 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Falling through the floor (Posted on 2007-05-18) Given uniformly randomly chosen x on the interval (1,5) and y on the interval (1,5) find the probability of each:

[x] + [y] = [x+y]

[x] - [y] = [x-y]

[x] * [y] = [x*y]

[x] / [y] = [x/y]

Where [x] is the floor function, the greatest integer less than or equal to x.

 No Solution Yet Submitted by Jer No Rating Comments: ( Back to comment list | You must be logged in to post comments.) solutions for first three | Comment 1 of 4

[x] + [y] = [x+y]

If the fractional parts (portions after the decimal) add up to less than 1, the two will be equal, otherwise unequal.  The probability is zero that they'd add to exactly 1, so that can be ignored. If the fractional parts are plotted on coordinate axes, the resulting unit square is divided by a diagonal line from (0,1) to (1,0) separating these two equal areas. The probability is 1/2.

[x] - [y] = [x-y]

This is the same as the addition case except that for the y, it's 1 minus the fractional part.  The probability is still 1/2.

[x] * [y] = [x*y]

When [x] = 1, and [y] = 1 this tests if xy<=2. Graphed, this is the portion of the unit square from (1,1) to (2,2) that is under the hyperbola y=2/x. That is 2*ln(2) - 1. We subtract 1 because this unit square has another unit square below it, i.e., its base is not the x-axis.

Similar calculations need to be done for each of the 16 unit squares with lower left corners from (1,1) to (4,4) (not necessarily equal x and y). In the general case the integration does not proceed all the way to the next integer of x, but only to where the hyperbola cuts the lower y value, b, which is at x = (a*b+1)/b = a + 1/b. For any pair (a,b) as the lower left corner of the square we need

Integ{a,a+1/b} (a*b+1)/x  dx - b/b = ((a*b+1)*ln x){eval a to a+1/b} - 1

= (a*b + 1)*(ln(a+1/b) - ln(a)) - 1

So overall we need:

(Sigma{a=1 to 4} Sigma{b=1 to 4} ( (a*b + 1)*(ln(a+1/b) - ln(a)) - 1 )) / 16

as we need to average over the 16 squares.

`10   for A=1 to 420   for B=1 to 430      Term=(A*B+1)*(log(A+1/B)-log(A))-140      T=T+Term:print Term50   next60   next70   print:print T/16`
`a  b        1                    2                   3                    41  .3862943611198906187 .2163953243244931449 .150728289807123711  .11571775657104877712  .2163953243244931449 .1157177565710487771 .0790547587908081287 .06004732090745108943  .1507282898071237086 .0790547587908081279 .0536051565782630099 .04055519975597353244  .1157177565710487771 .0600473209074510894 .0405551997559735331 .0306185708793923204`

It's a good sign that when a and b are interchanged we get the same result.  The average of these values is 0.1194520715913994681.

Edited on May 18, 2007, 11:41 am

Edited on May 18, 2007, 11:41 am
 Posted by Charlie on 2007-05-18 11:26:33 Please log in:

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