View New Posts
123
1. I've been all over the forums but I can't quite seem to locate Part III of this series. Does it exist?

2. Originally Posted by CatPulp
I've been all over the forums but I can't quite seem to locate Part III of this series. Does it exist?
No

3. Over 7500 views...for this?

4. Ganch had exhausted us after this one, then he took his slide rule and went back to CR.

5. Originally Posted by Flying Dutchman
Ganch had exhausted us after this one, then he took his slide rule and went back to CR.
Damn, this is a pure shame. But if you would be so kind as to excuse my ignorance; what is CR?

6. Civilized Return (on investment).

SBR Founder Join Date: 12/14/2005

8. If you grow your bankroll to 105%, from \$100 to \$105, do you adjust your bet size so that 1% becomes \$1.05 vs \$1 before. And then lower it to \$.95 if you shrink 5%?

9. Yes.

10. I am trying to find the answer to this question (and to avoid thinking myself What if I would like to make 1500 simultaneous bets, that are mutually exclusive (ie either all of them loses, or exactly one wins). How can I use the kelly criterion to figure out the stakes?

thanks for helping out.

F

11. I believe you want to compute the kelly stake (for one of your 1500 bets), remove the amount from the bankroll, and repeat w/ the next bet.

*edit* above is wrong for simultaneous bets *edit*

and while this thread is bumped...

I just realized that I had read this exact Ganchrow-post years ago during my college years. The bankroll management parable has been stuck in my mind for all these years; thanks for well-explained words of wisdom Ganchrow!

12. Originally Posted by Blax0r
I believe you want to compute the kelly stake (for one of your 1500 bets), remove the amount from the bankroll, and repeat w/ the next bet.
You don't even do that for independent simultaneous wagers, much less for mutually exclusive ones. His problem is theoretically possible to solve with Kelly but good luck solving a 1500 bet simultaneous, mutually exclusive Kelly stake.

13. every time i reread a ganch post i feel so stupid, then i go over to pt and feel much better about myself

14. Originally Posted by donjuan
His problem is theoretically possible to solve with Kelly but good luck solving a 1500 bet simultaneous, mutually exclusive Kelly stake.
As per the problem description, this can be treated as one (likely large) bet at (likely short) odds. This bet will either win or lose and calculating Kelly stake for this bet is trivial. I would not pollute this thread with such triviality but could not pass by without wishing you good luck on working on your reading comprehension skills.
Points Awarded:
 the_fredrik gave Data 2 SBR Point(s) for this post.

15. Originally Posted by Data
As per the problem description, this can be treated as one (likely large) bet at (likely short) odds. This bet will either win or lose and calculating Kelly stake for this bet is trivial. I would not pollute this thread with such triviality but could not pass by without wishing you good luck on working on your reading comprehension skills.
Thanks, Data.

For the case where all the 1500 bets have the same probability and payoff and thus equity this is obviously true.

For the case where many of the bets have very different probability and payoff and equity, i.e. some bets have a much larger probability to win than others, it is not obvious to me that treating them as one bet is exactly right.

Let's say we have a 10,000-sided die that is crooked, some sides have a 1/300 chance of winning while paying +400, other have a 1/100,000 chance of winning but paying +200,000 (and so on, 1498 more variants) - can you still treat them together as one large bet then? How would you do that?

16. Originally Posted by the_fredrik
Thanks, Data.

For the case where all the 1500 bets have the same probability and payoff and thus equity this is obviously true.

For the case where many of the bets have very different probability and payoff and equity, i.e. some bets have a much larger probability to win than others, it is not obvious to me that treating them as one bet is exactly right.
The bettor may utilize two different methods to make selections among those 1500 outcomes.

1) Divide all the selections between selections where he has an edge and selections where he does not have an edge.

In this case it is safe to assume that the edge is the same across all the bets and he just needs to add the probabilities of his selections and size bets proportionally to probabilities.

Originally Posted by the_fredrik
Let's say we have a 10,000-sided die that is crooked, some sides have a 1/300 chance of winning while paying +400, other have a 1/100,000 chance of winning but paying +200,000 (and so on, 1498 more variants) - can you still treat them together as one large bet then? How would you do that?
2) If the bettor is capable of identifying and calculating varying edges accross 1500 bets then he should have no problem maximizing EG as he already has those 1500 outcomes enumerated. He will use either his programming skills or Excel Solver to maximize EG.

The condition that only one bet wins ensures that the problem is not that complicated in either case.
Points Awarded:
 the_fredrik gave Data 5 SBR Point(s) for this post.

17. Originally Posted by Data

2) If the bettor is capable of identifying and calculating varying edges accross 1500 bets then he should have no problem maximizing EG as he already has those 1500 outcomes enumerated. He will use either his programming skills or Excel Solver to maximize EG.

The condition that only one bet wins ensures that the problem is not that complicated in either case.
I have thought about this and came up with the following formulation, for 3 simultaneous bets (that are mutually exclusive)

E = (1 + W1*F1)^(P1) * (1 + W2*F2)^(P2) * (1 + W3*F3)^(P3) * (1 - (1-F1-F2-F3))^(1-P1-P2-P3) - 1

W1 = odds-1 on bet 1, P1 = probability that bet 1 wins, F1 the stake on this bet in % of total BR.
W2 = odds-1 on bet 2, P2 = probability that bet 2 wins, F2 the stake on this bet in % of total BR.
W3 = odds-1 on bet 3, P3 = probability that bet 3 wins, F3 the stake on this bet in % of total BR.

