July 2007 Puzzle Solutions

Join Our Community of Science Lovers!

Solutions:

1. Let's consider the world from the point of view of some immortal X. We will prove the following statement:

Regardless of how many other immortals N there are, if their total power is Tot then X's chances of winning are P(X)/(P(X) + Tot) regardless of the order of duels.


On supporting science journalism

If you're enjoying this article, consider supporting our award-winning journalism by subscribing. By purchasing a subscription you are helping to ensure the future of impactful stories about the discoveries and ideas shaping our world today.


The statement is clearly true if there is just one other immortal (i.e. N =1). We will prove that if it is true for all immortals up to N = k, then it is true for all up to N = k+1. In general, before the last fight, X has faced some set J of immortals either directly or indirectly (i.e., among those J, some have killed others). At that point X has some power JPower which is the sum of P(X) plus the original power of those J other immortals. The remaining set have power [Tot + P(X) – Jpower]. Therefore X's probability of winning is (JPower/(JPower + Tot + P(X) - JPower) = JPower/(Tot + P(X)). Given this statement, order doesn't matter. Aggression doesn't contribute to overall safety.

2. The probabilities are such that the initiator and the first non-initiator have the same probability 0.36(0.5 * 0.8 * 0.9), the second non-initiator has probability 0.18 of surviving and the last non-initiator has probability 0.10.

3. The initiator should start from the least powerful and go to the most powerful. If he does that, his first chance of winning will be

(1/2) against non-initiator having power 1,
(1/2) against non-initiator having power 2,
(1/2) against non-initiator having power 4,
with a resulting probability for surviving all the duels of 1/8.

If the initiator instead starts with the non-initiator having power 4, his chances of winning that duel alone are only 1/16. The odds are less bad but in the same discouraging spirit if the initiator starts with the middle non-initiator.

4. It is better for the initiators to start fighting one another than for each initiator to start fighting a non-initiator. In the former case the probabilities are 0.36 and 0.36 for the two initiators, and (0.18)[(1/5) * (9/10)] and 0.10 for the first and second non-initiator, respectively. If each initiator starts to fight a non-initiator, the probabilities are 0.25 for each of the four immortals.

5. If you have a choice between weaker initiators and stronger ones, then choose weaker ones. If all initiators have the same initial power and work just as fast as one another, then it is better for them to face one another first.

Ivan Rezanka helped greatly with all aspects of this puzzle.

It’s Time to Stand Up for Science

If you enjoyed this article, I’d like to ask for your support. Scientific American has served as an advocate for science and industry for 180 years, and right now may be the most critical moment in that two-century history.

I’ve been a Scientific American subscriber since I was 12 years old, and it helped shape the way I look at the world. SciAm always educates and delights me, and inspires a sense of awe for our vast, beautiful universe. I hope it does that for you, too.

If you subscribe to Scientific American, you help ensure that our coverage is centered on meaningful research and discovery; that we have the resources to report on the decisions that threaten labs across the U.S.; and that we support both budding and working scientists at a time when the value of science itself too often goes unrecognized.

In return, you get essential news, captivating podcasts, brilliant infographics, can't-miss newsletters, must-watch videos, challenging games, and the science world's best writing and reporting. You can even gift someone a subscription.

There has never been a more important time for us to stand up and show why science matters. I hope you’ll support us in that mission.

Thank you,

David M. Ewalt, Editor in Chief, Scientific American

Subscribe