I think I see what you're doing.In post 3517, lilith2013 wrote:it doesn’t really work the same because there’s only two coin flips. let’s say there are 3 coin flips.In post 3514, Something_Smart wrote:Can you rephrase 3502 using "exactly" where you intend to have it so I know what you're arguing?In post 3510, lilith2013 wrote:i don’t think you’re looking at this right? for the coin flip scenario, it’s the probability of getting exactly 2 heads, not at least one?
the probability of having 2 or more coins flip heads, given that one has already flipped heads, is basically equivalent to ignoring the first coin flip and calculating the probability of one or more heads given that we are now flipping the coin twice. the previous flip has no bearing on the probability of the two other flips. the probability that i am a K, given that skitter is also a K, is equivalent to the probability that i am a K given that skitter is literally any role - the probabilities are independent
For the coin example, you're assuming that we have flipped one coin and observed it to be heads, then we flipped the others. This gives different results from just assuming "one of them is heads."
As an example, suppose I have a red coin and a blue coin. If I flip them both, then the probability of getting two heads given that one coin was heads is 1/3. But the probability of getting two heads given that the RED coin was heads are 1/2. It matters whether or not you name the coin that was heads.
And in this case, we have no way of naming WHICH letter skitter got (as in, the first, second, third, etc.). Therefore, the probabilities aren't independent.