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Bonus Level Donut Game Math

4 posts New member
Hello farmers!

I decided to delve into the math regarding whether or not to choose a second or third box from the bonus level up. Based on my own observations, it does appear to be a somewhat fairly random distribution. That is, if you always choose box 1 you'll get a roughly equal number of 1's, 2's, and 3's. There may be streaks where the same number comes up repeatedly for 3 or 4 levels, but over a long enough timeline the scores will average out.

So the question then is, what's the cost, long term, to open those extra boxes? Setting aside the best way to generate EXP, once you get to that million mark what should you do?

Let's say you suddenly got 300,000,000 exp and were able to level up 300 times.

Scenario 1: Always take the first box
Avg donuts/level: 2
Overall you would get 100x3 + 100x2 + 100x1 = 600 donuts total, or an average of 2 donuts per level, with zero money spent.

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Scenario 2: Never take 1 donut, never open all 3 boxes
Avg donuts: 2.5
Cost/extra donut: \$33,333
Cost to avg ratio: 13,333
In this scenario you will always take 2 or 3, but if you get 1 donut you always choose 1 more box. In this scenario you'd be getting 100x3 + 100x2 + (50x1>3) + (50x1>2) = 750 donuts, with a cost of \$5,000,000 for opening extra boxes. That works out to an average of 2.5 donuts per level with an average cost of \$33,333 for each extra donut beyond the 600 you'd get not opening anything in scenario 1.

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Scenario 3: Always get 3
Avg donuts: 3
Cost/extra donut: \$50,000
Cost to avg ratio: 16,667
In this scenario you open boxes until you get 3 donuts no matter what. The payout here is 100x3 + (50x2>3) + (50x2>1>3) + (50x1>3) + (50x1>2>3) = 900 dounts, with a cost of \$15,000,000 to cover 300 box opens. You're getting an average of 3 donuts per level at a cost of \$50,000 per donut vs scenario 1.

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Scenario 4: Always open 1 box
Avg donuts: 2.67
Cost/extra donut: \$50,000
Cost to avg ratio: 18,750
In this scenario you always open a second box if you get 1 or 2 on the first try. Half of these will get you 3, half will get you 2 (since 2>1 yields the original 2). 100x3 + (50x2>3) + (50x2>2) + (50x1>3) + (50x1>2) nets you 800 donuts total, at a cost of \$10,000,000. Here you get an average of 2.67 donuts per level, and the per donut cost is \$50,000.

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Scenario 5: Always turn 1 into 3
Avg donuts: 2.67
Cost/extra donut: \$37,500
Cost to avg ratio: 14,063
Similar to scenario 2, but any run that starts with 1 you end with 3
100x3 + 100x2 + (50x1>2>3) + (50x1>3) = 800 donuts, \$7,500,000

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Scenario 6: Always open 1 more on 2
Avg donuts: 2.17
Cost/extra donut: \$100,000
Cost to avg ratio: 46,154
In this scenario you always accept 1 or 3, but always open exactly 1 more box on 2:
100x3 + (50x2>2) + (50x2>3) + 100x1 = 650 donuts, \$5,000,000 cost.

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Scenario 7: Always turn 2 into 3
Avg donuts: 2.33
Cost/extra donut: \$75,000
Cost to avg ratio: 32,143
Similar to scenario 6, except you always turn 2 donuts into 3
100x3 + (50x2>1>3) + (50x2>3) + 100x1 = 700 donuts, \$7,500,000

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To sum up, the lowest cost/donut is, of course, to always take what you're given on the first try. Beyond that the question is whether you want to maximize donuts or keep your cost down. Scenarios 6 and 7 are presented only as examples of what not to do, as they maximize cost with the smallest increase in overall donut payout. Opening again on a 2 donut box has a 50% chance of increasing your cost with zero increase in donuts, so unless you're always driving for 3 donuts, the best play is to take your 2's as they are.

Which way you go depends on your play style. If you have tons of cash and nothing to spend it on, always going for 3 donuts is your best bet. If you only want to spend a little money but get a nice bump in total donuts, scenario 2 is the best ratio of cost to extra donuts.

Replies

• 4314 posts Member
Or you could use the KEM simulator which was posted by @alapalme a few years ago:
http://lwshost.com/tsto/simkem.pl
A link to this can be found in the ‘tips for players and other links’ thread in game discussion.