I encountered this simple problem in a book (I will disclose the name of the book in a later post along with the answer).

Here is the problem:

You need to call the throw of a dice a 1000 times. Like all dices, in each throw, this dice also gives you a number between 1 and 6. You are also told that the dice is slightly distorted / damaged – the probability of getting the six results is as follows: 1- 20%; 2- 10%; 3- 25%; 4-15%; 5-15%; 6-15%.

What strategy would you use to call the answers for the 1000 throws? Your objective is to get the right answer for a maximum of the throws.

Here are some answers that I have heard:

Call the number ‘3’ all the 1000 times – this is the most common answer I have heard.

Call the numbers in the same pattern as the probability: 1- 200 times; 2- 100 times; 3- 250 times; 4-150 times; 5-150 times; 6-150 times.

Call the numbers randomly, ignoring the distortion in the dice.

A variation of 2 above is to call the numbers in the same pattern, but also taking into account the answers to the past throws, so that we try and keep the probabilities similar to the expected patterns. So if in the first 100 throws, 1 has already rolled more than 20% and 2 has been rolled less than 10%, then in the 101st throw, call 2 instead of 1, and so on.

There are other possible answers too – and the right one may not be listed above (this is not a mutiple choice question 🙂 )

Work out the reasons for your choice, not just make a choice. The reasons are more important.

This is a simple question, and you should get the right answer.

The answer will be posted later.

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

Calling 3 a 1000 times might give me the highest probability for the right answer (probability of 250 correct calls).

Alternately, I will call 3 for 250 throws; check probability of occurences for the 250 throws (for 1,2,2,4,5,6); use this information to take a call for the next 250 (call with highest probability for the past 250 calls) and so on for the next 250 till I reach 1000 to maximize my probability of success.

Calling 3 a 1000 times might give me the highest probability for the right answer (probability of 250 correct calls).

Alternately, I will call 3 for 250 throws; check probability of occurences for the 250 throws (for 1,2,2,4,5,6); use this information to take a call for the next 250 (call with highest probability for the past 250 calls) and so on for the next 250 till I reach 1000 to maximize my probability of success.

Regards

Prakash