Probability/ Stats Puzzle – 1

This problem was presented to me by Swapna (my wife) on last Friday – I could not work out the right answer even after considerable struggle.

You are participating in a TV show contest. You have reached the last round. If you win this round, you get take home a pure gold brick of 5 KG (5KG = 11.02 lbs); if you lose you have to take away an equivalent quantity of stinking garbage.

Here is the problem in the last round:

There are 3 closed boxes (let us say B1, B2, B3). Inside two of the boxes is garbage. Inside one of the boxes is the gold. You have to open one box and take home whatever is in that box. You decide to open B1. The show-host/ quiz-master asks you to stop, and as a hint opens one of the other two boxes, and inside that box there is garbage. The show-host gives you the option of changing your choice. Would you still go for your original choice or switch to the other unopened box?

Three Boxes

 

Here are some relevant assumptions/ hints/ guidances:

  1. Most important: you would prefer to take home the gold instead of the garbage :-).
  2. You will not be able to smell the garbage or gold without opening the boxes, or in any way be able to “know” what is inside the unopened boxes.
  3. You do not know the show host’s motivation. The show-host may be trying to help you or trick you, or trying to increase hir (his/her) popularity rating, or just following a script. So, do not consider the show host’s motivation in trying to solve the problem (when Swapna presented me the problem, I went on the motivation track, and could not approach it as a problem of probability, even after she told me to ignore the show host’s motivation 🙁 ).
  4. There is no “trick” in the problem or the solution – so, approach it as a problem of probability/ statistics.
  5. Do not be lazy and search the internet to find a solution. That is cheating. I have changed some things in the problem so that is not easy to search. However, this is not a test of how quickly and ingeniously you can search the internet.
  6. You will have to work out the reasons for your choice, not just make a choice. The reasons are more important.

Don’t feel bad if you don’t get the answer right, many renowned statisticians have got it wrong.

The answer is available in another post here.

Please share your views in the “comments” feature available.

 


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.

17 thoughts on “Probability/ Stats Puzzle – 1”

  1. The odds of picking the gold box has gone up from 1/3 to 1/2. So i will continue to hold onto the belief in the box picked and pray.

    Not sure what complex probability computations are at play here.

    Will await the golden solution

    Regards
    Prakash

  2. 1. don’t go with the show host’s because he will try to make you to chose the wrong one only, but still the probability is very less if you go with the show host’s.
    2. Depending upon your hear and brain decision this can be taken but this may or may not be the right box.
    According to me this is the only solution depending on the situation given.

    1. Hi Sridhar,

      Thanks for your comment. Like me, I think you took the track of the show host’s motivation, rather than viewing this as a pure probability problem. Maybe you should attempt it again, asssuming that the show-host is neutral to whether you win gold or garbage.

  3. Hi Rajesh,
    Here is the solution.

    The intuition is that the quiz master has opened B3 because he knows that there is no gold in it, which increases the chances that B2 has the gold.

    With probability too, we can solve using Bayes theorem.

    Prob that B1 has gold given that B3 is opened = (Prob B3 is opened given that B1 has gold * Prob that B1 has gold)/Prob B3 is opened = (1/2 * 1/3)/(1/2) = 1/3. The chances that B2 has the gold is therefore 2/3. In the above calculation, the denominator 1/2 comes about because the Prob that B3 is opened by the quiz master is 1/2. Since the quiz master will not open a box containing gold, he will open B3 (which surely does not contain gold) or one of B1 or B2 (whichever does not contain the gold).

    There is another more intuitive way of understanding Bayes.

    We take the prior odds of gold being in each box which is 1:1:1. We then multiply by the Bayes factor to get the posterior odds (after additional info is available). The Bayes factor is the odds that the host would open B3 given that the gold is in each of the boxes. So the chances that he will open B3 if gold is in B1 is 50%. The chances that he will open B3 if gold is in B2 is 100% and the chances that he will open B3 if the gold is in B3 is 0%. The Bayes factor is therefore 0.5:1:0 or 1:2:0. Multiplying this with the prior odds (1:1:1) gives the posterior odds as 1:2:0. This means that the chances that gold is in B2 are twice that of the gold being in B1. Better to switch his choice.

    Regards
    Amit

  4. A simple analysis can be as follows:

    Probability of Gold being available in all the three boxes

    P(B1) = P(B2) = P(B3) = 1/3

    The show contestant identifies and indicates one box say B1, based on some intuition but does not open it. But the show host knows in which box the gold is present but opens Box 3 which contains garbage and the contestant is asked if he would like to change his choice between 2 boxes, B1 to B2 and now the contestant has another option to change his option to B2.

    Now, the probability that the contestant finds Gold in either of these 2 Boxes, B1 and B2; Since B3 is opened by the show host.

    P(B1) = P(B2) = 1/2

    Answer: The probability of choosing the Box of Gold = 1 / 3 / 1 / 2 = 2/3, that would translate to an approximate 66.67 %.

    This can also be derived using the concepts of conditional probability and Baye’s Theorem. I will post this solution as well very soon. Watch out…

    Regards,
    Satish K Mariyappagoudar

  5. In Attempt 1, I have 33% chance of getting gold.

    After host opening one box.
    Not knowing past performance / outcomes of host motivation.

    Based on pure probability in Attempt 2, I have 50% chance of getting gold (nothing more nothing less).

    So my approach will be to toss my favorite coin and pick a box among B1 and B2.

  6. I will stick with my choice. the host gives me an option to change my choice – there is only 1 left now, the chance that the host is helpful / misleading is 50:50 assuming the host will mislead given the quantity of gold, it is better to stick with one’s own choice.

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