*** Probabilty Conundrum *** Advice wanted
Brad Thompson
Thanks for reading. Can anyone solve this conundrum please?
The directors of an oil company have requested yr assistance in determining
the *maximum* cash bid that should be made to obtain an exploration permit.
Two prospects have been identified and will be drilled one after the other.
They have colected/estimated the following data by evaluating the prospects
*independently*.
Cost of drilling PV of cash flows Prob. of Success
if Successful
Prospect A $10m $30m 0.32
" B $10m $40m 0.32
The directors understand how a geologist arrived at the probability
estimate. through consideration of the probty of success of each of the 3
main attributes of oil discovery.
Overall Prob = Trap(0.8) x Reservoir(0.8) x Source(0.5) = 0.32
What they don't understand is how to deal with the geologist's comment that a
successful first well would prove the presence of a source, and that this
would alter the ex-post probability estimates of the 2nd prospect as follows:
First Well Outcome Source Prob. Overall Prob.
Success 0.8 0.512
Failure 0.5 0.32
Assist the directors in their determination and indicate:
i) which prospect should be drilled first
ii) whether the 2nd should be drilled if the 1st was a failure.
This is then evaluated using a probability decision tree:
Solution:
Prospect A would be rejected
NPVA = $30m x 0.32 - $10m = -$0.4m
Prospect B would be accepted
NPVB = $40m x 0.32 - $10m = $2.8m
If prospect B was successful, then the probabilities on prospect A would
change, thus its value would be:
NPV = $30m x 0.512 - $10m = $5.36
which would happen with a 0.32 probability, thus the 'expected' value of
prospect B would be: $5.36 x 0.32 = $1.715m
The total value of the 'permit' containing the two prospects is therefore:
$2.8m + $1.715m = $4.515m
However, if these were drilled in the ***reverse order*** , the value would
be $4.858m as follows:
[($40m x 0.512 - $10m) + 30m] x 0.32 + 0.68 x [$40m x 0.32 -$10m]
-$10m = $4.858
Logic says that you would drill the more attractive 'prospect B' first, but
the above suggests otherwise. Where is the flaw?
Cheers
-----------------------------------------------------------------------
"Merde!" Count Cambronne's (General of the Old Guard Chasseurs)
reply to the Allies' offer of surrender at Waterloo. Direct & succinct.
Brad Thompson | Internet: bra...@csuvax1.murdoch.edu.au
Finance | Tel +61 09 360 6275 Fax +61 09 310 5004
Murdoch University, Perth, West Australia 6150
T. Scott Thompson
>Hi everyone,
>Thanks for reading. Can anyone solve this conundrum please?
>The directors of an oil company have requested yr assistance in determining
>the *maximum* cash bid that should be made to obtain an exploration permit.
>Two prospects have been identified and will be drilled one after the other.
>They have colected/estimated the following data by evaluating the prospects
>*independently*.
> Cost of drilling PV of cash flows Prob. of Success
> if Successful
>Prospect A $10m $30m 0.32
> " B $10m $40m 0.32
>
> The directors understand how a geologist arrived at the probability
> estimate. through consideration of the probty of success of each of the 3
> main attributes of oil discovery.
>Overall Prob = Trap(0.8) x Reservoir(0.8) x Source(0.5) = 0.32
We really have to consider the joint distribution of 6 variables:
Trap, Reservoir and Source at site A, and Trap, Reservoir, and Source
at site B. (This means that there are 2^6 = 64 possible "states of
nature".) You seem to treat the two sites symmetrically, and also
assume that the Trap, Reservoir and Source variables are independent
at each site. This does not completely determine the distribution,
however. I assume (because I think this is what you probably
intended) that all of these variables are jointly independent, except
for the two Source variables, which are dependent on each other, but
independent of the other variables. Without some assumption like this
the problem is not well posed.
>What they don't understand is how to deal with the geologist's comment that a
>successful first well would prove the presence of a source, and that this
>would alter the ex-post probability estimates of the 2nd prospect as follows:
> First Well Outcome Source Prob. Overall Prob.
