Let's see if I can sort this out:
When rolling two six sided dice you have to roll 18 times to
have a 50% chance to get a twelve. (right?)
No. The probability of rolling a 12 is 1/36, which means the probability of not rolling a 12 is 35/36. The probability of not rolling a 12 in n trials is (35/36)^n, so the probability of rolling at least one 12 in n trials is 1-(35/36)^n. To get this above 50%, we need at least 25 rolls.
Now assume the year has six days then there is a total of 21
( (5*6/2)+6 ) ways to combine two days since the combination 1 and 2 is the same as 2 and 1.
Yes, this is the same as saying the number of ways to choose 2 days out of 6 where order does no matter is to calculate 6c2 (6 choose 2) and add 6 more combinations to cover all matched pairs (11, 22, etc...). However, it's not relevant to the question.
Out of these 21 combinations six are the same day twice (11 22 33 44 55 66) this means you would need to pick 21/6=3,5 to have a 50% chance to get at least one?
The probability of picking one pair out of 21 combinations is 6/21, the probability of not picking a pair is 15/21, so the probability of picking at least one pair in n trials is 1-(15/21)^n. To get this above 50% we need n to be at least 3.
Back to orignal numbers:
If the dice has 366 sides. You would have to roll them 66978 times to have 50% chance of getting 732. (right?)
No. 732 is the maximum roll with two 366-sided dice. The probability of rolling a 732 is 1/366^2, the probability of not rolling a 732 is 1-1/366^2, and the probability of rolling at least one 732 in n trials is 1-(1-1/366^2)^n. To get this above 50%, we need n to be at least 92851.
There are 366 results with the same day twice out of 67161 ((366*365)/2 +366) you need then 67161 /366 =183,5 to get 50%
and now magicaly I'm back at (19*20)/2=190
First, using the same procedure as above, the number of trials required to bring the probability of picking at least one of the 366 pairs out of 67161 possibilities to above 50% is 127. But I think I'm seeing the problem with your reasoning more clearly now.
That is 20 people to get 50%.
And As I have said I KNOW this is wrong. But why?
Please I'm going Mad And since I'm only examining pairs of dates 3 or more matching dates is not an issue.
You are trying to find the probability of picking two random people and seeing if they have the same birthday, then doing that again and again until you find them, just like rolling dice. That is not the question we are trying to answer. We are trying to find the probability that any two people will have the have birthday, and we compare all birthday dates simultaneously (that is, no specific pair is required to have the same birthday, any pair or more will do). This is another application of the Pigeon Hole principle. Imagine a calendar; as people mark their birthdays on the calendar, there are fewer spots available for the next person to plot a clean day because more and more spots are taken. This makes it more and more likely that two people will have the same birthday as the number of people increases, until you get to 367 people at which point it is guaranteed
that two people's birthdays will coincide.
I believe you estimation technique of dividing the number of possibilities by the number of desired outcomes to find the number of trials required for 50% success was misleading you, as it's just plain wrong but it's resoponsible for giving you numbers closer to the real answer then you would otherwise have.
Does that help?