What are the odds of winning two 50/50s with only 4 pulls in between??

I saw someone on twt who got venti, and then Durin 4 pulls later, and I was wondering what the odds of that were, and I'm insanely bad at math and I can't even begin to think of what to ask google or literally anyone that DOESN'T play genshin

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peruanToph2026-01-02 15:47

The chance of winning a 50/50 is 0.5. Back to back is 0.25

The chance of getting a 5 star within 4 pulls is 0.0238

Chance of getting a limited 5 star within 4 pulls then is 0.0119

Chance of getting a limited 5 star within 4 pulls after having gotten a limited 5 star is 0.5*0.0119

= 0.00595 = 0.595%

atabar932026-01-02 19:13

I feel like you divided by 2 twice?

peruanToph2026-01-02 19:27

I never divided by 2 here

atabar932026-01-02 19:29

1- Chance of getting a limited 5 star within 4 pulls then is 0.0119

2- Chance of getting a limited 5 star within 4 pulls after having gotten a limited 5 star is 0.5*0.0119

The chance of getting 2 limited chars in 4 pity is just the chance of getting 1 limited char in 4 pity

We have 0 info about the first limited character. For all we know he could've been on 90 pity at a guaranteed

The chance is just P(getting another limited at 4 pity/we got one)=1-P(not getting a limited in 4 pulls)

=1-((1-P(getting a limited))^4

=1-(1-0.003)^4

=1-0.997^4

=0.01194610791

~1.19%

peruanToph2026-01-02 20:06

I thought i had to multiply the chances of two independent events happening to learn the chamce of them happening together?

Like:

P(AnB)=P(A)*P(B|A)

We know A and B are independent from each other (winning 50/50’s reset the 50/50 count) so:

P(B|A)=(P(B)*P(A|B))/P(A)=P(B)

Then P(AnB)=P(A)*P(B)=0.5*0.0119 = 0.595

Also I asked a friend about this and he told me that I did it wrong cuz I had to make sure I didnt get a 5 star in the first 3 pulls and only won once reached the 4th (so not the chance of winning *within* 4 pulls, but the chance of winning on the *4th pull*) so idk if this would be correct?:

(1-0.006)^3 *(0.006 *0.5)= 0.00294= 0.294%

atabar932026-01-02 20:14

The calculations seems right but the result feels wrong (I think it is correct but I got so weirded by how much it differed)

What we previously calcuated is the probability of at least 1 limited 5* in 4 pity

What you are calculating is the probability of exactly 3 non 5* and then the limited. But I didn't realize this would lower the odds so much!!

Shahadem2026-01-02 20:15

Those are pretty good odds

lostn2026-01-02 23:30

>0.595%

That is way too high a probability. 0.595%, remember it's 0.3% to get a 5 star and win the 50/50. How can the odds be higher than base odds?

If I understand your explanation, you're saying the chance of getting 2 5 stars within 5 pulls total, and both of them won the 50/50 is 0.595%? Those odds sound way too high.

peruanToph2026-01-03 02:27

No, i calculated the chance of getting a limited five star within 4 pulls after having gotten a limited five star, but I did not care about how much pulls it costed the first five star because OP didn’t say it, because it wasn’t important to their question.

So the question wasn’t what’s the chance of winnign two 50/50’s back to back within 5 pulls, it was whats the chance of winning in 4 pulls after just having done it

Edit: to clarify. It sounds high because the two events are independent from each other from happening, which means that winning the first 50/50 doesnt affect the chances of winning another 50/50 after 4 pulls, and thats why the chances of them happening one after the other gets multiplied

lostn2026-01-05 06:06

I think you interpreted his question wrong then. At least at the time he posted it.

Because the way you interpreted it, the first 5 star is not even relevant. The question could be rephrased to what are the chances of a 5 star win in 4 pulls? That's lucky but not extraordinarily lucky. OP mentioned the luck as memorable because it was two 5 stars not one.

OP asked what are the chances of getting a 5 star win 4 pulls after another 5 star win. If you aren't including the odds of the first 5 star win, then the math would not change if OP never even mentioned the first 5 star win. He mentioned it for a reason. A second win in that short a span makes it extremely lucky. This is the difference between getting one 5 star in a 10-pull or two 5-stars in the same 10-pull. He wouldnt' mention the first 5 star if it didn't matter to him.

TBF OP did edit his post later to clarify that the first one was at pity, so your math ended up being correct. But at the time you calced it, he did not mention that. Without that info the fair interpretation would be that two 5 stars in the span make it luckier than only one. Otherwise, it could just be any 4 pulls, even if the 5th pull before the second 5 star was just a 3 star and the math would be the same.

