Steve,
Really appreciate your input here, but just want to get picky about
one point. Probability theory tends to break down when it encounters
the infinite. Statements such as the ones you make above only apply
in the infinite. Once we limit ourselves to plausible real-world
scenarios, we get different results.
This simply isn't true in general (or, really, in your example.) It is a
common misconception that probabilistic principles only apply "in the
long run".
Take the following two games - each costing $1.
Game A -
999,999/1,000,000 returns $0
1/1,000,000 returns $1,000,001Game B -
9/10 returns $1.11
1/10 returns $0Any reasonable definition of lower-risk-game would pick game B as
having lower risk.
I disagree with that statement on several levels, so I'm not quite sure
how to respond. But your example is perhaps an excellent one for
talking some about what "risk" is and isn't.
Here's a few possible responses:
1) I think you may be confusing "risk" with "volatility." Clearly game A
has higher volatility, but that isn't the same thing as risk.
2) The math doesn't "break down" as you claim. If the math didn't
apply to real-world situations, it would be worthless. But it applies
quite well, and allows accurate (probabilistic) predictions to be made
regarding these games.
3) If you say that you prefer game B, then I certainly believe you, in
spite of the fact that game A has higher EV and lower risk. EV isn't
everything. Risk isn't everything. In general, the answer to "which
game is best" depends on what objective you wish to achieve by
playing the game. If your sole objective is "maximize the probability
of starting with one unit and playing more than once" then game B
is certainly better, but that I think most educated players expect more
from a game.
4) Risk is a counter-intuitive. I think people tend to attach a "risk"
label to any concept that doesn't feel like EV. In particular, there
is a tendency to associate risk and volatility, but they are very
different concepts.
It could be argued that the only way to show that game B has
"lower risk" is to compare them in a way that is unfairly biased against
game A. If you compare risk in a context where both games are trying
to make a big win, game A will come out ahead.
Let's do a risk computation to compare these games. This comparison
will actually be reasonably finite in duration -- the worst case will be
less than 150 games.
Suppose the objective is to parlay $1 into $1,000,000. Game A gives
a one-in-a-million shot at achieving this, and either succeeds or fails
on the very first trial.
Let's see how game B performs. I'm going to modify the conditions
for game B slightly, to help compensate for the "unfair" EV advantage
of game A, by allowing the player to wager any amount at any time.
This makes a "bet it all" strategy the best way to proceed, until such
time that the $1 million goal can be reached by betting only part of
the bankroll.
If we bet everything on a 1.11:1 payoff, and repeat this to shoot for
$1 million, then after 132 wins we have:
1.11^132 = 960800
This isn't quite to our goal yet, but making bet #133 just large enough
to reach the goal allows us to get there after 133 trials. The probability
of winning the first 132 bets on game B is:
0.9^132 = 9.1203 x 10^-7 or 1 in 1096450.
So, getting within one bet of winning with game B is less likely than
actually winning by playing game A. The overall probability of a win
is even more of a long shot. That is what it means to say that game A
has lower risk than game B.
Now, you might try to claim that this happens only because game A
has positive EV, but that isn't the case, as I'll now demonstrate.
Let's change game A so that it returns 999,000 for a win, and has
a probability of winning that remains 1/1,000,000. Now game A
and game B have exactly the same EV, yet the risk for game A
is still the same with a 999,999/1,000,000 chance of losing.
Now game B only has to reach 999,000 in order to win. But wait,
the same 132 trials still only reach to $960,080, and still give
game B less of a chance of success. EV isn't the problem, game A
simply has lower risk, when shooting for a big win. This is a general
property of negative EV games -- if two N-for-1 games have the same
EV, the one with a lower probability of a larger payoff will have lower
risk of ruin. This is counter-intuitive -- the "long shot" has lower risk,
when both games have the same (negative) EV.
Notice that there is nothing "infinite" about these comparisons.
Yet the math shows that game A has lower risk.
Of course, you can lower the target, and say "which game is better
if you only want to parlay $1 into $1000?" But I will then claim that
you have unfairly biased the contest against game A by making it
shoot beyond the target by $999,000. Lowering the target below
$865585 allows game B to win in 131 trials with odds that are
986805:1. Now the odds are pretty close, but game A pays off
better than game B by about $135,000. To make these two games
about equal, you have to reduce the payoff for game A to about
$866,000. In other words, if you want to parlay $1 into a large
sum, a game with an EV of about 87% might be better than a
slow 1.11:1 growth rate game with an EV of 99.9%. From this
perspective, risk seems to me to be fairly important.
Real people playing these games for plausible
amounts of time will have lower RoRs for game B than game A. It is
only when we allow the infinite that it switches.
I think what you're really getting at here is that game A gets boring
really fast because you virtually always lose, while game A is more
interesting because you almost always win. But, that really doesn't
have anything to do with RoR, or with risk in a contest that doesn't
require one to "play forever" in order to succeed. I think perhaps
it only has to do with volatility, and so your example may be a
good way to show the difference between risk and volatility.
ยทยทยท
On Tuesday 23 August 2005 08:19 am, Bill Cook wrote: