I Juicy game probability calculation

[benorin]

TL;DR Summary

This Q was asked in a chat room: Find the number of pulls (draws from the distribution) required to have a 50% chance of getting a full set of event 5*s (*'s denote rarity in the game) from the Magic Tower Summon (of the Empires & Puzzles mobile game) which has 5x event 5*s each at a 0.2% drop rate. My first dilemma is binomial distribution or geometric distribution?

This Q was asked in a chat room: Find the number of pulls (draws from the distribution) required to have a 50% chance of getting a full set of event 5*s (*'s denote rarity in the game) from the Magic Tower Summon (of the Empires & Puzzles mobile game) which has 5x event 5*s each at a 0.2% drop rate.

My first dilemma is binomial distribution or geometric distribution?
I imagined that since the stated odds of pulling an event 5* is 1% but that includes 5x unique 5*s assumed weighted equally, I might have events A thru E be the pulling of the very first unique event 5* thru the fifth unique event 5*, where the associated drop rates are 1% thru 0.2% stepping down 0.2% each event after A. But since the unknown is the number of pulls (trials) don't I want to use geometric distribution? Please advise or solve this if you're inclined to (I won't take credit if you do and will cite you with permission from you of course)?

 

[Dale]

Summary:: This Q was asked in a chat room: Find the number of pulls (draws from the distribution) required to have a 50% chance of getting a full set of event 5*s (*'s denote rarity in the game) from the Magic Tower Summon (of the Empires & Puzzles mobile game) which has 5x event 5*s each at a 0.2% drop rate. My first dilemma is binomial distribution or geometric distribution?

My first dilemma is binomial distribution or geometric distribution?

Neither, it is the negative binomial distribution.

https://en.m.wikipedia.org/wiki/Negative_binomial_distribution

 

[benorin]

Ok, so negative binomial distribution it is. I get probability mass function

$f(n)= \binom{n-1}{4} \cdot 0.02^5 \cdot 0.98^{n-5}$

$\text{cmf}(n) = \sum_{k=5}^{n}f(k)$ and setting this equal to $0.5$ I get $n=245$ approximately which seems very low? Did I make a mistake? I used Wolfram|Alpha to do the sums. We were thinking the answer was closer to $1,000$.

 

[Office_Shredder]

This isn't a precise answer to your specific question, but I think is an easier way to think about it. When you still need all 5, you have a 1% chance of drawing one of them on each draw, so on average it takes 100 draws to get your first 5*. Then you have a 0.8% chance on each draw, so on average it takes 1/.008 = 125 draws to get your second 5*. Then 1/.006 + 1/.004 + 1/.002

This all adds up to 1142 draws on average before you get all 5, which is pretty close to your original guess

 

[Dale]

Ok, so negative binomial distribution it is. I get probability mass function

$f(n)= \binom{n-1}{4} \cdot 0.02^5 \cdot 0.98^{n-5}$

$\text{cmf}(n) = \sum_{k=5}^{n}f(k)$ and setting this equal to $0.5$ I get $n=245$ approximately which seems very low? Did I make a mistake? I used Wolfram|Alpha to do the sums. We were thinking the answer was closer to $1,000$.

You typed in $p=0.02$ which is a 2% chance. If you really meant a 0.2% chance then that would be $p=0.002$. $E(NB(n=5,p=0.002))=2495$

However, the answer by @Office_Shredder makes me think that I may have misunderstood the situation. I calculated the probability for all 5 being the same. Does the probability change as he suggested?

 

[Office_Shredder]

You typed in $p=0.02$ which is a 2% chance. If you really meant a 0.2% chance then that would be $p=0.002$. $E(NB(n=5,p=0.002))=2495$

However, the answer by @Office_Shredder makes me think that I may have misunderstood the situation. I calculated the probability for all 5 being the same. Does the probability change as he suggested?

Now I think I probably did it wrong after reading this. I read the first post as there are like, a list of things you can get, each with some probability, and there are five things in the list that each have a 0.2% probability, but reading the first post again I'm not confident that's right.

