Amount of attempts for something to happen

[Roen]

Hello!

To start off, this is my first post here, math is not my strongest subject and so I do not know if this is pre university or university math.
Apologies if this is the wrong sub forum.

I have created a google doc where I am currently trying to solv my problem, link below.
https://docs.google.com/spreadsheets/d/102W_V674wH3wHw13EhS50hhi5IM410Rb74uUkQYkvUQ/edit?usp=sharing

I am trying to figure out how to calculate how many attempts on average it would take to achieve success, when every failed attempt increases the chance of succeding.

Example:
Base chance to succeed is 20%, every failed attempt increases chance to succeed by 2,5%.
How many attempts would it take on average, to succeed, when taking into concideration that the chance of success increases every failed attempt?

Thank you for any help you can give!

A major bonus would be if you could write your answer in a way that I can use excel to autopopulate cells with rows of higher base chance.

-Roen

 

[Greg]

Hello Roen and welcome to MHB! :D

We ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

Can you post what you have done so far?

 

[I like Serena]

Hello!

To start off, this is my first post here, math is not my strongest subject and so I do not know if this is pre university or university math.
Apologies if this is the wrong sub forum.

I have created a google doc where I am currently trying to solv my problem, link below.
https://docs.google.com/spreadsheets/d/102W_V674wH3wHw13EhS50hhi5IM410Rb74uUkQYkvUQ/edit?usp=sharing

I am trying to figure out how to calculate how many attempts on average it would take to achieve success, when every failed attempt increases the chance of succeding.

Example:
Base chance to succeed is 20%, every failed attempt increases chance to succeed by 2,5%.
How many attempts would it take on average, to succeed, when taking into concideration that the chance of success increases every failed attempt?

Thank you for any help you can give!

A major bonus would be if you could write your answer in a way that I can use excel to autopopulate cells with rows of higher base chance.

-Roen

Hi Roen! Welcome to MHB! ;)

How does the level of the enchantment factor in?

You list for enchantment level 8 that 5 attempts would be needed.
Doesn't that mean there would have been 4 failures?
Can you clarify?

Btw, your table shows that enchantment level 16 would suddenly be possible with a much lower number of attempts.
What's happening there?

 

[Roen]

The reason for enchant +16 becomming easier is due to the base % chance of success and the incremental increase per failure attempt being higher than +15.

I am able to calculate the average amount of attempts at a given already attained amount of failures.
For example:

At "enchant level" 9, which has a base success chance of 17,5%
With 8 accumulated failure attempts.
the average amount of attempts would be 3.
=100%/(17.5%+2%*8)

But my formula does not take into account the increase of 2% chance to succeed every time it fails to succeed.

This is where I am a bit lost.

How can I write this to properly calculate the correct average amount of attempts, taking height for the chance increasing every time you don't succeed.

Thank you!

-Roen

 

[I like Serena]

The reason for enchant +16 becomming easier is due to the base % chance of success and the incremental increase per failure attempt being higher than +15.

That didn't make any sense to me.
Can you clarify?

I am able to calculate the average amount of attempts at a given already attained amount of failures.
For example:

At "enchant level" 9, which has a base success chance of 17,5%

I'll stop here.
How did you get that at enchant level 9 the base success chance would be 17.5%?

 

[Roen]

This is for use in a game.
There is already a stated base success chance at each level of enchant.
Base chances for success are as follows.
At +8: 20% base, increase of 2,5% per failure.
At +9: 17,5% base, increase of 2% per failure.
Etc.
All of these base values the game already informs you of.

 

[I like Serena]

This is for use in a game.
There is already a stated base success chance at each level of enchant.
Base chances for success are as follows.
At +8: 20% base, increase of 2,5% per failure.
At +9: 17,5% base, increase of 2% per failure.
Etc.
All of these base values the game already informs you of.

Ok, so let's focus on +8 for now.

As I understand you now, it works as follows.
There's a 20% chance that we have a success on the first attempt.
And an 80% chance that we fail the first time, after which we have a 22.5% to succeed in the second attempt.
The chance that the second attempt also fails is then 77.5%, after which we have 25% to succeed in the third attempt.

That brings us to the following table.
\begin{array}{|c|c|c|} \hline
\text{Attempts} & \text{Probability formula} & \text{Probability} & \text{Probability} \times \text{Attempts}\\
\hline
1 & 20\% & 20\% &0.2 \\
2 & 80\% \cdot 22.5\% & 18\% & 0.36 \\
3 & 80\% \cdot 77.5\% \cdot 25\% & 15.50\% & 0.465 \\
4 & 80\% \cdot 77.5\% \cdot 75\% \cdot 27.5\% & 12.79\% & 0.512\\
\vdots & \vdots & \vdots & \vdots\\
33 & 80\% \cdot 77.5\% \cdot ... \cdot 0\% \cdot 100\% & 0\%& 0\\
\hline
\text{Sum}& 100\% & 100\% & 3.871 \\
\hline
\end{array}
It shows that the expectation (average number of required attempts) is $3.871$ instead of the $5$ that you currently have in your table.

Is this the number you are looking for?

 

[Roen]

yes! exactly.

There is one more element to this though.

for the +8, the incremental increase of 2,5% stops after failing 13 times.

so the maximum chance would be 20% + 32,5% chance of success.

How will this affect the table you made?

Thank you so much for your help so far.

 

[I like Serena]

yes! exactly.

There is one more element to this though.

for the +8, the incremental increase of 2,5% stops after failing 13 times.

so the maximum chance would be 20% + 32,5% chance of success.

How will this affect the table you made?

Thank you so much for your help so far.

