Operation Gumball Tutorial – Advanced Strategy.
On this site, I have two strategies. The first one is fairly simple, but will only take you so far. I could routinely get to level 17 or 18, although I sometimes had problems on level 9 and 10. Many times, I was able to get past level 20, and get to level 25 or 26.
So, I knew I needed a more complex strategy that works for me.
The basic tenet is to never introduce more than 2 unknown numbers in your guess  and once you know that one of the numbers is in the answer, you only want to introduce one new number per guess. When you introduce 2 unknown numbers, you get the indication that one is part of the answer, and then you can test that on the next move.
So, I start with a sequence of 121. The reason that I do 121 instead of 112 is just ergonomic. It is easier to repeat the numbers like that, and when I get to later levels, it just takes a little less thinking. I used to do 112 on the first levels and 1122  11122  111222 (which is not wrong), but I found that doing the 121, 1212, 12121, 121212 was just easier on the fingers and the brain.
This first part is very tedius, but it tries to explain the basics of the strategy in words. If you are more visually oriented, skip dowm to look at the picture tutorial. If you are able to process text, then read through this first part and understand the foundation of the strategy.
Here are the rules. First, it might be helpful to write the numbers 1 2 3 4 5 6 7 8 9 0 on a sheet of paper and cross them out as you go (at least until the processes become automatic).
At the beginning, you obviously don't know the presence of any of the numbers in your answer and you do not know their position. Throughout this tutorial, I will use the terms presence and position. Here are their definitions:
Presence: Is the number in the answer
Position: Where in the answer is the number
On each turn (except the first and possibly second), I will test one number for presence (is it in the answer) and one number for position.
Throughout the process, I refer to the number list (although it is mentally referred at this point), and if I don't have any numbers that I don't know the presence, I go to the next 2 numbers on the list. Since at the beginning, I don't know anything, I start out with 121, 1212, 12121 or 121212 (depending on the level).
If you can mentally wrap your mind around this next concept, you will have no problem. It doesn't matter if you are on level 1 or level 10, you are normally only really examining 3 spaces at a time. Hopefully, this will become clear as the tutorial goes along.
Level 13
Turn 1  start by alternating the top 2 numbers  121
Turn 2a  if you have nothing, go to the next set, repeating Turn 1
Turn 2b  if you have a red, take the number you used the least amount of times in the earlier turn (2 in this case), and put it first. Then fill in the other spaces with the next number on your list. (233)
Turn 2c  if you have a green, take the number you used the greatest amount of times in the earlier turn (1 in this case), and put it first. Then fill in the other spaces with the next number on your list (133)
Turn 2d  if you have a green and red, you know the number you used least is incorrect (or else you would have 2 greens). You also know the number you used the most is one of the greens. So, the answer is either going to be 1x2 or 2x1 (the x is the unknown number).
Turn 3  The best case is that you get a green and a red. That tells you that the number you are testing for position (the lower number) is not in the correct position, but has a valid presence in the answer. In any case, if you get any reds, you know that the number that you are testing for placement is incorrect in that place. So, you move the lower number to the next spot, and then fill in all of the blanks with the next number on your list. For example, if you get a red on 233, you fill in the next spot with 442, because you know the 2 cannot be in the second spot (because it was red in the first turn). If you get a green and a red, now you know the 2 is definitely in the last place (same reason as Turn 2d above). If you get nothing for 233, you know the 1 is the correct number in the second place (because the 2 did not pass the presence test in this turn, and the 1 cannot be in position 1 or 3).
Turn 4  At this point, most likely, you know where the 1 or the 2 goes. If you get a return on turn 2, statistically speaking, it is likely that you are testing the correct number (either the 1 or the 2). There are some cases where you could be wrong, though For instance, if your final answer is 315, when you test 233, you will get a red (which above, I suggested means the answer is probably xx2). So, when you test the 442, you will get nothing  which tells you that the 3 and 1 are now correct (meaning that the answer is 3x1). Even more peculiar is if the answer were 341. When you do 442, you get a green, which seems to confirm that the 2 is correct (when it really isn't). To combat this, the next step is to take the next number in line  552. This will either yield 1 green (meaning the 2 is indeed correct), or 2 greens (meaning the 5 is correct in the presence and the 2 is indeed correct), or you could get nothing  which means that your 2 is incorrect, and the results from the earlier tests that you assumed were for the 2 were actually for the other numbers (which is why we only test 2 numbers at a time  this eliminates the 2 from those answers and confirms the other numbers belong in the answer).
So, if you get to this point, and you only know where one number goes, then you just have to keep trying the next numbers on the list  462, 472, 482, 492, 402.
OK, let’s look at some pictures (since most of you probably didn’t read through that very well).
Essentially, you ought to examine these examples one by one, trying to understand why the combinations are played in each one. You might also want to have this window open (or printed out) while you play, so you can see if you are getting returns that are similar to what you see here (in terms of the number of reds and greens). In some cases, I have used a grey highlight to block out the columns that we already know the answers – which really shows how simple it can be, reducing the numbers you are testing.
