Or to put it simply, how would one select random point  (e.g. in a circle) 
using Polar Coordinates??

Roger

> On Sep 3, 2020, at 8:17 AM, Roger Guay via use-livecode 
> <use-livecode@lists.runrev.com> wrote:
> 
> Jerry,
> 
> You’ve done a very nice job of describing what’s actually(?) happening in my 
> code, but I think you missed the point of my question. 
> You agree that if you simply sample random pixels then the ratio of a random 
> pick inside the smaller circle will depend on the area of the circles.
> And, if I pick a random x and y within the concentric circles of radius R and 
> 2R, ¼ of the time they will lie in the smaller circle and ¾ of the time in 
> the bigger.
> So, pick any random x and y and convert to radial coordinates. Everything 
> should work! 
> In my code I pick a random angle and a random radius (radial coordinates) 
> within the limits of the larger circle, thus picking random points within the 
> area of the larger circle, yet I get ½ (which you say is the right answer). 
> My intent was to pick random points (using radial coordinates) for which the 
> result should be ¼!
> 
> What’s wrong with my code?
> 
> Thanks,
> 
> Roger
> 
>> On Sep 2, 2020, at 8:27 PM, Jerry Jensen via use-livecode 
>> <use-livecode@lists.runrev.com> wrote:
>> 
>> 1/2 is the right answer.
>> 
>> Take your drawing of the circles. Cut a verrrryy thin radial slice from the 
>> center to the outside circle. So thin that it is just a line. 
>> 
>> Now think of how likely a random point on that line will be in the part of 
>> the line that was in the smaller circle. The part that was from the smaller 
>> circle is HALF as long as the entire line.
>> 
>> Now add up all the possible positions of that line. Why would that change 
>> the answer?
>> 
>> Congratulations, you understand integrals!
>> .Jerry
>> 
>>> On Sep 2, 2020, at 7:38 PM, Roger Guay via use-livecode 
>>> <use-livecode@lists.runrev.com> wrote:
>>> 
>>> Your chance to be Genius du Jour:
>>> 
>>> If I construct 2 concentric circles, one being half the radius of the 
>>> larger, then simple math shows that the smaller circle has an area ¼ the 
>>> area of the larger.
>>> Now if I generate a random point within the radius of the larger circle, I 
>>> should expect that the probability of it landing in the smaller circle to 
>>> be ¼.
>>> But, I must be doing something wrong because I get ½ !
>>> 
>>> Here is my script:
>>> 
>>> on mouseDown
>>> 
>>>     getStuff
>>> 
>>> end mouseDown
>>> 
>>> 
>>> local tR, tTheta, tX0, tY0, tX1, tY1, tTotCount, tL, tLongCount
>>> 
>>> on getStuff
>>> 
>>>     put item 1 of the loc of grc OuterCircle into tx0
>>> 
>>>     put item 2 of the loc of grc OuterCircle into tY0
>>> 
>>>     put "" into tTotCount
>>> 
>>>     put "" into tLongCount
>>> 
>>>     emptyFlds
>>> 
>>> end getStuff
>>> 
>>> 
>>> on mouseUp
>>> 
>>>     lock screen
>>> 
>>>     repeat 1000
>>> 
>>>             put random(200) into tR -- 200 is half the width of the larger 
>>> circle
>>> 
>>>             if tR > 1 then
>>> 
>>>                     ## put random(2*pi) into tTheta1
>>> 
>>>                     get random(360)
>>> 
>>>                     put it*pi/180 into tTheta1
>>> 
>>>                     put tR*cos(tTheta1) into tX1
>>>                     put tR*sin(tTheta1) into tY1
>>> 
>>>                     set the loc of grc Ptgrc to tX0 + tX1, tY0 - tY1 --- 
>>> grc Ptgrc is a 2 pixle oval
>>> 
>>>                     if intersect(grc Ptgrc, grc InnerCircle, "opaque 
>>> Pixels") then add 1 to tLongCount
>>> 
>>>                     add 1 to tTotCount
>>> 
>>>             end if
>>> 
>>>     end repeat
>>>     put tTotCount into fld "totcountFld"
>>> 
>>>     put tLongCount into fld “LongCountFld"
>>> 
>>>     put tLongCount/tTotCount into fld "RatioFld"
>>> 
>>>     unlock screen
>>> 
>>> end mouseUp
>>> 
>>> 
>>> Apparently, this does not generate a random point within the larger circle! 
>>> Can someone please tell me what’s wrong here?
>>> 
>>> Thanks,
>>> Roger
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>> 
>> 
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