Question 365744
<pre>
Here are all the ways two dice can fall

(1,1)  (1,2)   (1,3)  (1,4)   (1,5)  (1,6)
 
(2,1)  (2,2)   (2,3)  (2,4)   (2,5)  (2,6)
 
(3,1)  (3,2)   (3,3)  (3,4)   (3,5)  (3,6)
 
(4,1)  (4,2)   (4,3)  (4,4)   (4,5)  (4,6)
 
(5,1)  (5,2)   (5,3)  (5,4)   (5,5)  (5,6)
 
(6,1)  (6,2)   (6,3)  (6,4)   (6,5)  (6,6)


I will color the ones red which have sum 6.

(1,1)  (1,2)   (1,3)  (1,4)   <font color = "red">(1,5)</font>  (1,6)
 
(2,1)  (2,2)   (2,3)  <font color = "red">(2,4)</font>   (2,5)  (2,6)
 
(3,1)  (3,2)   <font color = "red">(3,3)</font>  (3,4)   (3,5)  (3,6)
 
(4,1)  <font color = "red">(4,2)</font>   (4,3)  (4,4)   (4,5)  (4,6)
 
<font color = "red">(5,1)</font>  (5,2)   (5,3)  (5,4)   (5,5)  (5,6)
 
(6,1)  (6,2)   (6,3)  (6,4)   (6,5)  (6,6)

So the probability of having sum 6 is 5 ways out of 36 or {{{5/36}}}

So you will roll a sum of six {{{5/36}}}ths of the time, so

{{{5/36}}} of 200 times is {{{27&7/9}}} or about 28 times.

Edwin</pre>