Can you win money on a bad bet?

Mute

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?

mfb

Well, it does. It reduces the risk a lot, so you can increase the total amount of money you bet.
Without horse 3, you have a 17%-probability to lose everything you spent on that round. With some money on horse 3, you will still lose some money, but the risk to lose most of it is just 2%.

techmologist

Yes, it has been improved. Thanks for boiling the problem to its essence for me, Norwegian. That was a very helpful explanation!

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

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.

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, A New Interpretation of Information Rate, 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.

f₁+ f₂+...+ f_n + b = 1

Suppose you start out with a bankroll of B₀. If horse 1 wins the first race, then your new bankroll after one race is

[tex]B_1 = (1+o_1f_1-f_2-...-f_n)B_0[/tex]

where o₁ 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 [itex]o_1[/itex] is the normal odds on horse 1, then [itex]\alpha_1 = o_1 + 1[/itex] is the for-odds on horse 1. So assuming horse 1 wins,

[tex]B_1 = (b+\alpha_1f_1)B_0[/tex]

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:

[tex] B_N = (b+\alpha_1f_1)^{W_1}(b+\alpha_2f_2)^{W_2}...(b+\alpha_nf_n)^{W_n} B_0 [/tex]

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
[tex] \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 [/tex]

which tends to

[tex] G = p(1)\ln(b+\alpha_1f_1)+p(2)\ln(b+\alpha_2f_2)+......p(n)(b+\alpha_nf_n) [/tex]

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 f₁+ f₂+...+ f_n + b = 1, each term being nonnegative. Using the Lagrange multiplier method, this is the same as maximizing


[tex] 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)[/tex]

with the bet fractions nonnegative. This leads to

[tex]\frac{\partial G}{\partial f_j} = p(j)\alpha_j/(b+\alpha_j f_j) = k[/tex]

for those f_j that aren't 0, i.e. the ones horses we bet on. Also

[tex]\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[/tex]

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

[tex]\frac{\partial G}{\partial f_j} = p(j)\alpha_j/b \leq k[/tex]

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

[tex]p = p(1)+...+p(d)[/tex]

and

[tex]\sigma = \frac{1}{\alpha_1}+...+ \frac{1}{\alpha_d}[/tex]

From the equation for [itex]\partial G/ \partial f_j[/itex] we get

[tex]p(j) = k(f_j + b/\alpha_j)[/tex] for j = 1, ..., d

Summing over j = 1,...,d we get

[tex]p = k(f_1+...+f_d + b\sigma) = k(1-b + b\sigma) = k - kb(1-\sigma)[/tex]

or

[tex]kb = (k-p)/(1-\sigma)[/tex]

On the other hand, from [itex]p(j)/(b+\alpha_j f_j) = k/\alpha_j[/itex] and the equation for [itex]\partial G / \partial b[/itex], we get

[tex]k\sigma + (1-p)/b = k[/tex]

or

[tex]kb = (1-p)/(1-\sigma)[/tex]

Therefore k=1, [itex]b = (1-p)/(1-\sigma)[/itex], and [itex]f_j = p(j) - b/\alpha_j[/itex] for the horses we bet on. I'll will cover the procedure for determining which horses to bet on in a later post.

ImaLooser

OK, we are on the same page. The theory would work if the horse race example were realistic. However the horse race example has no connection with reality, so it won't work in real life.

techmologist

Yes, you are right. I am starting to see what's going on now, mostly thanks to Norwegian's very clear explanation. Obviously if you go bust, your longterm growth rate is zero (EDIT: actually, minus infinity), so you have to reduce that probability to maximize your growth rate. Even if you don't go completely broke, it is probably not good for your growth rate to have frequent large drawdowns.

Well, yeah. When have you ever known a mathematical model to be realistic? If it were realistic, you couldn't solve it. Math is a big scam. But I like it. And I'm going to keep doing it because it makes me feel better.
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Naturally I made about a thousand mistakes in my previous post. It is too late to edit it, so I tried to clean it up here a little.

Norwegian

Good work! I am waiting for the rest. How to determine d, that is which horses to bet on. Would the following work: Choose the (hopefully) unique largest subset of horses, for which the formulas give non-negative values for all f_j and b? And I assume there is some additional trick to avoid checking all possible subsets. Obviously you include all the positive EV horses, but is there a natural way to order the rest?

