Monte Carlo/Crystal Ball question (not a technical question)

mach3

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Joined
Mar 24, 2002
Messages
245
Note: This was previously posted on the wrong board...


I've run 3 monte carlo simulations using Crystal Ball for a product in development. Each of the 3 simulations represents a different patient group that could potentially use this product.
Typically, I would run the results of these 3 simulations (mutually exclusive of each other) through yet another simulation (kind of like a simulation of the entire portfolio for this product).

Here is the atypical situation. Instead of running this last simulation of the 3 together, I'm going to add together the results for the best case revenues for the 3 simulations, the 3 base cases, and the 3 worst case revenues. What this 'roll-up' essentially does is make the best case extremely high and the worst case extremely low. It's really NOT a realistic scenario; it's what I call the Perfect Storm scenario. If you've seen the movie or read the book, you'll know what I mean.

My question is: is there a technical term for what I've done with the 3 scenarios? I'm sure there is a statistician out there who can tell me what to call my 'perfect storm' scenario. I need to put in on a slide for presentation purposes. Thanks!




Update: Since my original post, I've gotten some more insight. In my original 3 scenarios, what I called the 'best case' above was really the 90%-tile case, and the 'worst case' was the 10%-tile case; 'base' was 50%. When I added the 10%-tile stream of revenues for all 3 cases to create this perfect storm scenario, someone pointed out that this is NO LONGER a 10% case. Since each patient group is mutually independent of each other, what I've really created is a 10% * 10% * 10% = 0.1% case!!! In other words, the chance of the best case (or worst case) actually happening or exceeding is NOT 10%, it's really 0.1%.
 

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It's hard to know for sure without really understanding what the analysis you're doing is, but be careful with adding expected values like you seem to be doing...

For example, the probability of rolling a 6 on a die is 1/6. The probability of rolling a 6 on each of 2 dice is 1/36. It's not 1/6 + 1/6 = 2/6. Your "Update" section hits on this idea.

What is it that you're really trying to calculate? I'm struggling with your claim that the three patient groups are independent while at the same time wanting to combine their simulations... how are the patient groups defined?
 
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