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. 

Challenge of Father to Son

A man told his son that he would give him $1000 if he could accomplish the following task. 

The father gave his son ten envelopes and a thousand dollars, all in one dollar bills. He told his son, "Place the money in the envelopes in such a manner that no matter what number of dollars I ask for, you can give me one or more of the envelopes, containing the exact amount I asked for without having to open any of the envelopes. If you can do this, you will keep the $1000."

When the father asked for a sum of money, the son was able to give him envelopes containing the exact amount of money asked for. 

How did the son distribute the money among the ten envelopes?

THIS is how son accepts the challenge!

Son's Response to the Father's Challenge

What was the challenge?

For a moment, let's suppose father had given $30 to son and provided 5 envelopes and put the same challenge.
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Here, use of the binary number system helps in matter.

The son would distribute 15 dollars into 4 envelops like - 

2^0 = 1, 2^1 = 2, 2^2 = 4, 2^3 = 8.

Now, for any amount asked between 1 to 15, son can produce some of these 4 envelops wherever 1 is there in envelop column for that particular amount.

For example, if father asks for $10 (Binary - 1010), son would give envelop 4 and 2 (8+2=10).

After putting $15 dollar in 4 envelops, he puts remaining $15 in 5th envelop so that he can cover rest of amount between 16 to 30.

If father asks amount greater than 15 then he would take envelop of $15 first and depending on how much the amount asked is greater than this $15 he would pick some of those 4 envelops.

For example, if father asks for $24, then he picks envelop 5 having $15 and envelops for amount 24 - 15 = 9 (Binary - 1009) i.e. envelop 4 and 1 (8+1=9)  i.e. total of 15 + 9 = 24.

So, what we observe from this is that the number of envelops needed for such arrangement is equal to the number of binary bits needed to represent the amount itself or nearest power of 2 greater than the amount.

In above case, to represent 30 in binary we need 5 bits or nearest power of 2 greater than 30 is 32 which needs 5 bits for representation.

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Now, let's turn to the actual challenge where father has asked son to distribute $1000 rightly in 10 envelopes. 

The reason for selecting 10 as a number of envelops is clear now as 1000 needs 10 bits in binary or nearest power of 2 greater than 1000 is 1024 which needs 10 bits for binary representation.

So, the son puts 256, 128, 64, 32, 16, 8, 4, 2, 1 dollars in 9 envelops (envelop numbered as Envelop 9, Envelop 8.........Envelop 1 in order)
 and 1000 - 511 = 489 dollars in 10th envelop.

First 9 envelops will cover amounts from 1 to 511 and for amounts greater than 511 inclusion of 10th envelop having 489 dollars is mandatory.

Again selection of envelops for the amount 511 to 1000 depends on how much the amount exceeds the $489. The binary representation of that difference and selection of envelop accordingly is all that needed.
 
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Let's make sure this distribution with couple of examples.

If father asks for amount of $109 (binary - 1101101) then son picks 

Envelop 1($1) + Envelop 3 ($4) + Envelop 4 ($8) + Envelop 6 ($32) + 
Envelop 7 ($64) i.e. having amount = 1 + 4 + 8 + 32 + 64 = 109 dollars.

If father asks for $525 then son gives $489 via Envelop 10 and rest of amount 
530 - 489 = 40 (Binary - 101000) in form of Envelop - 6 ($32) and Envelop 4 ($8).   

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The Apple Conundrum

Two women are selling apples. The first sells 30 apples at 2 for $1, earning $15. The second sells 30 apples at 3 for $1, earning $10. So between them they’ve sold 60 apples for $25.

The next day they set the same goal but work together. They sell 60 apples at 5 for $2, but they’re puzzled to find that they’ve made only $24.

What became of the other dollar?

Here, could be that lost dollar! 

Behind The Apple Conundrum

What is the conundrum?

They sell 60 apples at 5 for $2, that means 12 such sets of 5 apples. Suppose, out of each such set, 1 woman takes out $1 for 2 apples and other takes $1 for 3 apples. So, first woman earns $12 by selling 24 apples and second woman sells 36 apples for $12.

In short, first woman gives away 6 apples (from her 30 apples) to second woman increasing her count to 36 reducing her own count to 24. First woman would have made $3 from those but second woman only made $2 from those 6 apples. And there is that lost dollar in earning.

So, 60 apples can't be divided equally to find the earning as they had sold apples at different rates on previous day.

Other way, if they wanted to sell apples together with 30 apples each, then they should have sold apples at average of (1/2 + 1/3)/2  = $5/12 per apple (i.e. 12 apples for $5) instead of $2/5 per apple.

The difference in price per apple (5/12 - 2/5) = (1/60).

