1vs2 70 percent sucess they roll a 1d10 i take note if it was successful or not they do not know they only know they have 70 percent chance. They are allowed to repeat at a time sacrifice and add outcomes to a pool they can look over of 1 or 2 then they choose likely the most common number to appear. At what point of attempts would they have a 90 percent success rate from their initial 70 percent by looking at the numbers that came up?
Yeah, I'm not really sure what you mean. I have two possible interpretations. is it possible to get a 90% sample from a 70% population (not really) or how many times do you need to take a 70% chance to have at least a 90% chance of succeeding at least once ( twice).
If you are asking when the probability of choosing a success from a random sample that was taken from a set of possible outcomes that are 70% success will be equal to 90% then it's impossible to tell because as the sample size grows it will approach that of the total possible outcomes i.e 70% success. The appearance of a 90% chance in the sample will occur randomly and becomes less likely as the number of rolls increases.
If you are asking how many times you need to repeat a 70% success event until you have a greater than 90% chance of having at least once success then the answer is twice. The probability of getting a 30% chance event twice in a row is 9% so they have a 91% chance of succeeding if they can pick either of the two die after they roll.
Yeah, I'm not really sure what you mean. I have two possible interpretations. is it possible to get a 90% sample from a 70% population (not really) or how many times do you need to take a 70% chance to have at least a 90% chance of succeeding at least once ( twice).
If you are asking when the probability of choosing a success from a random sample that was taken from a set of possible outcomes that are 70% success will be equal to 90% then it's impossible to tell because as the sample size grows it will approach that of the total possible outcomes i.e 70% success. The appearance of a 90% chance in the sample will occur randomly and becomes less likely as the number of rolls increases.
If you are asking how many times you need to repeat a 70% success event until you have a greater than 90% chance of having at least once success then the answer is twice. The probability of getting a 30% chance event twice in a row is 9% so they have a 91% chance of succeeding if they can pick either of the two die after they roll.
Yeah, I'm not really sure what you mean. I have two possible interpretations. is it possible to get a 90% sample from a 70% population (not really) or how many times do you need to take a 70% chance to have at least a 90% chance of succeeding at least once ( twice).
If you are asking when the probability of choosing a success from a random sample that was taken from a set of possible outcomes that are 70% success will be equal to 90% then it's impossible to tell because as the sample size grows it will approach that of the total possible outcomes i.e 70% success. The appearance of a 90% chance in the sample will occur randomly and becomes less likely as the number of rolls increases.
If you are asking how many times you need to repeat a 70% success event until you have a greater than 90% chance of having at least once success then the answer is twice. The probability of getting a 30% chance event twice in a row is 9% so they have a 91% chance of succeeding if they can pick either of the two die after they roll.
Oh! Is that how Advantage calculation is done?
Pretty much
If you wanted to find the probability of a specific number of successes or failures instead of just 1 then you have to use the binomial distribution formula but when you only need to know if there is at least one success or at least one failure you can calculate the probability that they are all successes or all failures then what ever chance is left over is the chance of at least one.
Advantage is usually calculated as 1 minus the probability of failure multiplied by itself. This is based on some basic probability rules, to calculate the probability of two events occurring together you multiply them and to calculate the probability of the opposite occurring you subtract the probability from 1. So if you have a +4 and need to roll a 15 there are 10 numbers you can roll to fail {1,2,3,4,5,6,7,8,9,10} out of a possible 20 numbers so there is a 50% chance of failure. The probability of two failures is 0.5^2 which is also 0.25 or 25%. Then the opposite of that is not two failures or at least one success so 1-0.25 showing the probability of at least one success is 0.75 or 75%. For disadvantage you would do the same with success instead of failure.
Yeah, I'm not really sure what you mean. I have two possible interpretations. is it possible to get a 90% sample from a 70% population (not really) or how many times do you need to take a 70% chance to have at least a 90% chance of succeeding at least once ( twice).
If you are asking when the probability of choosing a success from a random sample that was taken from a set of possible outcomes that are 70% success will be equal to 90% then it's impossible to tell because as the sample size grows it will approach that of the total possible outcomes i.e 70% success. The appearance of a 90% chance in the sample will occur randomly and becomes less likely as the number of rolls increases.
If you are asking how many times you need to repeat a 70% success event until you have a greater than 90% chance of having at least once success then the answer is twice. The probability of getting a 30% chance event twice in a row is 9% so they have a 91% chance of succeeding if they can pick either of the two die after they roll.
Oh! Is that how Advantage calculation is done?
There was a video someone linked to here a while back done by a statistician. According to it, advantage is basically like getting a +3 in terms of the difference over thousands of rolls. Though that didn't account for DCs and trying to hit specific target numbers and a lot of other things that go into an in-game roll. But basically, the average roll of a d20 is 10.5, and with advantage, the average came up to a 13 and change.
I'm not sure if that was the question; it was just pretty interesting.
