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Not really relevant to the point of the article, but worth mentioning: The article's commentary on the _Asian Disease_ problem which states "the number of cert

by ivarv 13y ago

Not really relevant to the point of the article, but worth mentioning: The article's commentary on the _Asian Disease_ problem which states "the number of certain deaths is the same in both versions of the problem.." is incorrect. In case A, the number of certain deaths is 400,000, in case B it's 0. Or am I reading it incorrectly ?

emiliobumachar 13y ago
You are correct, it's the expected number that is the same.
- spuz 13y ago
  As I understand it, there is not enough information to determine the expected number of deaths. In the first situation, the expected number of deaths of choosing medicine A must be greater than or equal to 400k and in the second situation (again choosing medicine A) the expected number of deaths must be less than or equal to 400k. My interpretation was that in the first situation, there is a possibility that everyone dies if you choose medicine A, therefore B was the right choice. Similarly, in the second situation, there is a possibility that everyone lives if you choose medicine A, therefore A was the right choice. Maybe I'm misunderstanding but there does not appear to be enough information in the question to determine the exact probabilities.
  - cortesoft 13y ago
    I think you are misunderstanding. Medicine A will always save 200,000 people, with 400,000 dying. Medicine B will save all 600,000 people 33.3% of the time, but everyone will die 66% of the time. The expected number of deaths (the 'expected value' in game theory) is the same for both cases (200,000). Medicine A is easy to calculate (200,000 * 1.00 = 200,000). For Medicine B, the calculation is 600,000 * 0.33 + 0 * 0.66 = 200,000. The idea is that the average number of deaths is the same for both cases; you are just trading the certainty of Medicine A (where you know you will save exactly 200,000) vs the chance to save everyone, but the risk of losing everyone.
    - spuz 13y ago
      I understand that the wording and the numbers chosen by the experimenters are intended to create two equivelent sitations, but the fact is, they do not. The sentence: "If you choose Medicine A, 400,000 people will die." Tells you that 400k people will die if you choose medicine A. It says nothing about what will happen to the remaining 200k. I believe it is illogical to assume that the sentence implies the remaining 200k live. Note that if you were to ask me in real life, I'd probably be happy to make that assumption but that doesn't change the fact that it is an assumption based on missing information.
      - cortesoft 13y ago
        I understand your point in terms of formal logic (that saying 400k people will die does not necessitate that the other 200k can't also die). However, the initial sentence that starts the experiment is "Recently, a dangerous new disease has been going around. Without medicine, 600,000 people will die from it. In order to save these people, two types of medicine are being made." When you follow that by saying "If you choose Medicine A, 400,000 people will die", it is not illogical to assume that they don't mean "oh yeah, the other 200k will also die, too" This isn't a Mitch Hedberg joke. In these scenarios, we are expected to make many basic assumptions that aren't included in the scenario. We are safe to assume that there isn't an unmentioned side effect of Medicine B that will leave everyone paralyzed, or that Medicine A won't turn the survivors into zombies.
        bittercynic 13y ago
        Are you referring to this? "I used to do drugs. I still do, but I used to, too." (It's a lot funnier when delivered with good timing than when you read it.) His comedy was very hit-and-miss, but some of it was spectacularly funny.
        cortesoft 13y ago
        Yes, yes I was.
      - 13y ago
  - logfromblammo 13y ago
    A: 100% probability of 200k cures; expected value = 200k survivors B: 1/3 probability of 600k cures, 2/3 probability of 0 cures; expected value = 200k survivors Of course, as software writers, we have probably been trained to analyze algorithms for expected performance and for the best and worst cases. Real-time programmers should always choose option A, as it has the best worst-case performance. Some programmers will choose A because it has lower computational complexity. Others will choose option B because it has better best-case performance, and in the worst case is no worse than doing nothing at all. The scenario is supposed to be mathematically ambiguous, but the people who designed it obviously did not consider factors beyond just the average case and whether the numbers are framed positively or negatively.
darnton 13y ago
The way I read it was that "both versions" refers to the two tellings of the problem, one in terms of deaths the other in terms of survivors, not the two different medicines.
losvedir 13y ago
> Or am I reading it incorrectly ? You're reading it incorrectly. It's saying that the number of certain deaths is the same, whether it's phrased as "400,000 deaths" or "saving 200,000 people".

