Probability of Halloween Candies Got some candy last night during Halloween, and upon opening some today, I got curious.
Starburst candies come in a pack of 2. There are 4 flavors: Cherry, Orange, Lemon, Strawberry. Both flavors per pack are random (e.g. Cherry and Orange can be in one pack, and Cherry/Cherry can be another)
I opened one pack and it had 2 strawberry-flavored candies.
I opened the second pack and it also had 2 strawberry-flavored candies.
I opened the third, and guess what... 2 strawberry-flavored candies.
So, my question is if there is a difference in probability between having 3 packs of 2 same-flavored candies, vs drawing 6 in-a-row of the same flavor from a big pile.
i.e. Is packing of candies into a 2-pack, and then selecting 3 of those packs having all the same flavor more rare vs getting 6 of the same flavor from single packs?
I know I'm talking about candies, but I've had the same thoughts around die rolls.
THANK YOU!!!
EDIT FOR DICE EXAMPLE:
What are the odds of rolling "snake eyes" 3 times in a row, vs rolling a 1 six times in a row (using 1 die)?
 A: @saulspatz is correct about the count ($10$ possible packets) but we do not actually know if each of the $10$ is equally likely.  


*

*If each of $10$ packets is equally likely, i.e. prob $1/10$, then prob of $3$ double-strawberry packets in a row is $1/10^3$.  

*Prob of $6$ single strawberries is of course $1/4^6$.

*However, my guess is that the packets are made like this: there is a machine with $2$ slots, say Left and Right, each slot is filled independently, then the machine makes them into a packet.  In this case, prob of a double-strawberry pack is $1/4^2$, while (e.g.) prob of a lemon-strawberry packet is $2/4^2$ since the slots could have been filled as (lemon, strawberry) or (strawberry, lemon) and they would result in the same packet (as far as you can tell).

*If my machine model is correct, then prob of $3$ double-strawberry packets is $1/4^6$.
I don't know if the $1/10$ model or my machine model is correct, but my guess is my machine model.  If you have many packets, maybe you can test this?  :)  Just count the number of each of the $10$ combos.

The case for dice is easier: it is equivalent to my machine model.  So prob of $3$ snake-eyes = prob of $6$ ones = $1/6^6$.  The two dice are two separate entites, and rolling $11$, i.e. $(5,6)$ or $(6,5)$, is twice as likely as rolling $12$, i.e. $(6,6)$, even though (with good dice) you cannot tell the difference between $(5,6)$ and $(6,5)$.  
If you prefer, think of it this way: suppose the dice are different colored, red and green.  Then surely you agree that rolling $11$ (Red $5$ + Green $6$, or, Red $6$ + Green $5$) is twice as likely as rolling $12$.  Now suppose your friend is colorblind and looking at the same dice.
