## Friday, August 28, 2009

### Mathematical explanation of entrepreneurial bug

Not all things in life are quantifiable (you know, things like justice, love, or your experience of the color purple), but the proverbial "entrepreneurial bug" is not one of them.

As a corollary to a well-known math problem called the two-envelope paradox, it is perfectly rational for an individual to want to start one company after another and expect a higher reward after each iteration, even if the expected outcome associated with each of these endeavors is completely random (as I would argue is the case every time you start a new company)! To understand that mathematically, this is how the two-envelope paradox works (taken from the latest analysis of this paradox on PhysOrg.com):

In the two-envelope paradox, a player must choose between two envelopes, one of which contains twice as much as the other. The player can open the envelope they choose, and then they have the option of switching envelopes. The other envelope, of course, has either twice the money or half the money as the first envelope, but the player does not know which.

It may seem that, since a player has a 50-50 chance of choosing either envelope, they have an equal chance of gaining or losing money whether they decide to switch or keep the original envelope. However, probability theory seems to confusingly show that it’s always better to switch.

For example, say the first envelope you pick has \$10, so that the other envelope has either \$20 or \$5. Then you can calculate the expected value (i.e. the probability-weighted sum of the possible values) of the second envelope, assuming that each possibility has a 50% chance: (0.5 x \$5) + (0.5 x \$20) = \$12.50. Since \$12.50 is more than \$10, it makes sense to switch. No matter which numbers you use, you always get an expected value for envelope two that is 5/4 higher than the value for the original envelope: if c is the value of the original envelope, the expected value of the second envelope is (0.5 x [0.5c]) + (0.5 x [2c]) = 5/4c.

This counter-intuitive mathematics can explain very well why so many first-time entrepreneurs try again, regardless of the success or failure of their first attempt. And why it is perfectly rational for investors to expect a higher payoff from a not-so-successful serial entrepreneur than a first-time entrepreneur.

1. And if the first-time entrepreneur gets a \$0 envelope the first time, subsequent logic fails! LOL

Jokes aside, this is a very interesting mathematical angle on serial entrepreneurialship, all things being equal (experience, timing, etc). I bet if you factor in more experience, better name recognition, and better connection the second time around it'll at least a magnitude greater than 5/4X.

2. Interesting angle! Fun article!

Its same probability for both players as they exchange the envelopes, thats where the math gets fuzzy and explains some of the reality of the startup world.

3. Touraj,

This "paradox" bugged me a few years back so I thought I'd tell you what I learned. I asked the guys I knew from Yale who were a lot better than me at math...

The expected value is the probability density-weighted integral of the possible values. The expected value computation only makes sense if the integral is finite.

Obviously, if I hand you an envelope and say it can contain any amount of money with equal probability then there is no meaningful expected value. This is because the integral of the density-weighted possible values is not finite.

The problem with the two envelope "paradox" occurs in the assertion that one envelope contains twice as much money as the other. This may be a true fact but it does not establish a meaningful probability density function. Therefore the expected value calculation is not meaningful.

It makes no difference whether you switch. This is consistent with one's intuition because the calculation which led one to switch did not depend on the dollar value of what was in the envelope; one didn't need to open the envelope to do it. One could as well have taken an envelope, used the calculation to decide to switch, and then used the same calculation to switch back. This is clearly nonsense and it's because the expected value calculation is not meaningful without a proper distribution function

Of course, if you have an envelope with \$10 and I offer to swap you for an envelope which contains either \$5 or \$20 depending on the flip of a coin then you should take the trade. But that is of course a probability density function whose integral is finite.

It may be rational for an individual to expect a higher reward after each consecutive company but the two-envelope "paradox" doesn't explain why.

I hope this helps.

Micah

P.S. a much more interesting and legitimately counter-intuitive problem in my opinion is the Monty Hall problem (http://en.wikipedia.org/wiki/Monty_Hall_problem) where it is in fact to your advantage to switch doors.

4. Thanks for the comments Kevin, Su and Micah!

Micah, the math in this is way beyond the intuition of myself or dare I say, most other entrepreneurs, especially when you start talking about probability density functions and integrals etc. But you are right, if the player was given the choice to PLAY AGAIN and randomly choose one of the envelopes, as in the Monty Hall case, I think you would do that as repeat plays should get you closer to the expected value, which is higher than the first pick (and therein lies the entrepreneurial bug as well). Your wikipedia link led me to the wikipedia page for the two-envelope problem, which has some interesting comments, explanations and variants of this at http://en.wikipedia.org/wiki/Two_envelopes_problem
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