You ended up being correct, but only because OP did not clarify.

lostn2026-01-02 23:29

probability to win a 50/50 and then win another 50/50 4 pulls later (5 pulls total) is [0.00895%](https://www.omnicalculator.com/statistics/coin-flip-probability?calculatorResult=H4sIAAAAAAAAA8VYbVPiOhT%2BK72dO7PujrJtKRQd9Q6rrhcU30CZ9c4OE2hos5smJUkRdPzvN%2BkLFK0Ku%2Bp%2B0TY5OS9PznlOD3c6JB4isC2AgPrWnT4AeBBh%2BeJeAYZAH0N%2BAgLI9a3%2F9JDRfi%2BEbACJ0NfjV9BHGIlpz%2Bwhsrikf1%2FXBz4gXpEm0htiFPL0RLHOIgtKJyUuEogSrtxFJIzEwoMPgctLhnocAxzJoJIlbXdHM6ROFwo4yIen%2FIkllPYACuACAdRxwDkdoDwUmS6pBvF9xEMMptDVt4YAc6gOcw68GMYIyWU9ql83924GJ3uGOR6RPf7jou6cljKxdd0XAZZi2%2BFux4caiYI%2BZBodarQvgLwUV0s8DyIutD7UgNAwBPLZKG1%2FDnf1%2B%2Fv7FEA67CVxm%2Fm4F7aejX9BcikcFk8sicc4hO0vm0bYiQKPeM2wap5b9Ek8crevQPGgEIh4q2NiPY3J9o5m%2FmFMkLUxubit14%2Fx7dVJOerbX63J7a9hMgCE0BgVj0HpG9OELEDNzCGTFl6pnMckXdR2n06QrGCXgWFe3EsB0EYnYZ%2FB87PmpFO%2B9XukWT1CSxSJoJzDeSKElEtaGMOCYO2iYNXVG8afCHhCosvqAAdO5aLT8Fv858TeGBUG3KU36%2BoOxQeuAQ3Lu5Vxx7b%2B0rr%2BVHMp%2BSC0G6gJNtU4lY76Kht4ADCG7J85EkkhVAovfUfL8vUZalxfCZCVCuDYGl5Ho4Oryw0Df%2FXPDb%2FCTXt1knwi%2BUVBzhSkSPW9oFkxV%2BpH1UkH2IenzHD7YowO%2By3aXQKcJICsNp5BJAEvw7L0MGWcPC6IN4iAHmRr8ebHd%2B6m5r%2Fdm1ErdC66Pw%2BbtYPqxkUZ8yWwSEL0AZfXH7cMoqEkjoJEqBUHnG4%2FGfIbpsBoPOoGo2%2FNg0aXnp3Zw9YBC86XToFnw5Zx00g8%2FnB6%2FlZTdASLpLuCIU%2BqlLSTCauPPOmPJ%2F%2F1WKQOF3%2BlLNd0f9%2BY9bbGFjrqS7mxnIHswMMW9jraF93PdYZlOe6Xg6i%2BqZUcYb1%2B%2Fi7QwyvfsirDCOWHmHxdxtty6BlDxuXU85WyQM5n8fCzMDup2W0m1cjkpJ21NfOz%2FNL5%2BGmNwSFkkAzglXLwoyKz%2BZHTSMzPSPlPa4%2FlY1dkT6FBkx8wRpmK%2FHtubRJSFgMj5XkSCJT6915zqIQTwcB%2BSrCI190xkC7OCRTxFpXScLYQccgSeCIiP5T2qKvgr1%2Bq8COmApzG7%2Fu5hfY06FNFq3%2BrrARj5AFB2bEcZyPJuh3gyS1INi7beqy%2FEcgIEMCXBAk%2BMxxK0exIahWSGMRhgrQS34chJK6y2QJhAhmmg59FUOVScr1gPAEkArihUijOjpx03Eukdcu5ur08P234RzatnJ9dT9v%2Bt1acBYE8lbSnO6lI%2BD%2FkIf0L8k7idqLPclq39EcDVoLt0hqMkmGU9TTn529ZOaymrDJTVJlxwIoqrJmKJLYExxfQzwjr4ciXY4Gkgk8JnuZ5IqOHN7%2Bi34bila7kDVLlMe%2BtpNEwapt21bRM27BqC9of7eSp570MJbS3mjXTNJ1aSeqyNzdtu%2BJUpH75UrHKNcepOI5Rsy2jXLFqhm2almNbZfnHic2bFcuoVWfevYKmlCDGyIVs9rsiiOTsBcYwbRgUYxBy6DbctCh8iMOOpHYlO8u7lEfT1MlYdTGhcquLAMYbMzZ9mT85xHFXjzm8jSOvsMeGDAWAqYahFuOP6P8Be4lZ2EQVAAA%3D). This is assuming no pity on the first one.

Or 1 in 11,178.

This isn't exact. The caveat with this math is that the two 50/50s can be anywhere within those 5 pulls. If the second one has to come exactly 4 pulls later and not 3 or 2, or the first one has to come on pull 1 and not later, the odds change and get even worse.

When we calculate probabilities, we don't normally go this exact. We're more interested in knowing the odds of 2 50/50 wins within 5 pulls than the chances of the wins being exactly in pull 1 and pull 5. In real life, we don't care what order they came in because it's immaterial.