# Probability/ Stats Puzzles – 2 & 3 (Solutions)

If you’ve not seen/ attempted the puzzles, the links are here: puzzle-2 and puzzle-3. These were presented in earlier posts.

Both these puzzles are adopted from a delightful little book by John Allen Paulos titled A Mathematician Reads the Newspaper.

I will provide more details about the book next week. For now, here are the solutions to the two puzzles.

## Puzzle-2

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.

Solution:

Call 3, 3, 3, 3…. all the 1000 times. This will get you aprroximately 250 right calls.

Or better still, tell the dice roller that your call is 3 all the thousand times, go for a coffee, or do something useful, come back after some time.

## Puzzle-3

Two contestants are to decide on the winner of 10 mn by flipping a coin. The winner will be the one who reaches six (6) correct calls first.

After 8 flips, contestant A has 5 correct calls, and contestant B has 3 correct calls. At this stage they agree NOT to continue with the flipping of the coin. Here are some proposals on how the money should be shared:

1. Contestant A says that since he is leading, he should get the 10mn.
2. Contestant B says that since the flipping was called off before the final result, the 10mn should be shared equally.
3. The show-host says that TV quiz program sponsors should retain the 10mn, since both the contestants agreed to call off the contest.
4. Someone from the audience suggests that the prize money be split in the 5:3 ratio (5 for A and 3 for B), in line with the number of right calls
5. A mathematician calls in to suggest that the money be split A7:B1 (try and guess the logic here, it is related to the probability of winning from this point, if the flipping had continued)

Solution:

The question on how the money is to be shared is not a mathematical /statistical problem at all! It is a matter of fairness and justice, and each solution proposed (and some yet to be proposed) has its own merit.

However, if you have not yet worked out the logic of why the mathematician proposed option # 5 above, here it is:

For contestant B to win 6 calls in a row, he/ she needs to call ALL of the next three calls correctly (even if he / she calls one incorrectly, A will reach 6 right calls. So the probability of B winning is (0.5) x (0.5) x (0.5) = 0.125; which means A has a probability of 0.875 – that is 7:1.

Next week, I will cover the source of these puzzles, a book titled A Mathematician Reads the Newspaper by John Allen Paulos.

You may also forward the link to this post to your friends, colleagues, and anyone else who may be interested.

Notes:

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.

# Probability/ Stats Puzzle – 3

I encountered another problem in the same book (I will disclose the name of the book in a later post along with the answer). Here is the problem:

Two contestants have reached the last round of a TV quiz contest and one of them is hoping to be the winner of a prize of 10 mn (currency deliberately left vague) via a tie-breaker. Even after the tie-breaker, neither of them has beaten the other.

The show-host offers to break the tie with a coin (my guess is that the show host did not have any more questions left :-)). However, to maintain the suspense and gain more TRP, he proposes that the winner will be one who reaches six (6) correct calls first.

After 8 flips, contestant A has 5 correct calls, and contestant B has 3 correct calls. At this stage both the contestants agree NOT to continue with the flipping of the coin (maybe the coin is lost or it breaks or falls into something disgusting – use your imagination). They have to decide on the winner based on result of the 8 flips.

Here are some proposals:

1. Contestant A says that since he is leading, he should get the 10mn.
2. Contestant B says that since the flipping was called off before the final result, the 10mn should be shared equally.
3. The show-host says that TV quiz program sponsors should retain the 10mn, since both the contestants agreed to call off the contest.
4. Someone from the audience suggests that the prize money be split in the 5:3 ratio (5 for A and 3 for B), in line with the number of right calls
5. A mathematician calls in to suggest that the money be split A7:B1 (try and guess the logic here, it is related to the probability of winning from this point, if the flipping had continued)
6. Any other…

It is interesting to note so many options to a simple situation.

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.

# Probability/ Stats Puzzle – 2

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:

1. Call the number ‘3’ all the 1000 times – this is the most common answer I have heard.
2. 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.
3. Call the numbers randomly, ignoring the distortion in the dice.
4. 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.
5. 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.