Using Decision Trees to Solve Complex Problems: An Example

Topics: Decision theory, Decision tree, Influence diagram Pages: 5 (841 words) Published: December 3, 2012
Using Decision Trees to Solve Complex Problems: An Example

Consider the following problem

The Disney Corporation is thinking about taking its hit Broadway musical, The Lion King on tour. It has made the following calculations ▪ The tour would cost $12 million
▪ There is a .50 chance that it will earn $20 million and a .5 chance it will earn $10

Draw the decision tree for this problem using the following conventions. (Let a box indicate a decision node and a circle indicate a chance node).


Don’t Produce

$8 million
Pr =.50

-$2 million

EV = .5x8-.5x2 = $3.0 million

Since the EV from the “Produce” node is greater than the EV from the “Don’t Produce” node, they produce But now consider the following complication

▪ Disney can do a preview showing of the production at some cost $P ▪ If the audience is favorable, the show is certain to be a success ▪ If the audience is unfavorable, the show has only a .1 chance of success. ▪ The prior belief is that there is a .50 chance that the preview audience will react favorably.

To decide whether to do the preview, look at this decision tree. Note that if they choose not to preview, they are essentially in the previous situation, they go ahead with the production and expect to earn $3.0 million.

No Preview $3.0


Pr=.5 Not Produce


EV = .5x8+.5x0-PNot Produce-P
Pr =.5

ProducePr=.1 High

Pr =.9
EV =.8-1.8-P=-1-P-$2-P

▪ The “trick” is to work backwards towards the start of the tree. Doing this, we see that ▪ If you get a favorable review, you produce ($8-P>-P). ▪ If get an unfavorable review, you don’t produce (-1-POne Week) = .4 (40% of all workers miss more than one week).

If a worker is a smoker, we know that Pr (Absent > One Week given Smoker)=Pr (Absent and Smoker)/Pr(Smoker) = .3/.5=.6. (60% of all smokers miss more than one week).

Bayes Rule and the Value of Information

(Formalities and General Application)

▪ Let “A” be interpretted as the some event (say, a worker absent more than one week) and “S” be interpretted as another event (the worker being a smoker). ▪ We’ve already seen that

Pr (A given S) = Pr (A and S)/Pr (S)

▪ By the same reasoning we could have written

Pr (S given A) = Pr (A and S)/Pr(A)

▪ Rearranging this expression gives

Pr (A and S) = Pr (S given A)xPr(A)
▪ Substituting this into our first expression, we can write


▪ Think about what this expression is good for
▪ Pr(A) is your assessment of the prior probability (if you don’t know whether S is true, it is how likely you think it is the event A will occur) In the example, Pr (A) = .40 ▪
▪ The term in brackets tells you how to revise your assessment of the liklihood of event A given the new information ▪ In the example Pr(S given A)=.75 (75% of those absent more than one week are smokers). Pr (S) = .50 (50% of the workforce smokes). Thus, the prior probability gets multiplied by 75/50=1.5 to get the new revised probability.

Can information have some value even if it doesn’t raise the expected value?

• Modify the previous example by assuming that you are evaluating an investment that has the same basic structure as described above, except that it is a “one-shot” bet with the odds described above. o If you don’t have any added information, you will make $10 million with pr = .74 or lose $10 million with pr .26. o This has an EV =4.8, but it is also quite risky

Variance = .5(10-4.8)2+.5(-10-4.8)2=76.96

• If you make the investment, your outcomes can be represented with this tree



EV = $4.8

• Now suppose that for...
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