Sorry Looser, but you are not on the same page as anyone else in this thread. The theory does work, and the horse race is a very good illustrative example. It is totally irrelevant for the math whether or not this has a connection to "reality", whatever that may be. And, in addition, you are just wrong. The theory does show that in any horse race including a +EV horse, and a very slightly -EV horse, it is Kelly-optimal to bet on the latter.

techmologist

Yes, that's it. I'm trying to pump myself up to show why that works. The rest is on the way (within an hour or two, hopefully). :)

Also, it would have made a lot more sense for me to define the long term growth rate as

[tex]\frac{\ln(B_N/B_0)}{N} = (W_1/N)\ln(b+\alpha_1f_1)+(W_2/N)\ln(b+\alpha_2f_2)+.....+(W_n/N)\ln(b+\alpha_nf_n)

[/tex]

rather than

[tex]\frac{\ln(B_N)}{N} = (W_1/N)\ln(b+\alpha_1f_1)+(W_2/N)\ln(b+\alpha_2f_2)+.....+(W_n/N)\ln(b+\alpha_nf_n) + \ln(B_0)/N
[/tex]

although the difference doesn't matter much for large N.

techmologist

I left off without showing how to determine which horses you bet on. Following Kelly, I'll call this set [itex]\lambda[/itex]. For all j in this set we have f_j > 0 and

[tex]\frac{\partial G}{\partial f_j} = p(j)\alpha_j/(b+\alpha_j f_j) = k = 1

[/tex]

which means that

[tex]p(j)\alpha_j > b[/tex]

On the other hand, for horses we don't bet on, i.e. j is not in [itex]\lambda[/itex], we have f_j = 0 and

[tex]p(j)\alpha_j/b \leq 1[/tex]

or
[tex]p(j)\alpha_j \leq b[/tex]

Recall that b is given by

[tex]b = (1-p)/(1-\sigma)[/tex]

where p and sigma were defined by

[tex]p = \sum_{j \in \lambda} p(j)[/tex]
[tex]\sigma = \sum_{j \in \lambda} \frac{1}{\alpha_j}[/tex]

Since we assume that b is positive, we must bet on horses such that [itex]1-\sigma > 0[/itex]

Next we renumber the outcomes if necessary so that

[tex]p(1)\alpha_1 \geq p(2)\alpha_2 \geq ... \geq p(n)\alpha_n[/tex]

Whatever value b turns out to have, there will be some d (possibly 0) such that

[tex]p(1)\alpha_1 > b[/tex]
.
.
.
[tex]p(d)\alpha_d > b[/tex]
[tex]p(d+1) \alpha_{d+1} \leq b[/tex]
.
.
.
[tex]p(n) \alpha_n \leq b[/tex]

Since these are exactly the requirements that determine membership in the set [itex]\lambda[/itex], we see that the ordering we chose forces the set of horses we bet on to be of the form

[tex]\lambda = \{1,...,d\}[/tex]

To determine d, set it at 0 and increase it (and adjust the value of b accordingly), testing each time to see if the above inequalities are satisfied and stopping when they are satisfied exactly. That's the basic idea, anyway. Here are the specifics.

Define

[tex]p_t = \sum_{j=1}^{t} p(j)[/tex]
[tex]\sigma_t = \sum_{j=1}^{t} \frac{1}{\alpha_j}[/tex]
and
[tex]b_t = \frac{1-p_t}{1-\sigma_t}[/tex]

So long as

[tex]p(1)\alpha_1 \geq p(2)\alpha_2 \geq ... \geq p(t)\alpha_t > b_t[/tex]

t is a possible value for d and b_t is a possible value for b. (SEE EDIT). But look what happens when we reach a point where

[tex]p(t+1)\alpha_{t+1} \leq b_t =\frac{1-p_t}{1-\sigma_t} [/tex]

(This could fail to happen if the denominator of b_t becomes negative first). For t+1 to be a possible value for d, we need to have [itex]1-\sigma_t > 1-\sigma_{t+1}>0[/itex] so that b is positive. Then

[tex]p(t+1)(1-\sigma_t) \leq \frac{1-p_t}{\alpha_{t+1}}[/tex]

[tex](1-p_t)(1-\sigma_t) -p(t+1)(1-\sigma_t) \geq (1-\sigma_t)(1-p_t) - \frac{1-p_t}{\alpha_{t+1}}[/tex]

[tex](1-p_{t+1})(1-\sigma_t) \geq (1-\sigma_{t+1})(1-p_t)[/tex]

[tex]\frac{1-p_{t+1}}{1-\sigma_{t+1}} \geq \frac{1-p_t}{1-\sigma_t} \geq p(t+1)\alpha_{t+1} [/tex]

So t+1 cannot be the value of d. Further, b_t continues to increase with t until the denominator becomes negative, so d couldn't be any higher than t+1. So the bottom line is that starting with t=0, you compute b_t until you find its minimum positive value. At that point, t = d.