Yeah, I'm not really sure what you mean. I have two possible interpretations. is it possible to get a 90% sample from a 70% population (not really) or how many times do you need to take a 70% chance to have at least a 90% chance of succeeding at least once ( twice).
If you are asking when the probability of choosing a success from a random sample that was taken from a set of possible outcomes that are 70% success will be equal to 90% then it's impossible to tell because as the sample size grows it will approach that of the total possible outcomes i.e 70% success. The appearance of a 90% chance in the sample will occur randomly and becomes less likely as the number of rolls increases.
If you are asking how many times you need to repeat a 70% success event until you have a greater than 90% chance of having at least once success then the answer is twice. The probability of getting a 30% chance event twice in a row is 9% so they have a 91% chance of succeeding if they can pick either of the two die after they roll.
Oh! Is that how Advantage calculation is done?
There was a video someone linked to here a while back done by a statistician. According to it, advantage is basically like getting a +3 in terms of the difference over thousands of rolls. Though that didn't account for DCs and trying to hit specific target numbers and a lot of other things that go into an in-game roll. But basically, the average roll of a d20 is 10.5, and with advantage, the average came up to a 13 and change.
I'm not sure if that was the question; it was just pretty interesting.
The average I get for picking the highest of two dice is 13.825 which is about 3.325 higher than just one dice. It's a little bit more complicated as the effect of advantage and disadvantage depends on the probability of success or failure. Basically as the probability of success/ failure gets closer to 50% the effect of advantage or disadvantage gets higher. At 50% advantage increases the probability by 0.25 or the equivalent of a +5 while at 95% or 5% chance it increases the probability by 0.0475 or slightly less than a +1. So your actual mileage out of advantage may vary but if the probability of you succeeding or failing in any test is essentially random then advantage acts like a +3 on average.
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So the question basically is
1vs2 70 percent sucess they roll a 1d10 i take note if it was successful or not they do not know they only know they have 70 percent chance. They are allowed to repeat at a time sacrifice and add outcomes to a pool they can look over of 1 or 2 then they choose likely the most common number to appear. At what point of attempts would they have a 90 percent success rate from their initial 70 percent by looking at the numbers that came up?
I’m having trouble following your post. If you can clarify it then maybe I can help
Yeah, I'm not really sure what you mean. I have two possible interpretations. is it possible to get a 90% sample from a 70% population (not really) or how many times do you need to take a 70% chance to have at least a 90% chance of succeeding at least once ( twice).
If you are asking when the probability of choosing a success from a random sample that was taken from a set of possible outcomes that are 70% success will be equal to 90% then it's impossible to tell because as the sample size grows it will approach that of the total possible outcomes i.e 70% success. The appearance of a 90% chance in the sample will occur randomly and becomes less likely as the number of rolls increases.
If you are asking how many times you need to repeat a 70% success event until you have a greater than 90% chance of having at least once success then the answer is twice. The probability of getting a 30% chance event twice in a row is 9% so they have a 91% chance of succeeding if they can pick either of the two die after they roll.
Oh! Is that how Advantage calculation is done?
Pretty much
If you wanted to find the probability of a specific number of successes or failures instead of just 1 then you have to use the binomial distribution formula but when you only need to know if there is at least one success or at least one failure you can calculate the probability that they are all successes or all failures then what ever chance is left over is the chance of at least one.
Advantage is usually calculated as 1 minus the probability of failure multiplied by itself. This is based on some basic probability rules, to calculate the probability of two events occurring together you multiply them and to calculate the probability of the opposite occurring you subtract the probability from 1. So if you have a +4 and need to roll a 15 there are 10 numbers you can roll to fail {1,2,3,4,5,6,7,8,9,10} out of a possible 20 numbers so there is a 50% chance of failure. The probability of two failures is 0.5^2 which is also 0.25 or 25%. Then the opposite of that is not two failures or at least one success so 1-0.25 showing the probability of at least one success is 0.75 or 75%. For disadvantage you would do the same with success instead of failure.
There was a video someone linked to here a while back done by a statistician. According to it, advantage is basically like getting a +3 in terms of the difference over thousands of rolls. Though that didn't account for DCs and trying to hit specific target numbers and a lot of other things that go into an in-game roll. But basically, the average roll of a d20 is 10.5, and with advantage, the average came up to a 13 and change.
I'm not sure if that was the question; it was just pretty interesting.
The average I get for picking the highest of two dice is 13.825 which is about 3.325 higher than just one dice. It's a little bit more complicated as the effect of advantage and disadvantage depends on the probability of success or failure. Basically as the probability of success/ failure gets closer to 50% the effect of advantage or disadvantage gets higher. At 50% advantage increases the probability by 0.25 or the equivalent of a +5 while at 95% or 5% chance it increases the probability by 0.0475 or slightly less than a +1. So your actual mileage out of advantage may vary but if the probability of you succeeding or failing in any test is essentially random then advantage acts like a +3 on average.