Then we'll get:
\begin{array}{|c|c|c|} \hline
\text{Attempts} & \text{Success} & \text{Previous attempts failed} & \text{Probability} & \text{Probability} \times \text{Attempts}\\
\hline
1 & 20\% & 100\% & 20\% & 0.2 \\
2 & 22.5\% & 100\% \cdot (100\% - 20\%) = 80\% & 18\% & 0.36 \\
3 & 25\% & 80\% \cdot (100\% - 22.5\%) = 62\% & 15.50 \% &0.465 \\
4 & 27.5\% & 62\% \cdot (100\% - 25\%) = 46.5\% & 12.79\%& 0.512\\
\vdots & \vdots & \vdots & \vdots & \vdots\\
13 & 50\% & 0.64\% & 0.32\% &0.042\\
14 & 52.5\% & 0.64\% \cdot (100\% - 50\%) = 0.32\% & \\
15 & 52.5\% & 0.32\% \cdot (100\% - 52.5\%) = 0.15\% & \\
16 & 52.5\% & 0.15\% \cdot (100\% - 52.5\%) = 0.07\% & \\
\vdots & \vdots & \vdots & \vdots & \vdots\\
\hline
\text{Sum}& & & 100\% & 3.872 \\
\hline
\end{array}

 

[Roen]

Perfect!

Thank you a million times.

I'll start working on how to get it into a forumla in excel.

Any ideas?

The kind of formula I'm looking for would be able to scale with the initial amount of failures.

Example:
+8, 4 failures

I'm sorry if it is a bit vague.

 

[I like Serena]

Perfect!

Thank you a million times.

I'll start working on how to get it into a forumla in excel.

Any ideas?

The kind of formula I'm looking for would be able to scale with the initial amount of failures.

Example:
+8, 4 failures

I'm sorry if it is a bit vague.

That is a bit vague. :)

I've already played around in Excel a bit and got this table:

Enchantment	Base success	Increase per failure	Nr Increases	Expectation
+8	20.0%	2.50%	13	3.9
+9	17.5%	2.00%	14	4.4
+10	15.0%	1.50%	15	5.0
+11	12.5%	1.25%	16	5.8
+12	10.0%	0.75%	18	7.3
+13	7.5%	0.63%	20	8.8
+14	5.0%	0.50%	25	11.3
+15	2.5%	0.50%	25	14.2
+16	15.0%	1.50%	25	5.0
+17	7.5%	0.75%	35	8.5
+18	5.0%	0.50%	44	11.3
+19	2.0%	0.25%	90	19.2
+20	1.5%	0.25%	124	20.5

I've put the sheet in Google docs here.

To "scale with the initial amount of failures" I'll need some more clarification of what you mean.

 

[Roen]

Amazing, I've made a copy for myself to study later.

By scaling I meant that.
Like in my original attempt, I want to see the expectation for starting the process at an already accumulated failure amount.

base + 1 failure = expectation
base + 2 failures = expectation
base + 3 failures = expectation
...
base + 13 failures = expectation

In a way that does not require me to do the individual calculation for every single one.

 

[I like Serena]

Amazing, I've made a copy for myself to study later.

By scaling I meant that.
Like in my original attempt, I want to see the expectation for starting the process at an already accumulated failure amount.

base + 1 failure = expectation
base + 2 failures = expectation
base + 3 failures = expectation
...
base + 13 failures = expectation

In a way that does not require me to do the individual calculation for every single one.

Ah okay, I've extended the sheet with the initial failures.

Expectation of the number of attempts is:

Enchantment	Initial number of failures
Level	0	1	2	3	4	5	6	7	8	9	10
+8	3.9	3.6	3.3	3.1	2.9	2.8	2.6	2.5	2.3	2.2	2.1
+9	4.4	4.1	3.8	3.6	3.4	3.2	3.0	2.9	2.7	2.6	2.5
+10	5.0	4.7	4.5	4.2	4.0	3.8	3.6	3.5	3.3	3.2	3.1
+11	5.8	5.5	5.2	4.9	4.7	4.5	4.3	4.1	3.9	3.8	3.6
+12	7.3	7.0	6.7	6.4	6.2	5.9	5.7	5.5	5.4	5.2	5.0
+13	8.8	8.5	8.1	7.8	7.5	7.2	7.0	6.7	6.5	6.3	6.1
+14	11.3	10.8	10.4	10.0	9.6	9.3	9.0	8.7	8.4	8.1	7.9
+15	14.2	13.6	13.0	12.4	11.9	11.4	10.9	10.5	10.1	9.7	9.4
+16	5.0	4.7	4.5	4.2	4.0	3.8	3.6	3.5	3.3	3.2	3.1
+17	8.5	8.1	7.7	7.4	7.0	6.8	6.5	6.2	6.0	5.8	5.6
+18	11.3	10.8	10.4	10.0	9.6	9.3	8.9	8.6	8.3	8.0	7.8
+19	19.2	18.6	18.0	17.4	16.9	16.3	15.9	15.4	15.0	14.5	14.1
+20	20.5	19.8	19.2	18.6	18.0	17.4	16.9	16.3	15.9	15.4	15.0

 

[Roen]

Great!

Thank you so much for your help.

I really appreciate it.

 

[Roen]

I was reviewing the information in the sheet, and I came across something I had forgotten about.

At +16, +17, +18, +19 and +20, the amount of failure increases per fail are higher than previously.

So at +16, every failed attempt gives you 2 failures.
+17 gives 3 failures.
+18 gives 4 failures.
+19 gives 5 failures.
+20 gives 6 failures.

I attempted to simply increase the amount each failure gives, but this ends up giving the wrong results.
As it does not match up correctly with the maximum amount of failures that way.

Might there be a way for it to account for the failure chance jumping 2/3/4.. amounts depending on the "enchant" level?

-Roen

 

[I like Serena]

I may not quite understand yet...
How about reducing the maximum number of failures as well?