Will we get any compensation for not working referrals rewards for missions?
I done over 82 missions (just since when I started to count it) with total over 400% together and got no referral.
While I do agree with you about compensation, you are aware that thats not how % works, right? 5% + 5% is not 10%, its still 5%.
Yes, I am aware of how propability works. I mentioned the sum because It gives together with missions count average chance per mission from which is possible to calculate approximate chance to get a referral per all done missions what is cca 98%. It's more precisely to use exact number of missions with its chances, but I didn't want to post here full list. The average value is enough.
400/82=4.87%
1-((1-.0487)^82)=0.98
5% + 5% isn't 5% but 9.75%.
Counted by the Complement Rule for Probability:
1-((1-.05)^2)=0.0975
This is valid for case of calculating getting at least one referral for all missions done.
PS: Even if I done 1000 missions then proprability will never be 100% but just very close to that. So I think if we will be compensated then it should be based on missions count and sum of their chances to get a referral.
If I am wrong, tell me how it should be right If you roll a dice its 16.66% chance to roll a 6. If you roll it again there is still a 16.66% to roll a 6. What I mean is that the chance of rolling a 6 doesnt increase every time you roll. However, the probability to roll a 6 at least one of those times is another thing! Roll the dice twice is 36 outcomes, 25 does not involve a 6, so the probability to roll 6 once would be 11/36 (30.56% probability chance).
Now, if the chance is 5% for the mission, you need basicly need to roll a 1 on a 20 sided dice (or 1-5 on a 100 sided dice), its going to be 5% chance every time you roll, but the probability for 3 missions with 5% to get a referal would be 1149/8000 (20^3 outcomes, 19^3 does not involve a 1-roll) making your probability chance 14.36%.
Now, its to early in the morning to do the math for a full set of complete missions. My point is, there is a difference in % chance and % probability chance
You are right (almost, you have there just small typo: 20^3-19^3=1141 not 1149, so then 14.26% not 14.36%),
but it doesn't disproves my calculations. I just used other way how to calculate it.
A proof based on your examples:
2 rolls of a 6 sided dice, at least one 6.
1-((1-0.1666666...)^2)≐0.3056=
30.56%3 rolls of a 20 sided dice, at least one referral:
1-((1-0.05)^3)≐0.1426=
14.26%Same results calculated by alternative way.