Solution to Puzzle #20: The Monty Hall Problem

Posted on June 15, 2013 by Alok Goyal

This is a very tantalizing puzzle. At the outset it is very difficult to see why the odds should be different than 50-50. Thanks to everyone who responded, and there were many who did.

The answer goes as follows:

Contestants who switch have a 2/3 chance of winning the car, while contestants who stick have only a 1/3 chance. One way to see this is to notice that there is a 2/3 chance that the initial choice of the player is a door hiding a goat. When that is the case, the host is forced to open the other goat door, and the remaining closed door hides the car. “Switching” only fails to give the car when the player had initially picked the door hiding the car, which only happens one third of the time.

If one would like a detailed discussion on this puzzle as this one has baffled mathematicians for a while, please go to: http://en.wikipedia.org/wiki/Monty_Hall_problem

Hope you enjoyed this!

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Puzzle #20: The Monty Hall Problem (Very Interesting – Must try!)

Posted on June 9, 2013 by Alok Goyal

This is a puzzle I picked up from the Movie “21” that I was watching 2 days back. The scene is a class in MIT, where the professor gives this problem to one of the students. Here it goes:

Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?

 Please send your answers directly to me on e-mail – alokgoyal_2001@yahoo.com

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Solution to Puzzle #19: Find the Average Salary

Posted on June 9, 2013 by Alok Goyal

This puzzle generated more responses than any of the other puzzles I have sent to date – thanks for the overwhelming response. What was more heartening was the creativity of the different solutions I received. In the solution, therefore, I will try to describe as many of the solutions as possible. Before, I do that, I first need to give credit to Gurmeet’s blog (http://gurmeet.net/puzzles/) where I took the puzzle from – he has mentioned that a friend of his asked him this puzzle in 1996.

For children, I have created a video answer on this link:

The simplest solution is the following: One person takes a large random number, adds his salary to it, passes on to the next person who adds his salary to the resulting number and the round goes on. When it comes back to the first person, he subtracts the random large number and gets the addition of all the four numbers and divides it by four to find the average. KJ was the first one to give the correct answer on the blog.

Similar solutions were sent by other folks – Rajwinder gave a small variation to this puzzle – wherein each person adds a random number in the first cycle and then subtracts that random number in the second cycle, thus arriving at the same answer.

Found another variation on Gurmeet’s blog site where I took the puzzle from:

Each engineer can break his salary in two parts(need not be equal) and whisper into both left and right neighbor’s ears. Then Each engineer will add the salary of his neighbor’s and announce it. They will add the total and divide by 4 to get the average. This way no one will be able to find the average of other three.

Tarun Gugnani gave a variation of the above by dividing into 3 parts instead of 2 and whispering the three parts to each of the three people and then doing the same.

Alok Mittal gave another interesting answer, though people end of revealing a very small part of their salary in this process:

this has been a real tease… does this work (though it does communicate more than an iota) – each of the engineers takes responsibility of totaling a certain “position of digit”, i.e. units, tens, hundreds etc – for a given position, the other engineers tell the totaler their numbers. So everyone knows some digits of the other people’s salaries but no one knows anyone else’s salaries… of course, one can use binary notation to make number of digits larger than what the decimal notation would provide…

Finally, I got another very interesting answer from Rajat Bhargava, whose answer is complicated, but very unique. It is a bit difficult for me to replicate here and hence attaching the answer directly. number 19

Thanks to everyone who tried – hope that you all will repeat this for the next puzzle which I think is even more interesting.

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Puzzle #19: Find the Average Salary

Posted on June 2, 2013 by Alok Goyal

This is one of the best puzzles I have posted, and would highly encourage all of you to try it – kids as well as adults.

Four honest and hard-working computer engineers are sipping coffee at Starbucks. They wish to compute their average salary. However, nobody is willing to reveal an iota of information about his/her own salary to anybody else. How do they do it?

 

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Solution to Puzzle #18: Who Will Win The Game

Posted on June 2, 2013 by Alok Goyal

I got two correct answers for the puzzle – no surprise that Tishyaa was one of them, and I also got an answer from Alok Kuchlous immediately after I posted the puzzle – thanks!

The answer was relatively simple – for a pile of n stones, you can make n-1 moves, therefore for the three piles, you can make 42 moves. Given that this is an even number, the one who plays first will lose. Here is the answer explained in a video for kids.

Hope you enjoyed this!

 

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Puzzle #18: Who will win the game?

Posted on May 26, 2013 by Alok Goyal

Here is a two player game – there are three piles of stones: one with 10 stones, one with 15 stones, and one with 20 stones. At each turn, a player can choose one of the piles and divide it into two smaller piles. The loser is the one who cannot do this. Who will win and how?

Source: Mathematical Circles, by Dmitri Fomin, Sergey Genkin and Ilia Itenberg

Age Group: For all ages

Note: please mail answers to alokgoyal_2001@yahoo.com

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Solution to Puzzle #17: Perfect Square with 0’s, 1’s and 2’s

Posted on May 26, 2013 by Alok Goyal

Surprisingly I did not get any correct answers to the puzzle.

