## Amazon Interview Question

Software Engineer / DevelopersI think that both results here are a little off. Even here is 2,4,6 while odd is 1,3,5. Here's my answer:

public static int rollDice ()

{

// Are we going to do even or odd?

int eo; // 0 means we're going to be odd, 1 means even

if ( Math.random() > 0.72 )

eo = 0;

else

eo = 1;

// Generate an odd number (1, 3, 5)

int result = ( (int)( Math.random() * 3 ) * 2 + 1 );

// If eo is 1, adding it to the result makes it even (2, 4, 6); otherwise, it remains odd

result = result + eo;

return result;

}

The simulation program would typically involves generating random number between and 0 and 1 and check in which range it falls in. For equal probability it should be in range of ( 1/6, 2/6, 3/6 ... ) to bias it we can shift the range. i.e for 72 percent we can have ( 0-28/300, 28/300 - 100/300, 100/300 - 128/300 ,...). In this way we can have biased no for even no generator.

when u generate a random number n between 0 and 1, for equal probability, u assign the following way

0 - 1/6 = 1; 1/6 - 2/6 = 2; 2/6 - 3/6 = 3; 3/6 - 4/6 = 4; 4/6 - 5/6 = 5; 5/6 - 6/6 = 6

now for the even numbers to have 72% prob, we have to divide each 1/3rd into 1/3 * 28/100 and 1/3 * 72/100 i.e 28/300 and 72/300

now the ranges will be

0 - 28/300 = 1; 28/300 - 2/6 = 2; 2/6 - 128/300 = 3; 128/300 - 4/6 = 4; 4/6 - 228/300 = 5; 228/300 - 6/6 = 6

this way u get the biased probability of 72% for even numbers

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x=rand()*100;

if(x<=72)

x=(rand()*3)*2;

if(x>72)

x=(rand()*3)*2-1;

I think this works fine

Suppose we want to achieve a probability of p%, where p is an integer between 50 and 100.

Here, we can simply try to generate a random number between 0 and 99, and compare it against p.

There is a standard problem Amazon gives that asks that you generate a uniformly random number between 0 and 6 given a generator that produces numbers between 0 and 4. We can use a similar approach here.

Generate a 3-digit base-6 random number by rolling a die three times. That number will be between 0 and 215 inclusive (in base 10). If the number is larger than 199, throw it away and find another one. Otherwise take its modulo 100, and you have a random number between 0 and 99 to compare to p.

This is with a minimum of 3 die rolls. We need at least that many unless p is some very nice integer (say 75; we can do a probability of 3/4 in two throws).

- We can try by throwing the dice two times.

- Rinks September 19, 2008- Now we have 36 events.. How do we divide the events so that the probability of even number is more than that of odd numbers ?

It can be done like this ..

a. When two consequtive even number comes - consider it even.

b. When we get two odd - consider it odd.

c. When we get two even - taken the second even.

d. When we get one even and one odd - take this even ( this is main).

With the above strategy - there are 27 events with even numbers and 9 events with odd numbers..

So this turns out - 3/4 (Probability for even) and

1/4 (Probability for odd)..