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# Math.random()

Thomas Markl
Ranch Hand
Posts: 192
Does this return a random value between 1 and 6 or between 1 and 7??
private int throwDice(){
return 1+(int)(Math.random()*6);
//Math.random returns a double value between 0.0 and 6.0 and cast it
//to int and adds one. So the return value of throwDice() is an int
//number between 1 and 7.
}

Dirk Schreckmann
Sheriff
Posts: 7023
1+(int)(Math.random()*6)
Math.random() returns a number in the range [0,1). Multiply this by six and get [0,6). Cast it to int and get [0,5] which is the same as [0,6) when considering only whole numbers. Add one to it and get [1,6]. So, it returns a random number from 1 inclusive to 6 inclusive, which is not unlike the range from 1 inclusive to 7 exclusive when considering only whole numbers. The documentation lies. Who's ever seen that before?

John Lee
Ranch Hand
Posts: 2545
Originally posted by Dirk Schreckmann:
Cast it to int and get [0,5] which is the same as [0,6) when considering only whole numbers.

Is this true?
for an int, [0,5] and [0,6) are the same in term of range, but the frequency of each number's appearence is not the same. I believe for a random int in [0, 6), you will more often get 5 than a random int in [0,5].

Ellen Zhao
Ranch Hand
Posts: 581
Is this true?

This is True. You can find the rigorous mathematical proof in some discrete mathematic theory or probability theory paper.
Random values can have certain well-known distributions. When random values are uniformly distributed between a minimum and a maximum value( for example, between 0 and 1 ), every value has an equal probability of occuring. Random values that are normally distributed have probabilities that are represented by the familiar bell curve.
A well known algorithm for generating uniformly distributed random values is the linear congruential algorithm. This algorithm generates random integer values that range over all the possible positive and negative values of the integer type. It uses the formula
X(n+1) = ( m * X(n) + a ) mod d
Where the Xi are the generated random values, m is a constant multiplier, a is a constant addend and d is a constant divisor, which is usually a very big number and relatively prime to m( I think Don must know "da4 shu4 xiang1 chu2 ding4 li3" ). A seed value kicks off the sequence. The formula relies on the fact that integer arithmetic does not overflow but wraps around.
Regards,
Ellen

Ranch Hand
Posts: 284
why does the cast to int change the range? because of a rounding down?

John Lee
Ranch Hand
Posts: 2545
Originally posted by Ellen Fu:
This is True.

Hi Ellen:
It seems to me that we don't even need bell curve here. Because a random number is generated uniformly in [0, 1) (by definition). So when we multiply the random number by 5 and 6, we get [0, 5) and [0, 6) respectively.
For a random number in [0, 5), assume we round the random number to closest int, same below. We have 20% chance to get 1, 2, 3, and 4. 10% to get 0 and 5.
For a random number in [0, 6), we have 16.67% chance to get 1, 2, 3, 4, and 5. 8.34% to get 0 and 6.
I think if in other rounding scheme, the result distribution should not be the same for [0, 5) and [0, 6) either.

Ilja Preuss
author
Sheriff
Posts: 14112
Originally posted by Don Liu:
I believe for a random int in [0, 6), you will more often get 5 than a random int in [0,5].

Why? All values in [0,1) would get rounded down to 0, all values in [1,2) to 1 ... and all values in [5,6) to 5. So it seems to me that the probability for all of them would be equal.

John Lee
Ranch Hand
Posts: 2545
In this scheme, you can't even get an int of 5 for a random number in [0, 5), but you can for an int in [0, 6). So there is difference.

Dirk Schreckmann
Sheriff
Posts: 7023
Don, it would seem to me that you are arguing that by multiplying a uniform distribution by some constant (so to speak), the distribution suddenly becomes unevenly distributed. Why do you think so?

Dirk Schreckmann
Sheriff
Posts: 7023
why does the cast to int change the range? because of a rounding down?
I don't think that's an unreasonable way to understand this. Note that the cast to int doesn't change the range of numbers produced before the decimal point, it just chops off whatever might have been to the right.

John Lee
Ranch Hand
Posts: 2545
Originally posted by Dirk Schreckmann:
Don, it would seem to me that you are arguing that by multiplying a uniform distribution by some constant (so to speak), the distribution suddenly becomes unevenly distributed. Why do you think so?

Hi Dirk:
Thanks for your response. I didn't say multiplying a uniformly distribution will get an unevenly distribution.
If we go back to the original question:

Does this return a random value between 1 and 6 or between 1 and 7??
private int throwDice(){
return 1+(int)(Math.random()*6);

I think the answer is [1, 6]. I think you also agreed on this.

Jim Yingst
Wanderer
Sheriff
Posts: 18671
The answer is indeed [1, 6]. However you've also indicated you think the endpoints in this range (1 and 6) are half as likely as the other mid-range points (2, 3, 4, 5). This is incorrect - do you still believe this? Or did you follow Ilja's point about int arithmetic rounding down (or more accurately, towards zero) rather than towards the nearest integer? If you still think 1 and 6 are less likely, I recommend running the method a few hunderd times and counting the results.

Ilja Preuss
author
Sheriff
Posts: 14112
Originally posted by Don Liu:
For a random number in [0, 5), assume we round the random number to closest int, same below. We have 20% chance to get 1, 2, 3, and 4. 10% to get 0 and 5.

Casting doesn't round to the nearest int - it always rounds down (at least as long as only positive numbers are involved). So we get a 20% change for 0 and a 0% chance for 5 in the original question.

John Lee
Ranch Hand
Posts: 2545
Thanks, guys. Now I am clear about casting to int.