# How do I calculate the probabilities in this decision making scenario?

I have the following problem: A decision has to be made. Paradoxically, there is value in only being marginally correct in making this decision, say I want to be correct 60 percent of the time. That is $p(\text{adjusted correct})=0.6$. I know from experience that I'm actually able to make the decision correctly 90 percent of the time, say $p(\text{I'm correct})=0.9$. After making my decision, I can ask someone else to make the decision instead. This person is very naive, and is only able to make the correct decision 10 percent of the time, that is $p(\text{Alternative opinion correct})=0.1$. What percentage of decisions should I let the naive person make?

My solution:

Let $x$ be the proportion of decision made by the naive person.

$0.1x+0.9(1-x)=0.6$

$x=0.375$

Is this correct? If the two decisions are not independent of one another how will this change? That is, what if the accuracy of the naive person's choice somehow depends on the choice made by myself? Say if I make decision A, then the naive person is $30$ percent correct in their choice, if I make decision B, then the naive person is $20$ correct in their choice, and if I make decision C, then the naive person is $10$ correct in their choice.

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In the latter case we'd need to know what percent of times A,B and C are the right decisions, I think. – curious_cat Mar 5 '13 at 17:58
This is just an abstraction of a larger problem I have. For this example, any probabilities can be used as needed. The true probabilities I'll have to calculate as makimum likehood estimates or laplace estimates from historic data. – entropy Mar 6 '13 at 7:16
Why should you let the naive person make any decisions at all? I suspect your effort to abstract the larger problem may have lost some of its important characteristics. – whuber Jun 12 '13 at 20:03

Let's say there are only two choices A and B. If you make decision A, then the naive person is 30 percent correct in their choice; if you make decision B, then the naive person is 20 correct in their choice. Historically let 60% times A be the right answer. You making decision A happens $0.6 * 0.9 + 0.4 * 0.1$. Hence the other person being correct would be $(0.6 * 0.9 + 0.4 * 0.1)*0.3 + (1-0.6 * 0.9 - 0.4 * 0.1)*0.2$.

Then solve for x as before?

Just my 2 cents.

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From probability theory:

1. $P(T|You) = P(T \& You) / P(You)$

2. $P(T|Them) = P(T \& Them) / P(Them)$

3. $P(T) = P(T|You)P(You) + P(T|Them)P(Them)$

1. $P(You) + P(Them) = 1$

The $P(You)$ is the probability that you're answering the question and $P(Them)$ is the probability that they are answering the question, and P(T) is the marginal of the true answer. To answer both your questions:

$P(T) = P(T|You)(1 - P(Them)) + P(T|Them)P(Them)$ $= P(T|You) + P(Them)[P(T|Them) - P(T|You)]$

thus

$P(Them) = [P(T) - P(T|You)] / [P(T|Them) - P(T|You)]$

$P(Them) = [0.6 - 0.9] / [0.1-0.9] = 0.375$

So yes your calculation is correct. In the event that their choice is sequentially dependent on yours then:

$P(T|Them,X) = P(T \& Them | X) / P(Them | X)$

where $P(X)$ is the probability of you making some decision.

Hope that helps.

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