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3-ingredient Bake vs. 1-ingredient Bake

Aethelcatt VIII

Aethelcatt VIII

Active Member
Help! I need a math expert to help me sort my brains out on this: Mathematically, a 3-ingredient bake in the fall event minigame that has a 15% chance at the daily special gives you a better shot than 3 single-ingredient bakes that each has a 10% chance because, well, three shots at 10% is still 10% in total; but why does it always "feel like" that 3 shots at 10% is a better bet than a single 15% shot?

And truth be told--so far I have received more daily specials from those single-ingredient cookies and so on than from the fancy 3-ingredient cakes. Is that purely a coincidence? Full disclosure: I do make a lot of those single-ingredient bakes, not because I like the odds better but simply because I'm always trying to balance out the different ingredients I have left.

This is too much math for us felines to figure out...
 
P C C

P C C

Active Member
Divide the % chance by the number of ingredients to see which gives the best average return in daily specials, if that is what you are looking for. In your example, 0.05 dailies per ingredient used vs 0.10. (Three recipes with 10% chance gives 24.3% chance of 1 daily, 2.7% chance of 2, and 0.1% chance of 3.)

But as you both say, keeping a balance of ingredients is key
 
Aethelcatt VIII

Aethelcatt VIII

Active Member
P C C said:
Divide the % chance by the number of ingredients to see which gives the best average return in daily specials, if that is what you are looking for. In your example, 0.05 dailies per ingredient used vs 0.10. (Three recipes with 10% chance gives 24.3% chance of 1 daily, 2.7% chance of 2, and 0.1% chance of 3.)

But as you both say, keeping a balance of ingredients is key
Click to expand...
Thanks for that. Percentage of chance per ingredient...now that does make sense. But what mathematical principle dictates that "three recipes with 10% chance gives 24.3% chance of 1 daily, 2.7% chance of 2, and 0.1% chance of 3"? I keep thinking that for three shots at 10% odds each, you still have a 90% chance of not getting it--each time and in total. Am I missing something here?

Math is so hard...
 
P C C

P C C

Active Member
To not get any reward, you'd have to miss three times in a row. The chance of getting a daily first time is 10%, not getting one is 90%. And of the 90% of the times when you miss on the first, you will get one on the second 10% of the time (so 9% more total). And if you miss both of the first 2 times (81%), then you will get one on the third 10% of the time (so 8.1% more). So the chance of getting at least one is 10% + 9% + 8.1% or 27.1%.

If you either calculate probabilities for each possible sequence or check a binomial probability distribution, you get the numbers in my earlier post (which sum to 27.1%)
 
Aethelcatt VIII

Aethelcatt VIII

Active Member
P C C said:
To not get any reward, you'd have to miss three times in a row. The chance of getting a daily first time is 10%, not getting one is 90%. And of the 90% of the times when you miss on the first, you will get one on the second 10% of the time (so 9% more total). And if you miss both of the first 2 times (81%), then you will get one on the third 10% of the time (so 8.1% more). So the chance of getting at least one is 10% + 9% + 8.1% or 27.1%.

If you either calculate probabilities for each possible sequence or check a binomial probability distribution, you get the numbers in my earlier post (which sum to 27.1%)
Click to expand...
I see...I think. :) So in the context of three 10% shots in a row, the possibility of winning is more dynamic than I understood. These numbers are beginning to make sense to me, but let me sit on them a little longer and try to wrap my little feline brain around them. Thanks!
 

Pericles the Lion

Well-Known Member
P C C said:
To not get any reward, you'd have to miss three times in a row. The chance of getting a daily first time is 10%, not getting one is 90%. And of the 90% of the times when you miss on the first, you will get one on the second 10% of the time (so 9% more total). And if you miss both of the first 2 times (81%), then you will get one on the third 10% of the time (so 8.1% more). So the chance of getting at least one is 10% + 9% + 8.1% or 27.1%.

If you either calculate probabilities for each possible sequence or check a binomial probability distribution, you get the numbers in my earlier post (which sum to 27.1%)
Click to expand...
I don't think that this is correct. The events are independent of each other. If you miss on the first 10% chance the odds of winning on the second chance remains 10%. The result of the first chance has no impact on the result for the second, or subsequent, selections. Three chances, all 10%, equals a 30% chance of winning. If a player chooses 10 bakes, all with 10% chance, then the probability of a win is 100%, not 60.4% as your model suggests.
 
Sharmon the Impaler

Sharmon the Impaler

Well-Known Member
Pericles the Lion said:
I don't think that this is correct. The events are independent of each other. If you miss on the first 10% chance the odds of winning on the second chance remains 10%. The result of the first chance has no impact on the result for the second, or subsequent, selections. Three chances, all 10%, equals a 30% chance of winning. If a player chooses 10 bakes, all with 10% chance, then the probability of a win is 100%, not 60.4% as your model suggests.
Click to expand...
It doesn’t equal a 30% chance with 3 tries at 10% each anymore than it offers 100% chance with 10 tries.
 
P C C

P C C

Active Member
Pericles the Lion said:
I don't think that this is correct. The events are independent of each other. If you miss on the first 10% chance the odds of winning on the second chance remains 10%. The result of the first chance has no impact on the result for the second, or subsequent, selections. Three chances, all 10%, equals a 30% chance of winning. If a player chooses 10 bakes, all with 10% chance, then the probability of a win is 100%, not 60.4% as your model suggests.
Click to expand...
The first part of your statement is correct, and is indeed what I used for my calculations, but the part starting with "Three chances" is not. The second try still has 10% chance of winning a daily, but one tenth of those times you will already have won on the first time so it doesn't add to the chance of at least one, hence 9% added. With two tries at 10% you have 81% chance of zero wins, 18% chance of one, and 1% chance of two.
 

Pericles the Lion

Well-Known Member
P C C said:
The first part of your statement is correct, and is indeed what I used for my calculations, but the part starting with "Three chances" is not. The second try still has 10% chance of winning a daily, but one tenth of those times you will already have won on the first time so it doesn't add to the chance of at least one, hence 9% added. With two tries at 10% you have 81% chance of zero wins, 18% chance of one, and 1% chance of two.
Click to expand...
Yep. I agree. Not to split hairs but the original question was which was a better choice (1) three 1 ingredient bakes @ 10% chance of success or (2) one three ingredient bake @ 15% chance. I think the answer is (1).
 
P C C

P C C

Active Member
Pericles the Lion said:
Yep. I agree. Not to split hairs but the original question was which was a better choice (1) three 1 ingredient bakes @ 10% chance of success or (2) one three ingredient bake @ 15% chance. I think the answer is (1).
Click to expand...
Agreed. The simple calculation of dividing the % chance of winning by the number of ingredients used tells you which is better on average (higher expected value) for daily specials. On average, you get 0.3 DS from (1) and 0.15 DS from (2). Note that 0.243 + 2x0.027 + 3x0.001 = 0.3.
 

Pericles the Lion

Well-Known Member
P C C said:
Agreed. The simple calculation of dividing the % chance of winning by the number of ingredients used tells you which is better on average (higher expected value) for daily specials. On average, you get 0.3 DS from (1) and 0.15 DS from (2). Note that 0.243 + 2x0.027 + 3x0.001 = 0.3.
Click to expand...
I'm having flashbacks to SAT prep. :cool:
 
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