Probability/ Stats Puzzle – 1 (Solution)

If you have not tried to solve the puzzle, click here for the problem. The problem was discussed in an earlier post.

This is a famous puzzle, called the “Monty Hall Problem”. Monty Hall was a host in the early episodes of the game show Let’s Make a Deal.

The common version of the the puzzle used three doors (instead of 3 boxes) and a car and two goats (instead of gold and garbage).

The problem was originally posed by Steve Selvin and became famous when it was quoted by Marilyn vos Savant in Parade magazine in 1990.

The answer: You increase the probability of winning the gold if you change your choice of the box to open. The probability of winning the gold is only 1/3 if you continue with your original choice and 2/3 if you change your choice.

Here is a brief explanation of why:

When you initially selected a box, you had a 1/3 probability of being right. The host knowingly opened a box with garbage in it, so that eliminated one of the wrong choices.  You still have a 1/3 probability that you initially chose the right box; this means that the other unopened box has a 2/3 probability of containing the gold.

Amit Bhattacharjee, Satish K Mariyappagoudar, and Patrick OToole got it right.

Better explanations are provided on the wikipedia page here.

Or you can watch the youtube video.

If the video does not load, the link is

You can also search the internet for the keywords “Monty Hall Problem” – you will get lots of hits.

Please feel free to to share your views, experiences, and queries, using the “comments” feature available.
You may also forward the link to this post to your friends, colleagues, and anyone else who may be interested.


Nothing Official About It! – The views presented above are in no manner reflective of the official views of any organization, community, group, institute, country, government, or association. They may not even be the official views of the author of this post :-).

I am Rajesh Naik. I am an author, management consultant and trainer, helping IT and other tech companies improve their processes and performance. I also specialize in CMMI® (DEV and SVC), People CMM® and Balanced Scorecard. I am a CMMI Institute certified/ authorized Instructor and Lead Appraiser for CMMI® and People CMM®. I am available on LinkedIn and I will be glad to accept your invite. For more information please click here.

7 thoughts on “Probability/ Stats Puzzle – 1 (Solution)”

  1. Somehow I have not been able to convince myself with the purported solution which assumes that the probability of the first chosen box containing the gold still remains 1/3rd after removal of one box which clearly didn’t contain the gold. Once such a box is removed and probability relooked at afresh; it will be 50:50 for both the remaining boxes irrespective of the fact whether you you had chosen one box or the other (because factually you still don’t have any clue about which box has the gold).
    On the other hand it is doubtful to say that “You still have a 1/3 probability that you initially chose the right box; this means that the other unopened box has a 2/3 probability of containing the gold.” Actually nothing has changed for the probability of the chosen box as well as for the remaining (third box). All three boxes had probability of 1/3rd initially. It is not clear that why probability of remaining box has increased to 2/3rd while the probability of chosen box is static.

    1. Hi Arun

      Thanks for your reply and comments.

      There are a lot of explanations on the web – try the keywords “Monty Hall Problem” in Google. There are also solutions available using Bayesian theory.

      If none of them are able to convince you, you can try physically running it at home (or office) a hundred times when you switch the choice and another hundred times when you don’t.

      You may not want to use gold and garbage and boxes but you can use 1 chocolate and 2 chocolate wrappers and three cups.

      Of course, you will have to run the experiment a large number of times to see the difference (100 times should do the trick), and with someone else hiding the chocolate and wrappers in cups, and revealing which one has the wrapper, to allow you a switch.

      If the solution that I presented turns out right, please eat the wrappers. If my solution turns out wrong, I will eat the chocolate (you will have to courier it to me) 🙂



  2. The solution you’ve given to the Monty Hall Problem is dependent on 3 conditions being met, none of which are specified in the 3 box problem you described.
    In order to get an answer of 2/3 for switching doors:
    1) Monty has to offer the opportunity to switch whether you picked the prize or not (in your problem the host’s motivations are unclear)
    2) Monty knows where the prize is and intentionally opens a losing door(not mentioned in your problem)
    3) Given a choice of 2 losing doors (because you picked the prize) Monty picks 1 at random (again not mentioned in your problem)

    The only reasonable answer to the question as you posed it is: flip a coin for a 50% chance

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