23 replies to this topic

[mmiguel]

ah! haha
i suppose i should spend more time reading the prompt then, i guess i was solving the wrong problem

well... i guess my interpretation is not inconsistent with original problem statement, if 1-on-1 is interpreted as a one-to-one mapping i.e. a pair.

"Assume all encounters are 1-on-1 and random."

i can definitely see how one would see this and assume that in any "round" only 2 people are selected from the population, and said to have "met".


oh well, twas a good exercise of thought.

[Rob_Gandy]

well... i guess my interpretation is not inconsistent with original problem statement, if 1-on-1 is interpreted as a one-to-one mapping i.e. a pair.

"Assume all encounters are 1-on-1 and random."

i can definitely see how one would see this and assume that in any "round" only 2 people are selected from the population, and said to have "met".


oh well, twas a good exercise of thought.

Looking back at your answer posts

Spoiler for I also came up with

(n-k)/n for for chance of survival in a given round. k being number of killers and n being number of outcomes affecting survival.

I always enjoy seeing the way people work out problems. That's one of the great things about being a teacher, seeing how people understand.

[bonanova]

Spoiler for

can you share the derivation of P/(K+1)?

Spoiler for sure

I computed the encounter probabilities {kk kp pk pp} starting with K killers and P pacifists and noticed their form. Then I worked out K (even) killer, 1 pacifist case which gave me 1/(K+1). Knowing this is also the odd-K survival probability for killers, I devised the OP.

Tiring of the probability trees I simulated the general process, applying the encounter probabilities with their effects: kk-> reduce K by 2; kp or pk-> reduce P by 1; pp-> do nothing. I verified 1/(K+1) for the 1 pacifist case. Finally I verified (by simulation) that P proportionality holds. For wide ranges of K and P the result was always P/(K+1).

I simulated successive random encounters; not "rounds of encounters." Killers and pacifists were not individually tracked. At each stage, values of K and P yielded the probabilities of encounter types, and a RNG selected among them based on those probabilities. K then decreased by 2 for kk encounters; P decreased by 1 for kp or pk encounters. The values of P when K hit zero were averaged, over 100,000 trials, to get the expected number of survivors. Killers never survived.

I believe it can be derived; but, no, I didn't.

[joef1000]

Spoiler for

Whatever you are, you should hide in your house as long as you can. Eventually, the 10 killers that already live there eventually wipe each other out and if you are a killer, you are the last one left and can kill all the pacifists, and if you were a pacifist, you are safe because there are no killers left. The twist here is that anyone else coming into the town can register as a killer. This means you should always stay within earshot of the entrance, but not where you can be seen there. That way, if you ever here gunfire you can hide immediately and wait for another killer to come to town and wipe the other killer out. Another twist is that everyone might employ this strategy and hide in their houses. Then it will sound safe to come out, everyone might and the killers will kill everyone! A third twist is that the problem stated a killer must kill anyone they see, but if a killer killed another killer from behind, they do not die. Also, if it is part of your job you will have to go to work and not be able to hide forever. This means the best strategy could be to be a pacifist, and always travel with other pacifists to and from work, and hide at home whenever possible.


I think the best strategy is just to quit your job and find a safer one. Maybe sue your company for sending their workers top such a dangerous place!