Lets suppose that the month has 4 weeks and we will have a week to see the results, then number the 13 prisoners (1:13) and divide the wines into 6 groups (4 groups of 167 and 2 groups of 166 wines) and number them in each group (1:167/166).

(Note than 167/13 = 12.84 ~= 13 =~ 166/13)

Define strategy1(K): Give to each N prisoner the bottles in the range [(N-1)*13+1 : N*13] of the group K of wines

Define strategy2(K): Give to each N prisoner the bottles that satisfies N = BottleNumber%13 of the group K of wines.

On the day #1 we use the strategy1(1)

On the day #2 we use the strategy1(2) and the strategy2(1)

On the day #3 we use the strategy1(3) and the strategy2(2)

On the day #4 we use the strategy1(4) and the strategy2(3)

On the day #5 we use the strategy1(5) and the strategy2(4)

On the day #6 we use the strategy1(6) and the strategy2(5)

On the day #7 we use the strategy2(6)

So each wine is tested for only 2 prisoners, they will die in consecutive days during the last week that we have free, and its easy to get the bottle poisoned because in the strategies1/2(K) the groups of bottles share only one bottle

I know that we are using 3 prisoners more, but in this strategy you are not risking 9 of them, we know that exactly 2 of them will die

I know it is not the right solution, but I feel that if we change the problem a little, this one could be a better solution