 

[benorin]

You typed in $p=0.02$ which is a 2% chance. If you really meant a 0.2% chance then that would be $p=0.002$. $E(NB(n=5,p=0.002))=2495$

However, the answer by @Office_Shredder makes me think that I may have misunderstood the situation. I calculated the probability for all 5 being the same. Does the probability change as he suggested?

There are 5x unique 5*s and the drop chance for any of the 5x uniques is 1% or 0.2% each assuming they’re equally weighted. So… you’re smart, you get it now.

Edit: The answer I got via Sums done in WolframAlpha fixing my decimal place was $n=2350$. Now that you better understand how the distribution is am I still calculating this correctly?

 

[Office_Shredder]

There are 5x unique 5*s and the drop chance for any of the 5x uniques is 1% or 0.2% each assuming they’re equally weighted. So… you’re smart, you get it now.

And just to confirm, if you get a specific 5*, you can get it again before getting a new one, right?

 

[benorin]

Yes but we only require 1 set of 5x uniques

 

[benorin]

You wrote at some point $E(NB(n=5,p=0.002))=2495$ I think the $n=5$ part is wrong, should be $n=k+5$ where $k$ is the number of failures. Instead of using $E(\cdot )$ I just took partial sums of the pmf(n) and got close to 0.5 summing over n=5 to 2345. It was approximately true not exactly. Maybe I could have got closer to 0.5, not sure.

 

[Dale]

There are 5x unique 5*s and the drop chance for any of the 5x uniques is 1% or 0.2% each assuming they’re equally weighted. So… you’re smart, you get it now.

Ok, then it is not the negative binomial. It isn’t any simple distribution I think. Let me think how to approach this.

 

[Dale]

Ok, so I think that the way to approach this is analogous to what @Office_Shredder did. We will use the geometric distribution which is the number of failures before a success in repeated trials each with given probability.

So $A \sim Geo(0.01)$ gives the number of trials before getting the first 5*. At that point getting that 5* again is considered a failure so the probability drops. So $B \sim Geo(0.08)$ gives the number of trials after the first 5* to get the second unique one. And so on to $E \sim Geo(0.02)$ for the number of failures after the fourth success to get the last success.

Then the final distribution is simply $X=A+B+C+D+E$. This distribution doesn’t have a name, but it can be calculated in Mathematica. The mean is 1139 and the standard deviation is 604.

Note that this is the number of failures before getting the 5 successes. So the number of trials is $X+5$

 

[DaveC426913]

When you still need all 5, you have a 1% chance of drawing one of them on each draw, so on average it takes 100 draws to get your first 5*.

Shouldn't the average be 50 draws?

 

[benorin]

Ok, so I think that the way to approach this is analogous to what @Office_Shredder did. We will use the geometric distribution which is the number of failures before a success in repeated trials each with given probability.

So $A \sim Geo(0.01)$ gives the number of trials before getting the first 5*. At that point getting that 5* again is considered a failure so the probability drops. So $B \sim Geo(0.08)$ gives the number of trials after the first 5* to get the second unique one. And so on to $E \sim Geo(0.02)$ for the number of failures after the fourth success to get the last success.

Then the final distribution is simply $X=A+B+C+D+E$. This distribution doesn’t have a name, but it can be calculated in Mathematica. The mean is 1139 and the standard deviation is 604.

Note that this is the number of failures before getting the 5 successes. So the number of trials is $X+5$

Wouldn’t it be $E \sim Geo(0.002)$? Lol you made the same placement error I did a few posts ago. Can you please recalculate in mathematica for me? I only have WolframAlpha.

 

[Office_Shredder]

Wouldn’t it be $E \sim Geo(0.002)$? Lol you made the same placement error I did a few posts ago. Can you please recalculate in mathematica for me? I only have WolframAlpha.

Since he got almost the same mean as me I assume he put it in correctly. I'm surprised the mean isn't more accurate though (1141 and 2/3 is literally the correct answer).

Shouldn't the average be 50 draws?

No, it's 100. It might be a little unintuitive, but remember it can often take many more than 100 attempts.

 

[benorin]

Thank you both