Ludia Forums

Probabilities

What is the probability that someone got past dodge/evasion 5 straight times? Because it happened to me just now with indo twice, procera twice and erlidom 1x in battle that’s 25% chance success 5 straight times.

Anyone who’s good at math and probabilities able to figure that out?

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(0.75)^n

n = # of dodges

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Looks like you fell into the 25% chance window of not dodging… but 5 times in a row means it was more like a 5% chance window of that happening. About the same chance of you getting 100 DNA on a fuse.

My suggestion: Take those odd to Vegas!

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Haha gonna buy a lotto ticket then! Was mad but can’t complain. Lately evasive abilities seem more like 50% effective for me

It’s 1 in 1024 chances.

But assuming there are 20,000 players battling several times per day, thus would happen dozens of times every day to someone.

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Don’t you mean (0.25)^n?
Wouldn’t (0.75)^n be the chance of dodging n times in a row?
Multiplied by 100 for the percentage.

Anyway, (0.25)^5 leaves us with approximately 0.001 (it’s more like 0.000976525). That’s approximately a 0.1% chance (0.0976525%). That’s your chance of evasive failing 5 times in a row.

(0.75)^5 would be around 0.2 (0.237304687), so approximately a 20% chance (23.7304687%). That’s your chance of dodging 5 times in a row.

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Wow the stats are crazy thanks everyone for doing the maths!

This is the ludia math:

5(ludias) * .25(ludias) / Inifinte(ludias) = ludias^ludias

Seriously, 5 times is a lot. I have had it happen myself a few times. But, while the math seems covered enough already (I am not checking it, just at a glance it seems fine), it would be suspect if the system did not sometimes spit out many failures as ‘random’ means it should happen. There is a chance in flipping a coin that it will land on heads 5 times in a row…if it never did…I would think it was being manipulated in some way.

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If a Dino has a 75% of dodging once…
The odds of dodging twice back to back would be 0.75*0.75

And so on…

Calculating with 25% would be the odds of hitting through the dodge/evasive.

Hitting through a dodge twice back to back would be 0.25*0.25

And so on…

Doing this old school math instead of that common core mess: The first dodge has a 25% chance of not activating, so 25/100. The second in a row is 25% of that, so 6.25%. The third would be 1.5625%. The fourth would be 0.390625%. The fifth would be 0.09765625%

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I mean isn’t the chance reset every go. Each specific chance is it’s own. Every time its 75% chance to dodge…try 1, 2 3 and so on. The odds are only going to be 25/100 for you to hit everytime the dice is rolled

Well, you are correct in that it resets every time and each round has a 25/100… but were talking about the odds of 5 times in a row… so that’s what is being calculated.

Several recalculations above that still get to 1 in 1024.

Except for the couple that are completely wrong.

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Actually it would be 243/1024 ((3^5)/(4^5))or as Hersh said, 0.75^5. The odds of them dodging again after 4 dodges in a row is 75% or 3/4. The probability is ~23.7% chance of them dodging a 5th time in a row. You have a probability of ~76.3% of hitting them on that 5th turn but it’s actually only a 25% chance to hit them. Betting on probability is a losing bet most of the time.

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I don’t know. But I’m sure the probabilities of my 7 Critical hits in one battle all by one Dinosaur is much much lower.

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That isn’t what was asked, but the calculations check out. Are you sure that 76.3% is the chance of hitting through evasion on the 5th? Isn’t it the probability of any other combination of 5 turns of dodge/hit apart from 5 dodges? The complement of the one you calculated.

Good grief!

It’s a 25% chance of the creature successfully getting through the dodge. So the odds of success 5 times in a row is 0.25^5 or 1 in 1024.

Your testing the probability of the dinosaur dodging 5 times in a row (or the player not hitting 5 times in a row). 1/1024 is the probability of the dinosaur not dodging at all and getting hit 5 times in a row.

The total of an event happening plus the event not happening will always equal 1. The probability of a player dodging 5 times in a row is 243/1024 so the probability of them getting hit on the 5th attempt is 1 - (243/1024) or 781/1024.

Edit: a-ha… just went back and re-read the original post.

Yes, 1/1024 is the correct answer to the situation the OP asked about. As we should be testing for 5 hits in a row or 0.25^5. My apologies to misinterpreting the original question. There is less than a 1% probability that the 5th hit will land.

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So it wouldn’t be (0.75)^4 times (0.25) then?
My point is, 1 is the sum of all possibilities. Aren’t there more than 2 in this case?

Not dodging 5 in a row is 1/1024 as said before. Anything different is just wrong or they didn’t get the correct question. It’s very simple math actually.