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)?
@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.