What are the odds

[Select Two]

Sequential find just minutes apart.

https://www.nationstates.net/page=deck/ ... _history=1

[Edush]

That's impressing!

[Tinhampton]

Select Two answered five issues today.

Getting a Pack

There's a 20% chance of drawing a pack with every issue you answer. This is an independent event. There's a 98% probability it's going to be S3 (1% S2, 1% S1) so let's call it 19.6%.

The binomial probability formula for getting something k out of n times is [n! / k!(n-k)!] * p^k * (1-p)^(n-k).

We know that k = 2 and n = 5, so n-k = 3. Why? n! = 2x1 = 2. (n-k)! = 3x2x1 = 6. p = 0.196, as discussed earlier, so 1-p = 0.804.

Ergo, the probability of getting two packs in a set of five issues would be [2 / (2*6)] * 0.196^2 * 0.804^3 = 0.00332758... ≈ 0.33%.

The chance of getting two Season 3 packs in the first place is about 1 in 300, even if the cards are totally dissimilar. (The chances of getting two packs of any description are 1 in 293, but I won't go over that here.)

Getting a Card

All packs contain five cards. If my calculations are to be vindicated, there are 221,300 Season 3 cards. Cards also appear independently in packs.

The chance of getting any one card in any one pack is obviously 1/221,300, which I'll call c for short.

The probability of never getting that card is (1-c)(1-c) = 0.99999096... ≈ 99.9991%. This is so likely I won't even bother to express it in "1 in [xyz]" terms.
...of getting that card once is [c(1-c)]^2 = 0.000004563... ≈ 0.0004563% ≈ 1/221,301.
...of getting that card twice is c^2 = 1/48,973,690,000.

Conclusion

48,973,690,000 x [6 / (0.196^2 x 0.804^3)] ≈ 1 / 14,717,491,351,562

The odds of getting two Season 3 packs, and getting the same card in each pack, over a five-issue span is one in 14.7 trillion.

For a sense of scale: the odds of winning the British National Lottery twice are "only" one in 2,030tn. 59 choose 6 is 45,057,474. Squared is 2,030,175,963,260,676.

You're more likely to draw five legendaries in one (Season 3) pack - the odds of that are about one in 13.65tn. Derived from 0.1^5 * [6 / (0.196 x 0.804^4)], although the real S3 legendary rate is more like 1/992 - because there are 223 of them - and I don't know why.

[Improper Classifications]

Whoa

[Ostrovskiy]

this might just be the "luckiest" thing in ns history

[Select One]

Select Two answered five issues today.

Conclusion

48,973,690,000 x [6 / (0.196^2 x 0.804^3)] ≈ 1 / 14,717,491,351,562

The odds of getting two Season 3 packs, and getting the same card in each pack, over a five-issue span is one in 14.7 trillion.

For a sense of scale: the odds of winning the British National Lottery twice are "only" one in 2,030tn. 59 choose 6 is 45,057,474. Squared is 2,030,175,963,260,676.

You're more likely to draw five legendaries in one (Season 3) pack - the odds of that are about one in 13.65tn. Derived from 0.1^5 * [6 / (0.196 x 0.804^4)], although the real S3 legendary rate is more like 1/992 - because there are 223 of them - and I don't know why.

Wow! Just wow!

I answered all five issues one after the other. I don't read them as I (obviously) select two each time. So it was probably less than one minute to get through all five.

Thank you for the detailed analysis. That's really special. I'm going to print it out and put it in my box of interesting-things-in-my-life.

[Ioavollr]

...

Courageous effort, though I think you have made some incorrect assumptions and have set this up incorrectly.

You have an error in your binomial probability formula: it should be P(2)=5!/(2!(5−2)!) ⋅ 0.1962 ⋅ (1−0.196)^(5−2), you have confused some 2's and 5's in your formula. The correct probability of getting two packs in 5 issues is P(2)=0.19965..., about 19.7%. Since we would rationally include receiving 3, 4, and 5 packs, this probability increases to 0.25454..., about 25%

The duplicated card is not determined beforehand, so the first probability is 5/5, as any 5 cards will do. The assumption that we need to square c in your equation is incorrect. Further, the odds are not for that specific card twice, instead for any of the original 5 pulled in the second pack. The odds of finding any card twice in two packs would be 5/221,301 or 1 in 44,260.2.

Dividing by the 25% rate of getting two packs nets a final probability of pulling any card twice in 5 issues as about 1 in 173,877. This is about the same chance as being killed by lightning.

[Select One]

Dividing by the 25% rate of getting two packs nets a final probability of pulling any card twice in 5 issues as about 1 in 173,877. This is about the same chance as being killed by lightning.

Awww.

No matter. I'm still thrilled about it, and metaphorically, killed by a lightning bolt, as well.

Thanks for taking the time to refresh odds.

[Tinhampton]

My maths degree is 26 years old and hasn't been put to the best of use in my years as a managerial generalist, sorry

[Riemstagrad]

Code: Select all

3-10-2023 21:28:30 CEST: Found by Rariteiten VIII
3-10-2023 21:28:20 CEST: Found by Rariteiten VIII

on this card: https://www.nationstates.net/page=deck/ ... _history=1


twice on the same puppet, within 10 seconds.


(it was a puppet with 9 packs ready. found them two packs appart.)


about the probability: Since it is a non-CtE common in S3, the chance to pull it once was 1/27381.91 packs back in march when i compiled my statistics, based on the number of CtE and active cards back then. I guess that there are even more CtE cards in the S3 set today, so the chance for an active card will be a little higher now.

With these (old) numbers:
with 9 packs, for the chance for a double pull, each pack 'n' you open has a chance of (5*(n-1))/27381.91 to find a duplicate.
That makes 180/27381.91 = 1/152.12 to find a double pull on a puppet with 9 packs.

For a puppet with 2 packs the odds should be 5/27381.91 = 1/5476.38

Also, the probability to get at least two packs in 5 issues is more something like 26 % and not 1/300 dice roll probability

Edit: please correct me if i'm wrong. I like to hurt my head on statistics every once in a while, but i will never find it easy

[Select One]

Code: Select all

3-10-2023 21:28:30 CEST: Found by Rariteiten VIII
3-10-2023 21:28:20 CEST: Found by Rariteiten VIII

on this card: https://www.nationstates.net/page=deck/ ... _history=1

Wow. Another sequential find and within 3 days of each other.

I guess it's inevitable that this probably happens more than we notice. Kind of like how there's so many musicians on the planet today that bits of melodies from songs pop up in other artist's songs.

Order in the chaos?