> Success 0.8 0.512
> Failure 0.5 0.32
These numbers don't make sense. A Failure at the first site should be
informative about the Source prob. at site 2, but the second row of
your table indicates no change in the probabilities ex post when a
Failure occurs at the first site.
It's not too hard to work out what the correct numbers should be.
Since we are apparently treating the two prospects symmetrically (a
priori) except for their payoffs, the .5 prior probabilities of a
success at each site are consistent with the .8 posterior probability
given success at the other site only if the prior joint distribution
of sources is the following:
Source at B No Source at B
Source at A .4 .1
No Source at A .1 .4
This follows since
Pr(Source at A and B) = Pr(Source at B|Source at A)
* Pr(Source at A)
= .8 * .5 = .4
and from the fact that each row and column must sum to .5.
Now Pr(Source at B | Failure at A)
= Pr(Source at B|Source at A) *Pr(Source at A|Failure at A)
+ Pr(Source at B|no Source at A)*Pr(no Source at A|Failure at A)
= .4 * Pr(Source at A and Failure at A)/Pr(Failure at A)
+ .1 * Pr(no Source at A and Failure at A)/Pr(Failure at A)
= .4 * (1-(.8*.8))/(1-.32)
+ .1 * (1-.5)/(1-.32)
= 0.285294118
This is considerably lower than the figure of .5 you report above.
Multiplying this twice by .8 yields the overall ex post probability
conditional on an initial Failure. The second line of your table
should read:
Failure 0.285294118 0.182588235
rather than
> Failure 0.5 0.32
The distinction matters below.
>Assist the directors in their determination and indicate:
> i) which prospect should be drilled first
> ii) whether the 2nd should be drilled if the 1st was a failure.
>
>This is then evaluated using a probability decision tree:
>Solution:
>Prospect A would be rejected
> NPVA = $30m x 0.32 - $10m = -$0.4m
>Prospect B would be accepted
> NPVB = $40m x 0.32 - $10m = $2.8m
>If prospect B was successful, then the probabilities on prospect A would
>change, thus its value would be:
> NPV = $30m x 0.512 - $10m = $5.36
>which would happen with a 0.32 probability, thus the 'expected' value of
>prospect B would be: $5.36 x 0.32 = $1.715m
>The total value of the 'permit' containing the two prospects is therefore:
> $2.8m + $1.715m = $4.515m
>
>
>However, if these were drilled in the ***reverse order*** , the value would
>be $4.858m as follows:
> [($40m x 0.512 - $10m) + 30m] x 0.32 + 0.68 x [$40m x 0.32 -$10m]
> -$10m = $4.858
>
>Logic says that you would drill the more attractive 'prospect B' first, but
>the above suggests otherwise. Where is the flaw?
The flaw is that your reverse calculation includes the term
0.68 x [$40m x 0.32 -$10m]
This is presumably the probability of a failure at site A times the
expected payoff from site B. This term (which was not present in the
first scenario) gives the present value of the second project given
failure of the first. However, you have not correctly reduced the ex
post probability of success at B given failure at A, and are using the
ex ante probability of succes instead. The corrected expression is
0.68 x [$40m x 0.182588235 -$10m] = 0.68 x (-2.696470588)
As you can see, this is negative. This implies that one would not
drill at B at all given a failure at A, and therefore the entire term
should not appear in the present value calculation. When this term is
dropped it is seen that the present value of the "reverse" plan is:
[($40m x 0.512 - $10m) + 30m] x 0.32 -$10m = $2.9536m
and this is indeed less than the present value associated with
drilling at B first.
Aside: The "reverse" plan is not only less profitable on average, but
also has a must higher variance. With this plan one expects to lose
money initially, but is willing to take the chance because there is a
large expected payoff associated with the second well if we are lucky
on the first well.
--
T. Scott Thompson email: thom...@atlas.socsci.umn.edu
Department of Economics phone: (612) 625-0119
University of Minnesota fax: (612) 624-0209