OK, here is a typical sequence:
After the first 121 and 131 produced 1 green, I tested the 1 again with 145. Since this gave a green and a red, I was pretty sure the 1 was correct in position and presence, so I tested the 5 in the middle position with 156. If the 5 was correct, I would have gotten 2 greens, but since I got 1 green and 1 red, I knew the 4 was correct for presence (but not position) and the 6 was correct for presence, but not position – leaving 164.
This one shows a similar case – but you can see how when the 5 was correct, I got 2 greens in the 3^{rd} move.
This shows where I tested for the 1 in the second move, and upon getting a red, tested the one in the only other place that it could have been. When 561 turns up nothing, I now know the 2 is in the right place, and either the 3 or 4 are going to be in the answer. When the 3 turns up nothing, I knew the 4 was in the answer, and had to be in the first place. After that, the last number could have been 8, 9 or 0.
OK. On to 4 numbers. I had directions for this, but it just got too complicated, so let me use the pictures to show some of the most typical ways.
This one shows how I start with 1212 and when the 2^{nd} move reveals nothing, I know that the 2 has to be in either the 2^{nd} or 4^{th} position. Testing 4244 reveals that it is in the second position. Now, if you look closely at the rest of the game, I am only playing 3 spots – so it is the same rules as levels 13. In fact, when I black out the column with the 2, you can see how the 3 spot game plays out.
Here is a very typical example.
When 3434 reveals a green and 3555 reveals nothing, I am left with the 4 in the 2^{nd} or 4^{th} spot. The 6466 tells me that the 4 is in the 4^{th} spot, and the 6 occupies the 1^{st} or 3^{rd} spots. 6774 revealing 2 greens and 1 red tell me that the 7 is correct, but the 6 is to be in the 3^{rd} spot.
Again, if you look at how I have shaded it, once you get the one number right, you are playing a 3 spot game. 667 reveals a green and red – so you treat it with the same rules you would treat 112 (or 121) in levels 13.
OK, what if you get a red and a green in the first try?
This can be more complex, because you don’t know which one is green and which is red. I normally repeat one of them and then add the next number in sequence. So, here you have 3434 as red green, so I repeat the 3 and add a 5. When that reveals 2 reds, I know that in the 3434, the 3 was not the green (and the 4 was), so the final answer is going to be x3x4 or x4x3. In this case, I also know that the 5 is in the 1^{st} or 3^{rd} spot, so I try 6354. This will either get me 3 greens (which means the 3,4 and 5 are all right), 2 greens and a red, (which means the 3 and 4 are correct, but the 5 is not), 3 reds, (which obviously means the answer would be 54x3) or 1 green and 2 reds (which means the 5 is correct, but the 3 and 4 are wrong). That was the case here. After that, I could just go through the remainder of the sequence to get the first number.
Now, to 5 spots – level 79.
After 12121 revealed 1 green, I tested the 1 in the first spot, and filled in the remainder with 3 (an unknown number). If I get 1 green, then the 1 might be correct – but it is more statistically probably that the 3 is correct in one of those 4 spots. If I get 1 red, it is probable that the 2 is in the 3^{rd} or 5^{th} spots (but if that proves to not be the case, I would know that the 3 is correct in the first position and the 2 would be correct in positions 2 or 4). In this case, I got 2 greens, which tells me that the 1 is correct and the 3 belongs in one of the remainder spots. Now, I start testing for the 3 and ignore the 1. When 13444 gives a red, I know that 4 is not part of the answer and the 3 is not in position 2. When 15355 gives a green and a red, I still know the 3 is not correct, but the 5 is going to be in position 2, 4 or 5. When 16636 gives 2 reds (the best possible outcome), I now know the 3 has to be in the 5^{th} spot (the only one I haven’t checked) and the 6 can only be in the 4^{th} spot (otherwise it would have been green on that run). Also, I now know the 5 has to be in the 2^{nd} position (because it was green in the earlier turn, it cannot be in the 3^{rd} position). After that, I just run the sequence of numbers left.
Ok, this is the case I talked about earlier. We start with the same 12121, and when 13333 reveals a red, it is statistically more probable that the 1 goes in position 3 or 5 than the 3 going in the first position, but when 44144 reveals nothing, I now know the 3 is correct in the first place and the 2 is either in position 2 or 4. So, I check the 2 in the 2^{nd} position, and when it is not there, I know it is in position 4, and once again, I am only playing a 3 spot game – same as level 13, so when 676 reveals that one of those is green (as shown in the move 36726 which has 3 greens, 2 of which I already have in the bank), I try 688  the same way I would try 121 and then 133 in level 13).
Here is one that shows how getting 2 reds is the best possible scenario, giving you precisely the location of the new number you are testing, and it definitively rules out the position of the original number that you are testing for position.
Getting through the first 2 sequences without an answer is great, because it means that the 5 positions are going to be filled with 5 of the remaining 6 numbers (and we have a lot of spots to test). When 67777 shows 2 greens, I know the 6 is good. Then, 67888 shows 2 reds, which means the 8 has to be the 2^{nd} number and the 7 is either position 3,4 or 5. Then 67899 reveals 2 reds, again, now the 9 is in the 3^{rd} position and the 7 is either 4 or 5. When the 68970 reveals 2 reds, I just had to swap them to get the correct answer.
