6. Given that there are 15 chocolates arranged in a row on the table, some of which are milk chocolates and the rest are dark chocolates. Players A and B play the following game: The player eats a certain number of consecutive chocolates starting from the left, with the requirement that the number of chocolates of the same type as the first chocolate eaten must be odd. For example, if the remaining chocolates on the table are $\operatorname{MMDMD}(\mathrm{M}$ representing milk chocolate and D representing dark chocolate), the player can eat 1, 4, or 5 consecutive chocolates from the left. The player who eats the last chocolate wins. Question: If A goes first, out of all $2^{15}$ different arrangements of chocolates, how many situations will result in B winning?
6. Since 15 is an odd number, let there be $n=2k+1$ chocolates on the table.

Notice that if the number of chocolates with the same flavor as the first one is odd, then player A can win by eating all the chocolates in one move; if the number of chocolates with the same flavor as the first one is even, then there are $2^{2k}$ different chocolate arrangements.

Let $a_k$ be the number of arrangements where player A wins, and $b_k$ be the number of arrangements where player B wins.
It is easy to see that $b_0=0$, $b_1=4$.
When $k \geqslant 2$, $n=2k+1 \geqslant 5$.
First, consider the $2^{n-2}$ cases where the last two chocolates have the same flavor, i.e., MM or DD. In these cases, if a player can eat the first $n-2$ chocolates, they can also eat the last two, meaning there are $2b_{k-1}$ cases where player B wins.

Next, consider the cases where the last two chocolates have different flavors, for example, MD.
(1) If the third-to-last chocolate has a different flavor from the last one, i.e., MD or DM, then player A can eat the first $n-3$ chocolates and leave the last three to win.
(2) If the third-to-last chocolate has the same flavor as the last one, for example, when the third-to-last and the last are MM, player A cannot eat all the chocolates, and there will be a chance for player B to eat:
(i) If the remaining chocolates start with M, then player B can eat all the remaining chocolates to win.
(ii) If the remaining chocolates start with D, then when the remaining D is odd, player B can eat all the remaining chocolates in one move to win; when the remaining D is even, player B can leave the last three chocolates to player A to win.

The case where the third-to-last and the last are DD is similar.

In summary, if the third-to-last and the last chocolates have the same flavor and are different from the second-to-last, then player B wins. Since we only consider the cases where the number of chocolates with the same flavor as the first one is even, there are $2^{n-3}=2^{2k-2}$ different chocolate arrangements.
Thus, the recurrence relation is $b_k=2b_{k-1}+2^{2k-2}$.
Therefore, $b_7=8320$.
Note: From the above recurrence relation, the general term formula is $b_k=2^k+2^{2k-1}$.