So the difference in earning after selling 60 such apples = (1/60) x 60 = 1.

And there is that other dollar! 

What's wrong gone here on next day? Instead of averaging dollars per apple, apples per dollar are added directly which resulted reduced cost of each apple. 

"Save Your House From Me!"

Okay, I’ll ask three questions, and if you miss one I get your house. Fair enough? Here we go:

1.A clock strikes six in 5 seconds. How long does it take to strike twelve?

2.A bottle and its cork together cost $1.10. The bottle costs a dollar more than the cork. How much does the bottle cost?

3.A train leaves New York for Chicago at 90 mph. At the same time, a bus leaves Chicago for New York at 50 mph. Which is farther from New York when they meet?

You need little common sense in answering above!

Presence of Common Sense in Answers!

What where questions?

1. A clock strikes six in 5 seconds. How long does it take to strike twelve?

A: Not 10 seconds, it takes 11 seconds.

Here, interval between 2 strikes is 1 second i.e. if counter started at first strike, it will count 1 second after second strike, 2 seconds after third strike & so on. 

Hence, 11 seconds needed for strike 12.

2.A bottle and its cork together cost $1.10. The bottle costs a dollar more than the cork. How much does the bottle cost?

A: Not 1 dollar, it would cost $1.05.

If x is cost of cork,
x + (x +1) = 1.10
2x = 0.10
x = 0.05

Hence cost of bottle is $1.05 and cost of cork is $0.05



3.A train leaves New York for Chicago at 90 mph. At the same time, a bus leaves Chicago for New York at 50 mph. Which is farther from New York when they meet?

A: Obviously, when they meet at some point then that point must be at the some distance from New York. Hence, they are at the same distance from the city.

Profit Or Loss Or No-Profit No-Loss?

A man buys a horse for $60. He sells the horse for $70. He then buys the horse back for $80. And he sells the horse again for $90. How much profit did he make or did he loose in transaction? Or did he break even?

Confused? Find the right answer here!

And He Earned A Profit!

But what was the deal?

The confusion starts when he he buys same horse again. But if we look at it as two different transactions then it pretty straightforward.

At first he buys horse for $60 & sells it for $70. Here, he makes profit of $10. This is one transaction.

In next transaction, he buys same horse for $80 & sell the same for $90. Again, here he makes profit of $10.

In this way, the total profit he earns from these transactions is of $20.  

Mathematical Coincidence

Messi entered a candy shop and spent half of the money in his pocket. When he came out he found that he had just as many paise as he had rupees when he went in and also half a many rupees as he had paise when he went in. How much money did he have on him when he entered? (1 Rupee = 100 Paise just like 1 Dollar = 100 Cents)

Know the amount that Messy had initially!!

Source 

Money For Mathematical Coincidence

What was the coincidence?

Let X be the rupees & Y be the paise that Messy initially had in his pocket. That means he had 100X + Y paise initially. In shop he paid half of the amount he had i.e. (100X + Y)/2. 

When he came out of the shop he had Y/2 rupees & X paise i.e. 100(Y/2) + X paise.

Equating both,

(100X + Y)/2 = 100(Y/2) + X

Multiply by 2,

100X + Y = 100Y + 2X

98X = 99Y


Hence, X = 99 & Y = 98
 
Putting these values in 100X + Y = 9998. This is the amount in paise. Converting in rupees gives, 9998/100 = 99.98

 
To conclude, Messi had Rs.99.98 initially in his pocket.


 

Round Table Coin Game

You are sitting with one opponent at an empty, round table. Taking turns, you should place one dollar on the table, in such a way that it touches none of the coins that are already on the table. The first player that is not able to place a dollar on the table has lost. By tossing a coin, it has been decided that you may start.

Which strategy will you follow to make sure you are guaranteed to win?

  
Trick to win this game always! 

Never Loose Round Table Coin Game

What was the game? 

There is little trick with which you will always end on winning side in this Round Table Coin Game. Since you have got first chance to place the coin you should place the coin right at the center of the round table. Now for every next coin placed by opponent you need to place coin in such a way that it 'mirrors' opponent's coin.

Imagine line from the center to opponent's coin. Place your coin exactly opposite to that coin at distance equal to distance between center & opponent's coin. Or imagine a circle (with the center fixed at the round table) with opponent's coin lying on it's border.  And place your coin at diagonally opposite point of point where opponent placed coin on that imaginary circle. (Assume imaginary circle though it's not appearing perfectly in the image above)

In this way for every move of your opponent, you will have 'answer' in form of space for placing the coin. This will continue until last place left on the table with your turn of placing the coin in the end. 

This is how to make sure you always on winning side in this 'Round Table Coin Game'!