Question 893740
Jeff, most people's days off aren't random, and that makes things harder.  If they follow a pattern (same 2 days each week, for example) then it's easy to make predictions.  Most people want two days in a row, and in that case if there's any overlap at all, it's much easier to get several days in common.  
<br>So I'm going to make a few assumptions, even though to assume makes an "ass" of "u" and "me".  
<br>If the days off have been chosen in a <i>truly random manner</i>, then we can assume you have your 10 days separately from your ex, and so the chance of her getting exactly 6 of your days randomly assigned will be 
<br>{{{C(6,10)*C(4,25)/C(10,35)}}}
<br>Where the C(x,y) is the notation for the number of combinations of x choices out of y possibilities.  We use these because the order in which the days are chosen isn't important, we just want to know about 6 of the days being the same.  the expression says that 6 days match your 10 days off, 4 days match your 25 working days, and the bottom is all the ways to get 10 days off in that 5 week period.  
<br>The result comes out to be {{{210*12650/183579396 = 0.01447}}}, or about 1.45%.  In other words, not too likely.  
<br><i>HOWEVER</i>, like I said before, days off are <b>NOT</b> usually completely random.  Which means this calculation is pretty worthless for your situation.  In your case I'm thinking they guarantee 2 days off every 7 days, whereas with truly random assignments, you will often get clumps of days off, with some weeks where you only get one day off or even none.  This makes the situation a lot harder to compute, because she may have gotten one common day for four weeks and two in one week, or two common days for three of the weeks (which three?), or... well, you see the problem!  The complexity goes through the roof.  
<br>Good luck in court.  Most people (judges included) don't really understand probability anyway, so maybe you can throw a bunch of numbers around and no one will question it!  
<br>Sometimes all you need is to <i>sound</i> smart... ;-)