The Little Johnny's Dilemma

Little Johnny is walking home. He has $300 he has to bring home to his mom. While he is walking a man stops him and gives him a chance to double his money. 

The man says -

 "I'll give you $600 if you can roll 1 die and get a 4 or above, you can roll 2 dice and get a 5 or 6 on at least one of them, or you can roll 3 dice and get a 6 on at least on die. If you don't I get your $300."
 
What does Johnny do to have the best chance of getting home with the money?

THIS is the BEST he can do! 

The Loss Making Deal !

What was the deal?


The little Johnny should not accept the deal.

The problem is actually finding the probability of winning in each case dice throw.

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CASE 1 :  

Condition - Roll a die and get 4 or above.

The probability in the case is 3/6 = 1/2 (getting 4,5,6 from 6 possible outcomes). 

That is only 50% chances of winning the deal.

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CASE 2 :

Condition - Roll 2 dice and a 5 or 6 on at least one of them.

Let's find the probability that none get a 5 or 6. 

Probability that single die doesn't get a 5 or 6 is 4/6 = 2/3 by getting 1,2,3,4 in possible 6 outcomes.

Since, these 2 results are independent, the probability that neither dice gets a 5 or 6 is 2/3 x 2/3 = 4/9.

So, the probability that at least one of them gets a 5 or 6 is 1 - 4/9 = 5/9.

That is, about 56% chances of winning the deal.

By other approach, there are 36 possible combinations in 2 dice throw.

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) 

There are total 20 combinations (last 2 rows and 2 columns) where at least one die gets a 5 or 6.

Hence, the probability in the case is 20/36 = 5/9 i.e. about 56% of winning chances.

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CASE 3 : 

Condition - Roll three dice and get a 6 on at least on die.  

Again finding probability that none gets a 6.

Probability that single die doesn't get a 6 is 5/6 by getting 1,2,3,4,5 out of 6 possible outcomes.

Hence, the probability that three dices doesn't get a 6 = 5/6 x 5/6 x 5/6 = 125/216.

Then the probability that at least one of them get a 6 = 1 - 125/216 = 91/216 

That is only 42% chances of winning the deal.

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On the other hand Johnny can be 100% sure that he will take $300 to home if he doesn't accept the man's deal. 

The Mistimed Clock!

Andrea’s only timepiece is a clock that’s fixed to the wall. One day she forgets to wind it and it stops.

She travels across town to have dinner with a friend whose own clock is always correct. When she returns home, she makes a simple calculation and sets her own clock accurately.

 
How does she manage this without knowing the travel time between her house and her friend’s?


That's how she manages to set it accurately!
 

Correcting The Mistimed Clock!

 How it was mistimed?

Andrea winds her clock & sets it to the arbitrary time. Then, she leaves her house and when she reaches her friend's house, she note down the correct time accurately. Now, after having dinner, she notes down the correct time once again before leaving her friend's house.

After returning to home, she finds her own clock acted as 'timer' for her entire trip. It has counted time that she needed to reach her friend's house + time that she spent at her friend's house + time she needed to return back to home.

Since, Andrea had noted timings at which she reached & left her friend's house, she can calculate the time she spent at her friend's house. After subtracting this time duration from her unique timer count she gets the time she needed to reach to & return from her friend's house.

She must have taken the same time to travel from her home to her friend's home and her friend's home to her home. So dividing the count after subtracting 'stay time' she can get how much time she needed to return back to home.

Since, she had noted correct time when she left her friend's home, now by adding time that she needed to return back to home to that, she sets her own clock accurately with correct time.

Let's try to understand it with example.

Suppose she sets her own clock at 12:00 o' clock and leave her house. Suppose she reaches her friend's home and note down the correct time as 3:00 PM. After having dinner she leaves friend's home at 4:00 PM.

After returning back to home she finds her own clock showing say 2:00 PM. That means, she spent 120 minutes outside her home with includes time of travel to and from friends home along with time for which she spent with her friend. If time of stay at her friend is subtracted from above count, then it's clear that she needed 60 minutes to travel to & return back from friend's home.

That is, she needed 30 minutes for travel the distance between 2 homes. Since, she had noted correct time as 3:00 PM when she left friend's home, she can set her own clock accurately at 3:30 PM.

Grandma's Birthday And Troll Toll

You are on your way to visit your Grandma, who lives at the end of the valley. It's her birthday, and you want to give her the cakes you've made.

Between your house and her house, you have to cross 7 bridges, and as it goes in the land of make believe, there is a troll under every bridge! Each troll, quite rightly, insists that you pay a troll toll. Before you can cross their bridge, you have to give them half of the cakes you are carrying, but as they are kind trolls, they each give you back a single cake.

How many cakes do you have to leave home with to make sure that you arrive at Grandma's with exactly 2 cakes?

How many you should carry? Know here!

Source 

Cakes For Grandma's Birthday

What's between you & grandma's birthday celebration? 

No need to overthink on this. Just carry 2 cakes.

At each toll they would take 1 ( half of 2 ) & would give back 1.

So after each toll you would carry 2 cakes & last toll wouldn't be an exception.