Probability/Odds of multiple hits in a single instance

[LexBrawl]

Hey everyone. My first post here. I'm a YouTuber with over 500,000 subscribers and I have an idea for but I do not know how to do the math so I've come looking for some help from you all who are much more knowledgeable than I am!

Here is my specific scenario. I play mobile games on my youtube channel. Specifically a game called Brawl Stars. In the game you can buy loot boxes called mega boxes. These mega boxes are equivalent to 10 regular boxes. In each box you have a chance to pull a new character for the game. These odds are known (listed below) It is possible to pull multiple characters in a single mega box as it is essentially 10 regular boxes and the pull rate for the new characters is based off one single box. I've pulled as many as three new characters in a single mega box before (although only once... I'm hoping you can tell me the odds of that happening, as well as the odds of other combinations happening)

So here are the odds of getting a new character in a single box. They are broken down into rarity. Again a mega box = 10 chances (although it shows them all at the same time, not as individual box openings)

Rarity Pull Rates

Rare: 2.75%
Super Rare: 1.25%
Epic: 0.55%
Mythic: 0.25%
Legendary: 0.1%

So. With that all being said (I hope I've explained it sufficiently) What are the odds/probability/percentage of getting multiple items in a single mega box?

Using a specific instance I once got, in one mega box, One Super Rare, One Epic, and One Legendary character in one box. What are the odds of that happening?

What are the odds of getting 4 Legendary characters in one Mega Box? (I've never seen it happen but I think it's technically possible)

Sorry If I haven't explained something well. If there was a formula I could plug in the desired rarities and come up with the odds of that happening that would be wonderful as I could then give multiple scenarios in the video I'm planning. Thank you in advance for your help. It is very much appreciated.

 

[Cubist]

What are the odds/probability/percentage of getting multiple items in a single mega box?

The following statements are heading towards the answer (given in bold below)...

The probability of getting ANY character in a single box is (2.75 + 1.25 + 0.55 + 0.25 + 0.1)/100 = 0.049
The probability of not getting a character in a (single) box is 1 - 0.049 = 0.951
The probability of not getting any characters with a mega box is 0.951^10 ≈ 0.6050693712
The probability of getting exactly one character with a mega box is 0.049 * 0.951^9 * 10C1 ≈ 0.3117602438 or approx 1 in 3
NOTE 10C1 means the ways of choosing 1 item from 10. See this wikipedia page

The probability of getting more than one character (2,3,...,10) with a mega box is approx 1 - (0.6050693712 + 0.3117602438) = 0.0831703850 or approx 1 in 12

--

You don't ask this, but the probability of getting exactly 2 characters is 0.049^2 * 0.951^8 * 10C2 ≈ 0.0722851039 or approx 1 in 14

And the probability of getting exactly 3 (not just the specific ones that you once obtained) is:-
0.049^3 * 0.951^7 * 10C3 ≈ 0.0099319175 or approx 1 in 101

the probability of getting exactly 4 is 0.049^4 * 0.951^6 * 10C4 ≈ 0.0008955436 or approx 1 in 1117

What are the odds of getting 4 Legendary characters in one Mega Box? (I've never seen it happen but I think it's technically possible)

The probability of getting exactly 4 Legendary is:-
0.001^4 * 0.951^6 * 10C4 ≈ 0.00000000015534683010 or approx one in 6,437,208,917

Using a specific instance I once got, in one mega box, One Super Rare, One Epic, and One Legendary character in one box. What are the odds of that happening?

The calculation needs to be different due to specifying different types of characters
0.0125 * 0.0055 * 0.001 * 0.951^7 * 10 * 9 * 8 ≈ 0.000034823211254 or around 1 in 28,716
...but this probability seems overly specific. Wouldn't it be more informative to calculate the odds of getting this result OR better? But this would be harder to calculate!

 

[Cubist]

BTW The name "Mega Box" is very misleading since it implies a million standard boxes - but you only get 10. Scandalous! But I guess the name Deca Box wouldn't sound quite as cool

 

[LexBrawl]

Thank you Cubist! I got a couple questions once I wrap my head around the math, but first let me ask if you are OK with me giving you credit in the video?

haha.. never thought about the mega box moniker!

 

[Cubist]

but first let me ask if you are OK with me giving you credit in the video

I'd rather that you give any credit to this forum/ website, rather than myself. Its a great place for math students who need some help/ direction. (We're reluctant to just give out final answers, the idea is to help posters to answer their own question, which is probably why you've had a slightly slower response since you don't seem like you're a student yourself but were just interested in the final numbers)

 

[LexBrawl]

I'd rather that you give any credit to this forum/ website, rather than myself. Its a great place for math students who need some help/ direction. (We're reluctant to just give out final answers, the idea is to help posters to answer their own question, which is probably why you've had a slightly slower response since you don't seem like you're a student yourself but were just interested in the final numbers)

For sure I can do that! In fact the bulk of my audience are kids/young adults in school so it could be a very good resource for them! The video could get upwards of a million views so hopefully it will help a lot of people out too. Good call.

I have a question about the first part of your initial reply.

You said:

"The probability of getting ANY character in a single box is (2.75 + 1.25 + 0.55 + 0.25 + 0.1)/100 = 0.049"

I'm positive you're right however I don't understand. If the cumulative total is 4.9% of pulling a character wouldn't that be the percentage? I'm assuming I'm getting mixed up in the nomenclature.

 

[mmm4444bot]

For sure I can do that! In fact the bulk of my audience are kids/young adults in school so it could be a very good resource for them!

Hi Lex. If you reference this forum, please convey also the message that this is a tutoring forum. We expect students to show some effort, when posting requests for help. (Currently, we have enough members looking for people to do their schoolwork; we don't need any more.) Thank you!

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[Cubist]

You said:

"The probability of getting ANY character in a single box is (2.75 + 1.25 + 0.55 + 0.25 + 0.1)/100 = 0.049"

I'm positive you're right however I don't understand. If the cumulative total is 4.9% of pulling a character wouldn't that be the percentage? I'm assuming I'm getting mixed up in the nomenclature.

I'm not sure what you're asking. Probabilities in maths are usually represented by a number in the range 0 to 1. Zero means no chance of an event, up to one which means the event is certain. Sometimes a probability can be quoted as a percentage in which case the range is 0% to 100%. It's easy to convert from one to the other, simply divide or multiply by 100. So a probability of 0.049 is equivalent to 4.9%. This can also be written, approximately, as "a chance of 1 in 20.4", the latter number is calculated by 1/0.049

 

[LexBrawl]

I'm not sure what you're asking. Probabilities in maths are usually represented by a number in the range 0 to 1. Zero means no chance of an event, up to one which means the event is certain. Sometimes a probability can be quoted as a percentage in which case the range is 0% to 100%. It's easy to convert from one to the other, simply divide or multiply by 100. So a probability of 0.049 is equivalent to 4.9%. This can also be written, approximately, as "a chance of 1 in 20.4", the latter number is calculated by 1/0.049

That cleared it up! I was just not understanding/mixing up terminology. My last math class, and/or practical use of this sort of thing was 25 years ago Apologies for my ignorance.

Really appreciate your help.

 

[LexBrawl]

Hi Lex. If you reference this forum, please convey also the message that this is a tutoring forum. We expect students to show some effort, when posting requests for help. (Currently, we have enough members looking for people to do their schoolwork; we don't need any more.) Thank you!

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100%! I'll be sure to make mention of that.