So I want to create an app that would be used as a helping tool in an indie card game called the resistance. For this I need to make sure my math is good in order for me to start my project. A use case example would be as follows:
Out of a group of 10 people there are 6 good guys and 4 bad guys. As per the rules I want to have 3 people out of that entire group of 10 to go on a in-game mission, I want to know my probability of any randomly selected 3 people that they contain a bad guy.
So based on my understanding, I calculate that in that group of 3 (out of a total of 10) I have a 40% chance of choosing just one bad guy, 16% of choosing exactly 2 bad guys and 0,64% of all three of them being a bad guy. In other words there is a total of 62,4% chance of having at least 1 or more bad guys in that team.
So as per the rules of the game, the team of 3 goes on a in-game mission and the results (secretly) come in that there has been 1 out of 3 possible sabotages with 100% certainty that it was caused by a bad guy and I don't know who it was. So now I have a 33% certainty that one of those 3 is surely a bad guy.
Now, by the rules of the game I have to send 4 people on the next mission. Since I do not wish the next mission to fail, I know that I cannot take all 3 people from the first mission and include a fourth player because it will surely fail again.
I want to take into consideration the following continued scenario: Let's assume with absolute certainty that the first mission that failed actually had only 1 bad guy. That would mean that in first sub group of 3 people there is 1 bad guy and in the other subgroup (that haven't went on the mission) there are 3 out of 7 bad guys, because it totals out to 10 people.
I want to consider taking just 1 guy from the first mission and then 3 guys from the group of 7 that haven't went but I know that the group of 7 now contain 3 bad guys. This is where I get lost in calculating.
I don't know if it is better to actually take 2 guys from the first subgroup of 3 where I know there is only 1 bad guy and then 2 guys from the group of 7 where I know it has 3 bad guys.
Can I get some input on this please? I hope I have made this question clear.
Some formulas and thoughts would come in much appreciated on the exact scenarios.
$$P(B) = 1- P(\text{not }B) = 1- \frac{6\times 5\times 4}{10 \times 9 \times 8} = 1- \frac{120}{720} = 1-0.167=0.833$$
So you have an 83.3% chance of choosing at least one bad guy. As the first guy chosen has a 6/10 chance of being good, the second 5/9, and the last 4/8.
$$P(B) = 1 - P(\text{not }B) = 1 - P(1\text{ is good}, 2\text{ is good},3 \text{ is good})$$
Scenario 1: None from first group, all from second group
$$P(B) = 1- P(\text{not }B) = 1-\frac{5 \times 4 \times 3}{7 \times 6 \times 5} = 1-\frac{60}{210}=0.714$$
Scenario 2: 1 from first group, 2 from second group
$$P(B) = 1- P(\text{not }B) = 1-\frac{2 \times 5 \times 4}{3 \times 7 \times 6} = 1-\frac{40}{126} = 0.682$$
Scenario 3: 2 from first group, 1 from second group
$$P(B) = 1- P(\text{not }B) = 1-\frac{2 \times 1 \times 5}{3 \times 2 \times 7} = 1-\frac{10}{42} =0.761$$
Scenario 4: 3 from first group, 0 from second group
$$P(B) = 1$$
Since there's at least 1 bad guy in the first group.
So the optimal choice is to take 1 from the first group, 2 from the second, as it provides the lowest probability of getting a bad guy.
I hope that makes sense to you.