by ivarv 13y ago

emiliobumachar 13y ago

You are correct, it's the expected number that is the same.

spuz 13y ago
As I understand it, there is not enough information to determine the expected number of deaths. In the first situation, the expected number of deaths of choosing medicine A must be greater than or equal to 400k and in the second situation (again choosing medicine A) the expected number of deaths must be less than or equal to 400k. My interpretation was that in the first situation, there is a possibility that everyone dies if you choose medicine A, therefore B was the right choice. Similarly, in the second situation, there is a possibility that everyone lives if you choose medicine A, therefore A was the right choice. Maybe I'm misunderstanding but there does not appear to be enough information in the question to determine the exact probabilities.
- cortesoft 13y ago
  I think you are misunderstanding. Medicine A will always save 200,000 people, with 400,000 dying. Medicine B will save all 600,000 people 33.3% of the time, but everyone will die 66% of the time. The expected number of deaths (the 'expected value' in game theory) is the same for both cases (200,000). Medicine A is easy to calculate (200,000 * 1.00 = 200,000). For Medicine B, the calculation is 600,000 * 0.33 + 0 * 0.66 = 200,000. The idea is that the average number of deaths is the same for both cases; you are just trading the certainty of Medicine A (where you know you will save exactly 200,000) vs the chance to save everyone, but the risk of losing everyone.
  - spuz 13y ago
    I understand that the wording and the numbers chosen by the experimenters are intended to create two equivelent sitations, but the fact is, they do not. The sentence: "If you choose Medicine A, 400,000 people will die." Tells you that 400k people will die if you choose medicine A. It says nothing about what will happen to the remaining 200k. I believe it is illogical to assume that the sentence implies the remaining 200k live. Note that if you were to ask me in real life, I'd probably be happy to make that assumption but that doesn't change the fact that it is an assumption based on missing information.
    - cortesoft 13y ago
      I understand your point in terms of formal logic (that saying 400k people will die does not necessitate that the other 200k can't also die). However, the initial sentence that starts the experiment is "Recently, a dangerous new disease has been going around. Without medicine, 600,000 people will die from it. In order to save these people, two types of medicine are being made." When you follow that by saying "If you choose Medicine A, 400,000 people will die", it is not illogical to assume that they don't mean "oh yeah, the other 200k will also die, too" This isn't a Mitch Hedberg joke. In these scenarios, we are expected to make many basic assumptions that aren't included in the scenario. We are safe to assume that there isn't an unmentioned side effect of Medicine B that will leave everyone paralyzed, or that Medicine A won't turn the survivors into zombies.
      - bittercynic 13y ago
        Are you referring to this? "I used to do drugs. I still do, but I used to, too." (It's a lot funnier when delivered with good timing than when you read it.) His comedy was very hit-and-miss, but some of it was spectacularly funny.
        cortesoft 13y ago
        Yes, yes I was.
    - 13y ago
- logfromblammo 13y ago
  A: 100% probability of 200k cures; expected value = 200k survivors B: 1/3 probability of 600k cures, 2/3 probability of 0 cures; expected value = 200k survivors Of course, as software writers, we have probably been trained to analyze algorithms for expected performance and for the best and worst cases. Real-time programmers should always choose option A, as it has the best worst-case performance. Some programmers will choose A because it has lower computational complexity. Others will choose option B because it has better best-case performance, and in the worst case is no worse than doing nothing at all. The scenario is supposed to be mathematically ambiguous, but the people who designed it obviously did not consider factors beyond just the average case and whether the numbers are framed positively or negatively.

darnton 13y ago

The way I read it was that "both versions" refers to the two tellings of the problem, one in terms of deaths the other in terms of survivors, not the two different medicines.

losvedir 13y ago

> Or am I reading it incorrectly ? You're reading it incorrectly. It's saying that the number of certain deaths is the same, whether it's phrased as "400,000 deaths" or "saving 200,000 people".