This is the Follow up Question to A game with 52 cards. So it requires to understand previous question! I believe it is a bit harder to find!

what is the chance to know the positions of all the face-down cards in Alice's row in the worst optimal case?

Please indicate your answer as fraction.


closed as unclear what you're asking by f'', CodeNewbie, manshu, Hugh Meyers, Haobin Mar 4 '16 at 11:05


  • $\begingroup$ @CarlLöndahl yes, it is the worst optimal case. it has to require 51 tries to find it! but your answer is not right in my opinion. $\endgroup$ – Oray Mar 3 '16 at 18:44
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    $\begingroup$ It would be very useful to state all the conditions in the question. $\endgroup$ – Paul Evans Mar 3 '16 at 19:00

The chance of getting all of the cards right on your first guess is 1/52!, but if you get all but two, then you actually know the position of the last two (it's the opposite of the way you had them!). Three doesn't work, because you only know one of the ways that the three aren't arranged, but there are more than one other ways.

So what is the chance that you know all but two? If it was only the first two, then it would be a chance of 1/52! as well, because there is only one permutation that has them switched. Because it can be any two, and there are 52*51 possible positions for the two to be placed in the list, it is 51*52/52! = 1/49!.

But we need to remember the chance that we got them all right to begin with, so the final answer is:

(52*51 + 1)/52! = 2653/52!

  • $\begingroup$ This is the chance to know all of the cards on the first guess. If you are asking about the chance to eventually know all of the cards in the worst optimal case, then it is 100% : you always eventually know the cards. If you are asking about the chance that the worst case happens, then that is different. $\endgroup$ – Lacklub Mar 3 '16 at 18:56

1 / 52! If you try all possible guesses.

  • $\begingroup$ no actually, this is not right unfortunately :) $\endgroup$ – Oray Mar 3 '16 at 18:26
  • $\begingroup$ Then can you please provide more info. In the last one she told you if a card was right. In this one it's worded as if she just tells you yes or no to the whole thing. $\endgroup$ – Z. Dailey Mar 3 '16 at 18:28
  • $\begingroup$ imagine there were 3 cards, what would be the answer? 1/3!? think again. $\endgroup$ – Oray Mar 3 '16 at 18:31
  • $\begingroup$ Actually I'm pretty sure it works. 3 cards would result in 6 combos. If you want to talk probabilities then the average person would solve it in half the time of the total possible outcomes. $\endgroup$ – Z. Dailey Mar 3 '16 at 18:38
  • $\begingroup$ But you said worst possible not average @Oray $\endgroup$ – Z. Dailey Mar 3 '16 at 18:39