OK, now the hard one, level 10 – 6 spots.
Starting with 121212 and getting 2 greens, means that the 1 is either in position 1,3 or 5 AND the 2 is in either position 2,4 or 6. I will test those separately.
I got a lucky break with 133333 reveals 2 reds – in fact, using this strategy, 2 reds is the best possible case (as demonstrated above).
The next move reveals that the 1 is in the 3^{rd} spot (even if I would have gotten a red, I would have know where the 1 was, because it could only be in the 3^{rd} or 5^{th} spot).
Then, I start testing the 2. When I get a green and a red with the ones I am testing on the 4^{th} move (the 321555), I know the 5 is in the answer, and the 2 is either in position 4 or 6. The red in the next try confirms that the 2 is the last number, meaning that the 5 can be position 4 or 5. The next red tells me that the 5 is in position 5. Again, by this point, I am back to a 3 spot game, just like level 13.
Now, for a little more complex.
When I get the 1 green, I try to test the 1 in the first spot. When I get a red, it is more probable that the 1 is supposed to be in spot 3 or 5 than having the 3 the first number. When 441444 reveals a green, it seems to confirm that, but then the 551555 (which should have at least 1 green if the 1 is indeed correct) reveals a red, I now know that the 21 is not part of my answer. This means that the first number is a 3, and it means that the 3^{rd} number is a 5 (because both of the lines that introduced them revealed that they were correct in presence, but not position). After that, I try the 2 for position 2, 4 or 6. When it is confirmed in position 2, I go back and examine the 4 (knowing that it should be in position 4,5 or 6. The 2 reds derived from 325477 is great (again, 2 reds are always good), because I know the 7 is now position number 4 and the 4 is either 5 or 6. The next guess was proven correct.
Hopefully the shading helps you realize what you have to focus on (you have to mentally get rid of the columns that are correct, which shrinks the game down to less slots (and less numbers to fill them).
So, is it different for levels 10 – 20 or 20 – 30?
Not much – although there are some nuances. This strategy is really designed for those levels, because when you are testing a number that you know has presence, you add a new number to fill all of the other spots (which will give you a maximum return on the multiples (should they exist).
Here are some examples.
When 1212 reveals a green and 1333 reveals a red, I know that the 1 is only in the answer 1 time (or else I would have gotten more than one result on the first test) and if the number 3 is in the answer, it is only in one time. When I do 4514, I expected to get a green for the 1 ball (which I did), but I also got an extra green and red. Now, it is really a 3 spot game – as if I just played 454 and got a green and red. So, with 454 giving a green and a red, the answer was either going to be 4x5, 5x4 or 44x, x44 (the only combinations that would yield a red and green). So, when I tried 4615 and got 2 greens and a red, I immediately knew that the 4 had to be a duplicate, and that the 6 was part of the answer.
Here, when 34343 revealed a green and red, I tested the 3 first. When it was red in the first spot, I tried it in the second spot. When nothing came up, I immediately knew 3 things. First, the 3 isn’t part of the answer. 2^{nd}, the 5 goes in the first position.. 3^{rd}, the 4 was in the answer twice. Note: there are times where I get tripped up if the 4 were to be in the answer 3 times – just know that is possible. So, When 54777 turned up 2 reds (which is good news for me), I knew the 7 was supposed to go in the 2^{nd} position and I knew the 4 was in the 4^{th} position. After finding out the other 4 was not in the 3^{rd} spot, it was easy to just run the rest of the numbers.
With 12121 revealing a green and red, I repeat the 1’s, with some 3’s. When that gives me a green, the probability is that the 1 is the green and the 2 was a red. This means that the final answer can either be 1x2xx, 1xxx2, xx1x2, 2x1xx, xx2x1 or 2xxx1. When the 1 comes up green in the next try, I then tried the 2 in the 3^{rd} place, which left me with a 3 spot game (same as levels 13 (or 1012).
Here is one more example.
After getting a green and a red, then repeating the 1’s in their position and adding 3s, then getting a red, I was pretty sure the 1 was a red and the 2 was a green. So I tested the 2 (I could have easily tested the red instead, it was just a preference). When 4244444 revealed a green, again, I knew the 2 was correct, and the 1 could only be in positions 4 or 6 (since it was red before, and it could not occupy the same position as the correct 2). When 525155 revealed 4 balls, I knew the 5 was in the answer twice, and I knew one of those ties was in the 4^{th} position (again, getting 2 reds is always good). I also knew the 1 had to be in position 6. After that, you can see by the shading that it was only a 2 position game. By chance, it was the numbers next in my sequence (the 7 and 8), but I still had more than enough spots to figure out what the last 2 numbers were.
I hope the highlighting helps you focus on what you are asking and makes the strategy more clear. Good luck, and happy trophying.
So, here is your goal  get to level 30 and get a trophy.
