I had posted this once last year, I think….

a. Gin : Phillipines!!! (Say what???)
b. Beer : Czech Republic
c. Rum : India!!!!!!
d. Whiskey: France (what?????)
e. Vodka: Russia (ok, this I get)
f. Wine: Vatican City (this also, I can understand)
g. Tequila: US (not Mexico!!!)
h. Milk: Finland
i. Tea: Turkey (very surprised it is not China)
j. Coffee: Finland (most years) / Netherlands (2020)

Headed to DC on a Thursday. Which always reminds me of those puzzles I used to post every Thursday on my way out of DC. Posting puzzles after a long time. And since one good turn deserves another, I will post two…

First a warmup one…

1. Assuming you are as old as me (which is about 50), what is the probability that you have seen in your life somebody (human being) with more than average number of legs?

a) Certainly
b) Very likely
c) As likely as not (50-50)
d) Unlikely
e) Impossible

Now for an interesting one…

2. In my running trail, there is a spot where I have to climb up ten stairs. At any step, I can jump up one stair or two stairs. How many different ways can I climb up those ten stairs?

If you are reading this on Facebook, just send me personal message with your answer – I will let you know if you got it and post your name as somebody who got it right. There might be some time gap – I have a long night in front of me today.

Getting bored in the loooooong flight back. Suddenly remembered an email exchange I had with my friend Mita this week. It was actually a puzzle. In all honesty, I had heard a puzzle somewhat like that, so I was able to get it. Figured I will post it here.

The challenging part of the puzzle is its incredible simplicity. There are no catches in the puzzle. But the simplicity is bound to confuse readers. So here it goes…

“John called Mary. Mary called Tom. John is married but Tom is not”.

From the above, can you say whether the following statement

“A married person called an unmarried person” is

(a) Definitely True
(b) Definitely False
(c) Insufficient information to say whether True or False.

Send me personal message instead of putting in Comments and I will respond.

A couple of days back, after Sharmila had returned from India, we went out for lunch. We were discussing her experiences in India when the topic went to the new restrictions of driving in New Delhi. If you were not aware, New Delhi has promulgated new rules (I am not sure if it is effective still) that stipulates that cars with odd or even number license plates can be driven on alternate days. So, one day, if cars with only even number license plates can be driven, the next day it would be for the cars with odd number license plates. This, if I am not very mistaken, is to keep the pollution under control.

As a side story, I am told that women drivers are exempt from this rule. Which prompted me to suggest to Sharmila that we lost a great business opportunity to sell “burkhas” (the head to toe garb that many Muslim ladies wear) in New Delhi. Who is going to figure out whether it is a guy or a girl driving once they put a “burkha” on 🙂

In any case, another idea came to mind then. How about those guys with license plates that are reversible? Like if I had a license plate “6666”, I can easily take out four screws and put them back on with the license plate looking “9999”. In practice, that is not possible since there are other things written on the license plate that will clearly expose the trick but that did give rise to today’s puzzle:

How many license plates can be there which when reversed (and reattached to the car) gives a legitimate alternate parity (odd becomes even, even becomes odd) license plate?

Assume the following:

1. License plates are no more than 4 digits long
2. A license plate cannot start with 0
3. 0,1,6,8,9 are the digits that while reversed looks like legitimate digits
4. There cannot be two license plates which look the same on the roads ever

Here you go Debajyoti – since you were clamoring for another one. This one involves probability and can be considered very simple or very intriguing depending on your point of view…

I was having drinks with two of my friends last evening when the discussion meandered around our kids. All three of us have two kids each and no twins. Chris mentioned that his boy has really taken to high school football. The other child is less inclined towards sports and prefers to excel academically. Rick, the other friend – who is the youngest of the three of us – talked about his four year old son overjoyed with his grandparents’ visit during Christmas and all the gifts he got. Rick and his wife also just had a baby – and that was one more reason the grandparents had come – to help them around in the house.

So, now the question for you: Which of my two friends – Chris or Rick is more likely to have a daughter?

It has been a long time since I posted a puzzle. This is a variation of those category of logic problems where blindfolded people regularly get colored hats put on them and they have to guess the color of the hat on their own head 🙂

This particular one goes this way – Twelve sailors get marooned in an island. The local tribe finds them out and takes them to their leader. Their leader, not knowing what to do with the prisoners, come up with a novel idea. He tells them that next day morning, he was going to line them up in a straight line and put a cap on their head while they are blindfolded. The only colors he has his hats in are black and white. (I do not have the faintest idea why on God’s green earth the tribal leader had black and white hats, but that is what he had 🙂 ) After that, he would remove everybody’s blindfold. Then, every sailor could see the color of the hats of all the people in front of himself but not his own or that of the ones behind.

The leader would start from the last person in the rear and ask him to name the color of the hat he was wearing. He could only say “Black” or “White”. Any other words spoken would result in everybody getting killed. If the person got the right answer, he would be spared, else he would be killed. Then the leader would move to the next person who was ahead of the last person and go thru the same procedure… till he was done with all twelve of them.

You can assume that everybody can hear the answer given by the person being asked. All the people who were spared would be given a boat and shown how to go back to their land.

Tonight, the sailors have to come up with a strategy to save as many of themselves as they can.

What would be the best strategy to maximize the number of people who can survive? How many can survive?

When I met Lia’s mom (see a couple of posts back) last evening, Lia mentioned to her mom and her sister about my FB posts. And quickly added that she totally skips my puzzles. Which gave me an idea to form a puzzle including Lia and see what she does about it 🙂

Lia and I, both Saturday morning runners at Fowler Park, decide to run together tomorrow morning. We agree to meet up between 6am  and 7am. Both of us are definitely going to arrive in that timeframe. But, since we were likely to reach there at different times, we agreed on the rule that upon reaching the trailhead, if one does not find the other waiting there, he/she will wait exactly for 20 minutes for the other person before proceeding to run by himself/herself. Of course, in the middle of the wait, if it becomes 7 am already, the person just goes ahead for the run alone, (realizing that the other person came earlier and must have left after waiting for 20 minutes.)

Question is what is the probability that we will actually get to run together?

Last morning, I was having a rather humorous email exchange with some of my friends from my previous job when one of them, Gasan, sent a very interesting puzzle. This is not that simple. What is surprising is his kids got this problem as a homework problem. Also Raj let me know that the answers are all over the internet. So, if you want to exercise your brain, do not Google it up. After you have given it whatever time you want to give, feel free to look up the answer.

Meanwhile, just to show how much fun we had as a team – or rather the team had at my expense, I am going to copy Gasan’s articulation of the problem as is…

A line of 100 airline passengers are waiting to board a plane. They each hold a ticket to one of the 100 seats on the flight. (for some reference, let’s say that the nth passenger in line has a ticket for the seat number n.)
Unfortunately, the first person in line is a crazy bald headed guy named Rajib, and will ignore the seat number on their ticket, picking a random seat to occupy. All the other passengers including Gasan, Chris, Sunjay, even Raj and Karthik are quite normal, and will go to their proper seat unless it is already occupied. If it is occupied, they will then find a free seat to sit in, at random.
What is the probability that the last (100th) person to board the plane will sit in their proper seat (#100)?