Norwegian said:
Hi there!
OP describes a situation where the Kelly-optimal strategy includes making a bet with negative expectation, and seeks to get some intuitive understanding of this interesting fact.
Lets look at a somewhat simpler example. Suppose you can get 2 to 1 on an event that is 50% likely to happen. The Kelly criterion will tell you to bet a quarter of your bankroll on this (but it really doesn't matter how much).
Now suppose you also can get 0,9 to 1 on the opposite outcome. This bet will have negative expectation. But it is surely better, in the Kelly sense, to make a surebet with your remaining money on the two outcomes.
So we have a situation where the making a negative EV bet is obviously included in the Kelly strategy.
Now, you may say this is different from your example, since there was a "take" in your case, and you didnt have the option of making a surebet.
OK, let's assume in my example that the bookmaker also gives odds, say 3 to 1, on a third outcome which you happen to know will not happen. Will that change your betting strategy? (no) Is there now a "take"? (yes)
If you are still not happy, let's assume the third option has some microscopic chance of happening anyway. Will that change your betting strategy? (yes, but only very slightly)
Now we have essentially your situation, and hopefully your intuition has improved.
Yes, it has been improved. Thanks for boiling the problem to its essence for me, Norwegian. That was a very helpful explanation!
Mute said:
Ugh. The Black Swan was such an annoying book to read. It felt like nearly all of the legitimate points the author was making about how people ignore the risk of large, sudden events were obscured by his injection of his own ego into the narrative. It read less like a book trying to warn people about the dangers of rare, large events and more like a diatribe against everyone else for not recognizing how brilliant he is for looking at these important rare events that everyone else ignores.
Anyways, for the problem at hand, a negative expectation value is your expected payout if you can bet on the horses with the same odds a large (i.e., infinite) number of times. The odds aren't going to stay the same, though: they'll get readjusted over time, so shouldn't one's betting strategy have to take into account that you have a finite number of bets to make before the odds will be readjusted? Is this not relevant to the strategy at all?
If you think Taleb's ego is shows through in The Black Swan, you should see his more recent book of aphorisms called, I believe, The Bed of Procrustes. I recommend taking it in homeopathic doses. Taleb thinks of himself as a modern Heraclitus, I guess. I expect his next work will be a 30-40 page book of fragments. Stuff like
104) Taleb said, Economists [wear] neck[ties?] (undecipherable) nerds: The magnificent (remainder lost).
But most of the people I admire have large egos, so I will continue to pay attention to what he says. Bet on it.
You are exactly right about the odds changing in response to your bet and those of others. The problem I gave is highly idealized, but you have to start somewhere.
ImaLooser said:
Well, I just plain don't believe it. I think your intuition is correct and betting on horse three reduces your expectation. So if you want to contend otherwise you are going to have to show me this mysterious math. It only make sense if you introduce risk of bankruptcy.
Yep, the need to avoid bankruptcy is the reason why you need to "hedge" to maximize your expected longterm growth rate. I'm finally starting to get it now.
I will show the math for determining the Kelly optimal strategy for a problem like the one in my original post. It is somewhat involved and I have only recently acquired it myself, so the presentation may not be so smooth. And it might take several posts to develop it. In the meantime you could look at Kelly's 10-page paper from 1956, http://www.bjmath.com/bjmath/kelly/kelly.pdf, in which he derives the optimal strategy. I will follow his derivation, using mostly the same notation and filling in some steps.
You have an opportunity to make a series of bets on n mutually exclusive outcomes, where outcome j in our example is "horse j wins". You are allowed to bet a fraction of your money on any subset of outcomes. It is assumed that you don't borrow money to gamble with (thus ruling out most gamblers), so the fractions of your bankroll that you bet on horses 1, 2, etc., together with the fraction you don't bet, must add up to 1.
f
1+ f
2+...+ f
n + b = 1
Suppose you start out with a bankroll of B
0. If horse 1 wins the first race, then your new bankroll after one race is
B_1 = (1+o_1f_1-f_2-...-f_n)B_0
where o
1 is the odds paid on horse 1. This expression can be simplified by introducing what I call the "for-odds", namely the amount paid
for a one dollar winning bet. For example 1-to-1 odds is the same thing as "2-for-1", because you get your original dollar bet back plus the dollar you win. If o_1 is the normal odds on horse 1, then \alpha_1 = o_1 + 1 is the for-odds on horse 1. So assuming horse 1 wins,
B_1 = (b+\alpha_1f_1)B_0
where as before b is the fraction of your bankroll you hold back. Now consider your bankroll after N races, with the betting fractions held constant:
B_N = (b+\alpha_1f_1)^{W_1}(b+\alpha_2f_2)^{W_2}...(b+\alpha_nf_n)^{W_n} B_0
where the W's are the number of times each horse wins during the N races. The long term growth rate of your bank roll is
\frac{\lnB_N}{N} = (W_1/N)\ln(b+\alpha_1f_1)+(W_2/N)\ln(b+\alpha_2f_2)+...(W_n/N)(b+\alpha_nf_n) + \lnB_0/N
which tends to
G = p(1)\ln(b+\alpha_1f_1)+p(2)\ln(b+\alpha_2f_2)+...p(n)(b+\alpha_nf_n)
with probability 1 by the law of large numbers. p(j) is the probability that horse j wins. We want to maximize this subject to the constraints f
1+ f
2+...+ f
n + b = 1, each term being nonnegative. Using the Lagrange multiplier method, this is the same as maximizing
G = p(1)\ln(b+\alpha_1f_1)+p(2)\ln(b+\alpha_2f_2)+...p(n)(b+\alpha_nf_n) - k(b+f_1+...+f_n)
with the bet fractions nonnegative. This leads to
\frac{\partial G}{\partial f_j} = p(j)\alpha_j/(b+\alpha_j f_j) = k
for those f_j that aren't 0, i.e. the ones horses we bet on. Also
\frac{\partial G}{\partial b} = \frac{p(1)}{(b+\alpha_1 f_1)} + ... + \frac{p(d)}{(b+\alpha_d f_d)} + \frac{1}{b}(p(d+1) + ... + p(n)) = k
Here we have renumbered the outcomes so that we bet on horses 1,...,d but not on horses d+1, ..., n. We don't know what d is yet, but that's okay. We'll get to it later. Finally we have
\frac{\partial G}{\partial f_j} = p(j)\alpha_j/b \leq k
for those outcomes we don't bet on, namely j = d+1,...,n. First we solve for k. It will simplify things some to let
p = p(1)+...+p(d)
and
\sigma = \frac{1}{\alpha_1}+...+ \frac{1}{\alpha_d}
From the equation for \partial G/ \partial f_j we get
p(j) = k(f_j + b/\alpha_j) for j = 1, ..., d
Summing over j = 1,...,d we get
p = k(f_1+...+f_d + b\sigma) = k(1-b + b\sigma) = k - kb(1-\sigma)
or
kb = (k-p)/(1-\sigma)
On the other hand, from p(j)/(b+\alpha_j f_j) = k/\alpha_j and the equation for \partial G / \partial b, we get
k\sigma + (1-p)/b = k
or
kb = (1-p)/(1-\sigma)
Therefore k=1, b = (1-p)/(1-\sigma), and f_j = p(j) - b/\alpha_j for the horses we bet on. I'll will cover the procedure for determining which horses to bet on in a later post.
