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The Monty Hall problem

I’m looking for the better explanation to the solution of this riddle, because I always struggle to get people to accept the ones I know.


There are three cards facing down on a table and one of them is the 2 of clubs.

You are given the chance to point one of them to guess the 2 of clubs (1/3 chances of getting it right).

With the card you choose still facing down, the card dealer pick one of the other remaining two cards, and face it up, showing to you that it’s not the 2 of clubs.

You are now given the opportunity to change you card selection between the two that still facing down. What should you do to get the maximum probability of wining, keep your initial choice or change it to the other card? Why?


marked as duplicate by Ross Millikan, Thomas, Chris Eagle, Norbert, J. M. is a poor mathematician Oct 5 '12 at 13:10

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ See the Monty Hall problem. Short answer: You should switch to the other card. $\endgroup$ – Per Manne Oct 3 '12 at 10:39
  • $\begingroup$ View "Choose one card and always switch afterwards" as one single action, and see how likely you are to hit gold. $\endgroup$ – Arthur Oct 3 '12 at 10:40
  • $\begingroup$ You are right, but I was looking for an answer that most people could easly understand. I've tried to explaint the solution using different approachs (including the ones in the post you refer) but most people struggle to accept it. $\endgroup$ – Luis Oct 3 '12 at 10:45
  • $\begingroup$ @Luis That may be a problem of the people rather than of the explanations. Such probability tricks seem to be hard to communicate. Please don't take this as "arrogance" of mathematicians over "laypersons" - a lot of mathematicians also got upset over Savant giving the "wrong" (though correct) answer, so "we" have nothing to be arrogant about. $\endgroup$ – Hagen von Eitzen Oct 3 '12 at 13:00

You pick one card, but do not look at it. The chance that you have picked the 2 of clubs is $1\over 3$.

Let the dealer pick up the two other cards. The chance that he has picked up the 2 of clubs is $2\over 3$.

The dealer asks if you want to trade your one card against his two cards. Since that will increase your odds of getting the 2 of clubs, you say yes.

The dealer looks at his cards without showing them to you or telling you what he got. From your point of view, the chance that he has the 2 of clubs is still $2\over 3$.

The dealer reveals one card which is not the 2 of clubs. He can always do that, since he chooses the card himself, so the chance that he has the club of 2 is still $2\over 3$.

The dealer asks you again if you want to trade your one card against his two. You say yes, take his two cards, and discard the revealed card which is not the club of 2. The chance that you are left with the club of 2 is now $2\over 3$.


This is called the Monty Hall problem. The way I've heard it explained is that after your initial guess there is a two thirds chance that you have chosen incorrectly. When the dealer eliminates a card, the chances you are wrong don't change. Thus the one remaining card you didn't choose has a two thirds chance of being the $2$ of clubs.


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