The solution to the puzzle lies in divisibility rules – any number with 100 0’s, 1’s and 2’s will be divisible by 3 since the digits add up to 300. However, the number will not be divisible by 9, no matter what the sequence of these digits is. Therefore, the number can never be a perfect square. Argument remains the same with 10 0’s, 1’s and 2’s!

I got a couple of answers that argued that for any large square, there necessarily needs to be digits other than 0, 1 and 2. That is not necessarily true, for example square of 110 is 12100.

Hope you enjoyed this!

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Puzzle #17: Perfect Square with 0’s, 1’s and 2’s

Posted on May 18, 2013 by Alok Goyal

Since the previous puzzle was very tough, a relatively simpler problem.

Can a number be written with one hundred 0’s, one hundred 1’s and one hundred 2’s be a perfect square. If yes, what is that number, if not, prove it.

For Children: Do the same puzzle with ten 0’s, 1’s and 2’s instead of hundred.

Source: Mathematical Circles, by Dmitri Fomin, Sergey Genkin, Ilia Itenberg

 

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Solution to Puzzle #16: Crossing the Desert

Posted on May 18, 2013 by Alok Goyal

This is probably one of the tougher puzzles, and the responses indicate the same. I got only one answer for the puzzle, and once again, from Tishyaa – Thank you Tishyaa!

1. For the simpler puzzle, One needs to to set up a depot at 200 miles, leave the truck with a full load, empty 1/3rd at the 200 mile post and come back. In the next trip, leave again with a full load, reach 200  miles. By this time, 1/3rd of the fuel tank is empty as the total capacity is for 600 miles. At the 200 miles, therefore, it refills the 1/3rd of the full load available and again becomes full. With this load, it can cross the remaining 600 miles.

2. The solution to the tougher version is derived from an analysis presented in an issue of Eureka, a publication of mathematics students at The University of Cambridge. To make terminology simple, lets call 500 miles a “unit” and gasoline for 500 miles as a “load” and a “trip” is a journey of the truck in either direction from one stopping point to another.

Two loads will carry the truck a maximum distance of 1 and 1/3 units (as we have seen in the children’s version).

Three loads will carry the truck 1 and 1/3 plus 1/5 units in a total of 9 trips. The first deposit point (call it a cache) is 1/5 units from the start. Three trips put 6/5 in the cache (duming 3/5 load in each stop). The truck returns, picks the remaining full load and arrives at the first cache with 4/5 load in the tank. This, together with the fuel in the cache, makes two full loads, sufficient to take the truck another 1 and 1/3 as we have seen.

We are asked for the minimum amount of fuel required to take the truck 800 miles. Three loads will take it 766 and 2/3 miles (1 + 1/3 + 1/5), so we need a third cache at a distance of 33 and 1/3 miles (which is 1/15 units). In five trips, the truck can build up this cache so that when the truck reaches this cache at the end of 7th trip, the combined fuel of the truck  and the cache will be three loads. As we have seen, this will then be sufficient to take the truck the remaining 766 and 2/3 miles. Seven trips are made between starting point and first cache, using 7/15 load of gasoline. The three loads of fuel that remain are just sufficient to cross the rest of the way, and hence the total gas used will be 3+ 7/15 loads, which is ~3.46 loads.

As you can also see, the truck can successively cross 1, 1/3, 1/5, 1/7…and so on. This is a diverging series and hence the truck can cross any distance.

The problem has some more variations, will encourage the more mathematically oriented reader to read Martin Gardner’s My Best Mathematical and Logic Puzzles, puzzle #25. The original puzzle, btw, appeared in 1946/ 1947 in a few publications.

Apologies for a text only solutions, my iPad is not with me this weekend 😦

 

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Puzzle #16: Crossing the Desert, a Classic Puzzle!

Posted on May 12, 2013 by Alok Goyal

This is one of my favorites – I read this puzzle for the first time in 1992 on rec.puzzles (for those of you who remember these newsgroups!), and honestly have struggled with this ever since. I saw this puzzle again recently in one of the Martin Gardner books (more on sources when I send the answer). While the original puzzle is too tough for children, I have introduced a variant which the kids should be able to do.

Children’s version:

An unlimited supply of Gasoline is available at one edge of a desert 800 miles wide, but there is no source on the desert itself. A truck can carry enough gasoline to go 600 miles (this will be called one “load”), and it can build up its own refueling stations at any spot along the way. These caches (i.e. gasoline storage) can be of any size, and it is assumed that there is no evaporation loss.

What is the minimum amount of Gasoline (in loads) the truck will require in order to cross the desert?

Adult Version:

(i) In the previous version, change the size of the load to be enough to go for 500 miles and not 600 miles, and answer the same question.

(ii) Is there a limit to the width of a desert the truck can cross?

Source: My Best Mathematical and Logic Puzzles, by Martin Gardner, Puzzle #25

Please send the answers directly to me at alokgoyal_2001@yahoo.com

 

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