# 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 :-).

# 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.

# 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.

# 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.

Or you can watch the youtube video.

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

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 :-).

# 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?

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.