I recently came across this old paper about a meeting between Ed Thorp and Warren Buffett in the late 1960s. Buffett had recently closed his famous partnership after the bull market in the past few years left few undervalued securities to buy. Afterwards, former clients asked him to evaluate a money manager: Ed Thorp.
Compound interest is the first (popular) discussion that's worth repeating:
if the Manhattan Indians had been able to invest the $24 for which they sold Manhattan in 1626 at, say a net return of 8%, their heirs could buy it back now (1968) with all the improvements
This is just the beginning, however:
Buffett then brings up the "three very strange dice." Labeled as A = A=(2, 2, 2, 2, 5, 6), B=(1, 1, 4, 4, 4, 4) and C=(3, 3, 3, 3, 3, 3), two people can play a game where each chooses a dice to roll. The person with the highest number then wins that round. Interestingly enough, through repeated games (and deduction) it can be shown that A > B, B > C, but C > A. They dice are intransitive. As a result, the 2nd person to pick a die should always win in the long term by # of wins if he/she recognizes this. Just like rock paper scissors, no pick is universally best.
In other words, in such a game and indeed the market a strategy's success is often very much dependent on the others being employed. I would argue that the very reason for value's outperformance in 02-03 post-dot-com crash is because of the focus on speculative technology stocks. Similarly, active investors want more short term speculators and index funds because they would provide the other side to profitable buys. On the other hand, if everyone is preaching value or quality businesses a la nifty-fifty, value investors may want to stay away.
This phenomenon is indeed well-known already, but the examples above and detailed in Ed Thorp's paper show just how pervasive and powerful this quality is.
(icon source: http://www.zeitnews.org/node/7933)
wut
There's a quantitative answer to this problem, by law of large numbers: sum(A) = 19 while sum(B) = sum(C) = 18. So pick A if we're risk neutral. Otherwise if we are risk averse, C is preferred over B since it has less variance, and it depends on our utility function coefficients for whether A is higher than C.
Are you sure this is true? Aren't the probability pairings in A=(2, 2, 2, 2, 5, 6), and B=(1, 1, 4, 4, 4, 4) the same as in A=(2, 2, 2, 2, 6, 6), and B=(1, 1, 5, 5, 5, 5)? In this scenario, sum(A) = 20 and sum(B) = 22. If you choose any number in the first set, it's the same probability in both the first and second scenarios that the number will be higher than the number you choose in the second set. It seems like, as tt1254 says, the best set of numbers to choose is dependent on what your opponent chooses, not simply the law of large numbers?
Actually, all that said, it seems like B > C > A in terms of your probability of having the higher number, given that the number in your set that you choose is random.
No, this has nothing to do with utility functions or the sums of the eyes on the dice. I.e. a new die, Â = (2, 2, 2, 2, 5, 10000) would be no different than A in terms of the probability to win a round.
You need to pair the dice and look at the probabilities:
A vs. B: A has a 4/6 prob. of winning, thus this die is preferable. A vs. C: C has a 4/6 prob. of winning. B vs. C: B has a 4/6 prob. of winning.
This shows that the game is intransitive, making no die generally preferable to another.
Don't talk about maths to post BS like this....
The sum is irrelevant, as is risk-neutral nonnsense. You could have a dice with (0,0,0,0,0,151954959) It would be worse than all of them despite the sum.
I would pick Dice B. Here's why:
When I roll dice C I have a 33% chance to beat dice B and a 66% chance to beat dice A When I roll dice A I have a 33% chance to beat dice C and (what appears to be) a 55% chance to beat dice B When I roll dice B I have a 66% chance to beat dice C and (what appears to be a) 44% chance to beat dice A
Or
Dice A – 2/6 personal outcomes beats 12/12 opponent outcomes & 4/6 personal outcomes beats 2/12 opponent outcomes Dice B – 2/6 personal outcomes beats 0/12 opponent outcomes & 4/6 personal outcomes beats 10/12 opponent outcomes Dice C – 6/6 personal outcomes beats 6/12 opponent outcomes
What!? How do you get those fractions? That's like when Michael Scott claims to be 1/15th Native American.
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