Sharing Chocolate Bars

Step one is to find out how many sections of chocolate are on the edges of the chocolate bars. This can be done by dividing the total number of sections by the number of pieces along the top, so ${ L }_{ Milk }=\frac { 32 }{ 4 } = 8$ and ${ L }_{ Dark }=\frac { 35 }{ 5 } = 7$ .

Next, we need to figure out how many sections there are in total. That would be $35 + 32$ , which equals $67$ . Now, we need the number of corner pieces on the milk chocolate bar, and we also need the number of non-corner edge pieces on the dark chocolate bar. We can assumes that ${ Milk }_{ corners }$ equals $4$ , so we just need to figure out $Dark_{ edges }$ . Not including the corners, the top edges are 3 pieces across, and the sides are 5 across. $5 + 5 + 3 + 3 = 16$ , and that's what we were looking for.

Now, we need to find the probabilities of the events. ${ P }_{ Milk\quad Corner }=\frac { 4\quad corners }{ 67\quad sections }$ , and ${ P }_{ Dark\quad Edge }=\frac { 16\quad edge\quad pieces }{ 66\quad remaining\quad sections }$ , because Diamond already took a piece, so there are only 66 left to choose from. To get the probability that these events will consecutively happen, we need to multiply the probabilities. This gives us $\frac { 4 }{ 67 } *\frac { 16 }{ 66 } =\frac { 64 }{ 4422 }$ , or $\frac { 32 }{ 2211 }$ . That's a lot of trouble over chocolate, isn't it? :P