Note that b₀ = 1. So if there are no favorable bets at all, i.e. [itex]p(1)\alpha_1 < 1[/itex], then b_t only increases with t and has its minimum at t=0. We don't bet on any horses, as you would hope. But if there is at least one horse with positive EV, i.e. [itex]p(1)\alpha_1 > 1[/itex], then we keep adding horses until we minimize b_t. It is possible for some of those horses to have negative EV. I might add more on that point later.

God bless anyone who actually reads this.

EDIT: Of course, I forgot to point out that as long as [itex]p(t+1)\alpha_{t+1} > b_t[/itex], then [itex]b_{t+1} < b_t[/itex]. To see it, just change [itex]\leq[/itex] to > in the derivation that followed "(SEE EDIT)" alert. [itex]b_t[/itex] is decreasing here, and you keep adding horses to the set until it starts increasing again (or changes sign).

SteveL27

Like I said earlier ... the math is impressive; but it doesn't have much to do with horse racing.

mfb

So what?
It is a theoretical consideration. You could replace "horses" by anything else, it would not matter. It is a mathematics question, not a horse race question.

SteveL27

I asked the OP that question earlier. I asked if he was asking a math question, or if he was planning on spending real money gambling. He responded that real money might be involved. On that basis I regard this as a very misguided thread. OP should study pari-mutual betting, not make up fake odds and develop mathematical theories around them.

Here is the post where I posed that question and received the response that this thread is in part about real gambling.

http://physicsforums.com/showpost.ph...29&postcount=5

My suggestion is that anyone interested in making money betting horses should spend time hanging around the race track, not developing elaborate theories about artificial situations.

Of course if the question is purely theoretical, my advice does not hold. And the math may be interesting. But if OP is planning to spend real money gambling, he's going to find that the horses don't know math; past performances are deliberately manipulated by trainers and owners; and the condition of the track counts for a lot more than the mathematical expectation based on the odds.

techmologist

I appreciate your concern. I will put your mind at ease right now, as I should have done earlier. I have absolutely no intention of tracking horse (actually, greyhounds in my case) statistics to decide which ones to bet on. Every objection you made to that approach is solid. Instead, I hope to find and exploit certain market inefficiencies that sometimes show up in pari-mutuel markets, which were explored and tested by Asch and Quandt in the 1980's. You don't have to know the probability that any horse will win. It's a kind of arbitrage. You can't get rich doing it, but it might be intellectually stimulating :)

The situation I described in the OP will not occur in any situation that has to do with horses, I agree. But the logic of the situation is not impossible (this is what mfb pointed out). You could have a big wheel-of-fortune which is painted like a big pie chart with four sectors. The red sector takes up 55%, the blue sector takes up 28%, etc. A very generous bookie, who does charity work in his spare time, is prepared to offer you the odds I stated on every spin of the wheel. He is rich enough to cover any bet you have the stones to make. This is your big chance. You can turn your pitiful life savings into millions, or more. If only you had some math to tell you how to proportion your bets to make this outcome likely, even almost certain.

The math above provides a solution to this admittedly far-fetched, but not impossible situation. And there is reason to hope that it provides intuition for closer-fetched situations. In the same way that first year physics students solve problems about perfectly elastic collisions between perfect rigid spheres. Do you think their professors assign them these problems for no reason? But if you post that problem on this forum, carefully stating the assumptions, you will get one person helpfully pointing out that there are no elastic collisions, another person telling you that any sphere you try to make is not perfectly spherical, and still another telling you that you shouldn't be playing pool because shady characters hang out in pool halls.

Having said all that, I completely agree that a person intending to bet money at the track should spend a good deal of time there making small bets and taking in the scenery to get the feel of things before trying out any theories. That's what I plan to do.