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Problem with mathematics

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I for my schoolyear have good math grades but FE demands more and so I have a question, say I had a character with 90 hit in a 1rn FE game and he doubles, if I need him to hit twice to kill the enemy how do I figure out my probability of killing the enemy in queston? take in consideration both enemies have no crit so that doesn't affect anything. Would the CoS be 45%? If so I should feel very lucky because that has happened to me several times in FE5 and is for that exact reason I don't think it works that way by averaging them out so how do I get the CoS and if I had, in another battle, 10 crit how would that affect the method by having more than 2 variables and the CoS?

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Simple, take the the percentage, 90%, make it a fraction, 90/100. Because I'm lazy, and for simplicity, we'll make this 9/10. Same for the second 90%. Then we multiply it by itself, or square it. So 9/10^2 is 81/100. So it's about an 81 percent chance of both hits occurring in a sequence.

As for CoS, I just plugged that into Google and got that it was 68%, so that might be a better way of measuring probability. 

Now, as for adding critical hits into that, that adds some complication. For critical hits to occur, we need to hit first, obviously, so first we'll calculate the chance of a 90% hit getting a 10% crit. One explanation later we get there is a 9% chance of a hit and a critical hit. This makes sense due to the 10% critical being kinda low. 

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1 hour ago, DarthR0xas said:

Simple, take the the percentage, 90%, make it a fraction, 90/100. Because I'm lazy, and for simplicity, we'll make this 9/10. Same for the second 90%. Then we multiply it by itself, or square it. So 9/10^2 is 81/100. So it's about an 81 percent chance of both hits occurring in a sequence.

As for CoS, I just plugged that into Google and got that it was 68%, so that might be a better way of measuring probability. 

Now, as for adding critical hits into that, that adds some complication. For critical hits to occur, we need to hit first, obviously, so first we'll calculate the chance of a 90% hit getting a 10% crit. One explanation later we get there is a 9% chance of a hit and a critical hit. This makes sense due to the 10% critical being kinda low. 

Thanks

But if the chance of sucess (CoS) of two consecutive 90% is 81% why does it change here all of a suddedn? Unless you misinterpreted CoS for something else?

So I jut divide 90 by 10 to get the chance? That seems extremely simple, can you tell methe exact proceedings?

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Yes, the CoS in the first example is 81%.

8 hours ago, Critical Sniper said:

and if I had, in another battle, 10 crit how would that affect the method by having more than 2 variables and the CoS?

It's possible I mistook what was asked here, but I'll try.

Firstly, both hits landing is still a success and in such an event (not) activating a crit won't affect the outcome (so, 0.81).

Then you want to figure out the probability of only one of the hits connecting. There's a number of ways you can go about this, but we can just subtract the odds of both hits hitting/missing to get it: 1 - 0.81 - 0.01 = 0.18. Since only one hit won't result in a success on it's own, we need to land a crit as well, and that's intuitively 0.18 * 0.1 = 0.018.

Summing it up, you'd have a 0.81 + 0.018 = 0.828 (or 82.8%) chance of killing the enemy.

Edited by Topazd

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4 hours ago, Topazd said:

Yes, the CoS in the first example is 81%.

0.18 * 0.1 = 0.018.

 

Oh ok

Wait how did that 0.1 come into existence?
 

Quote

Use a calculator of course lol.

That's not fun.

Edited by Critical Sniper

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1 minute ago, Topazd said:

It denotes the 10% chance of a critical.

Oh!  So if it were 25% critical it would be 0.25 ahh ok so let me see if I got the math right (I'm also doing it on paper)

In a Battle with a double attack of 70% hit (1rn of course) with 15% crit and I need to crit once and hit my CoS would be: 28.86? YEah t's an extremely low chance but I'mnot sure if I did it right

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