$ result in a loss, $P(win) = \frac{2}{3}$.


Since there are 4 cards, there are 4! = 24 possible orderings of the cards.

You lose in the following cases:

1) Fire is the first card. There are 3! = 6 ways for this to occur. 2) Wind is the first card and Fire is the second. There are 2! = 2 ways for this to occur.

Hence, the probablity of losing = $\frac{6+2}{24}=\frac{1}{3}$
So your answer is correct.

  • $\begingroup$ What about the cases where you e.g. turn over earth first and then fire? $\endgroup$ Nov 27, 2016 at 11:08
  • 1
    $\begingroup$ @Henrik According to OP, you win the game if you turn over either water or earth. So, the only way you lose with Fire as the second card is if the first card is not water or earth (i.e. it is Wind). $\endgroup$
    – user137481
    Nov 28, 2016 at 19:31

Probablity of losing is not 1/3. It should be 2/3.

As Below, He wins if,

  1. First three cards are Water/Earth/Wind = 3!
  2. First two cards are Water/Earth and third is Fire = 2

so the total = 6 + 2 = 6

Probability of winning = 8/24 = 1/3 , so loosing 2/3.

  • $\begingroup$ $\frac{6}{24}=\frac14\neq \frac13$... $\endgroup$
    – Arnaud D.
    Nov 27, 2016 at 12:17

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