Maximize E (geometric growth) subject to 0<=F1<=1, 0<=F2<=1, 0<=F3<=1, 0<=F1+F2+F3<=1. (W and P constant of course)

(easy to generalise to 1500 simultaneous bets)

This is unfortunately a rather complex optimization problem to solve, especially as the number of bets grow large.

Are there a simpler way of expressing the problem that I am missing?
I suspect this formulation with 1500 variables will be impossible to solve in practice, definitely with excel but also with an optimzation program such as Cplex.

Any approximation that errs on the conservative side would be useful too.

18. I'm unaware of the existence of a closed form solution to the mutually exclusive outcome Kelly problem except in the very special case of bookmaker overround less than unity across all possible outcomes (for Full Kelly). Now I'm not saying it doesn't exist, just that I've never seen it (nor spent much time trying to suss it out for myself)

Specifically, given an unconstrained pure arbitrage optimal strategy would would be to wager a percentage of bankroll on each outcome equal to the probability of that outcome occurring.

What's most interesting about this is that in the face of a pure arbitrage, an unconstrained player would optimally wager without regard for individual odds.

For example, a fair coin prop offering +110 on heads and +100 would call for a 50% full Kelly wager on each outcome, while a fair offered at +20000 heads and -500 tails would call for the same pair of half bankroll bets.

Anyway, here's a plain-English description of the algorithm I used in the SBR Kelly calculator to calculate full Kelly stakes on mutually exclusive events.

1. Sort all bets by edge, from highest to lowest.
2. Calculate the fair implied probability for each bet. This is just the reciprocal of the decimal odds.
3. Starting with the highest edge bet, calculate a running total of the the implied probability and the actual probability. The running total for each bet includes the sum of the implied and actual probabilities for that bet and every bet with a higher edge.
4. If the sum of all the implied probabilities is less than 1 (i.e., a true arb exists), then for each bet the stake will be the actual probability. If this is the case, we can stop here.
5. If the sum of all the implied probabilities is greater than 1, then for each bet calculate the quotient (1 – the sum of actual probabilities) / (1- sum of implied probabilities).
6. Find the smallest value of this quotient that’s greater than zero. If no quotient is greater than zero then no bets will be made.
7. Then for each bet the stake will be the actual probability minus the minimum quotient from 6) above multiplied by the fair implied probability.

If it helps, following is the old skool JS I used in the Kelly Calculator for mutually exclusive outcomes with Kelly multiplier = 1.

Kind of embarrassing when I look at my old code now ...

Code:
```function calcMutExKelly(a_dEdge, a_dOdds, dKellyMult)  {
var lSingles,a_sParlayNames,a_dRealKellyStakes;

if (dKellyMult == undefined || dKellyMult <= 0 || isNaN(dKellyMult)) dKellyMult = 1;
dKellyMult = parseFloat(dKellyMult);

if (!isArray(a_dEdge) || !isArray(a_dOdds) ) {
var err = "calcMutExKelly: Odds and Edge arguments must both be arrays";
return err;
}
lSingles = a_dOdds.length;
if(lSingles != a_dEdge.length) {
var err = "calcMutExKelly: Edge (size=" + a_dEdge.length + ") and odds (size=" + lSingles + ") arrays are different sizes";
return err;
}
a_dRealKellyStakes = new Array(Math.pow(2,lSingles)-1);
a_sParlayNames = new Array(Math.pow(2,lSingles)-1);
var oSortedByEdge = new Array(lSingles-1);
var dTotProb = 0;
for (var i=0; i<=lSingles-1;i++) {
var mydProb = edge2prob(a_dEdge[i],  a_dOdds[i]);
dTotProb += mydProb ;
oSortedByEdge[i] = { n: i, dEdge: a_dEdge[i], dOdds: a_dOdds[i], dProb: mydProb};
}
if(dTotProb > 1 + 1e-6) {
var err = "calcMutExKelly: Sum of probabilities of mutually exclusive outcomes (" + dTotProb + ") may not be > 1";
return err;
}
var fnSortByEdge = function(a,b) { return(b.dEdge - a.dEdge); } ;
oSortedByEdge.sort(fnSortByEdge);
var dMinResult = 1, dOverround = 0, dSumProb = 0, dSumOddsRecip = 0;

for (var i = 0; i<=lSingles-1; i++) {
dSumProb += oSortedByEdge[i].dProb;
if ( dSumProb > 1 ) dSumProb = 1; // due to rounding error probability may erroneously be slightly > 1
dOverround += 1 / oSortedByEdge[i].dOdds;
var dProposedMinResult = (1-dSumProb) / (1-dOverround );
if (dProposedMinResult > 0 && dProposedMinResult < dMinResult) {
dMinResult = dProposedMinResult ;
}
}
for (var i = 0; i<=lSingles-1; i++) {
if (dOverround < 1 && dSumProb >= 1 - 1e-7 ) {
a_dRealKellyStakes[Math.pow(2,oSortedByEdge[i].n)] = oSortedByEdge[i].dProb;
} else {
a_dRealKellyStakes[Math.pow(2,oSortedByEdge[i].n)] = Math.max(0, oSortedByEdge[i].dProb - dMinResult / oSortedByEdge[i].dOdds);
}
a_sParlayNames[Math.pow(2,oSortedByEdge[i].n)] = ''+(1+oSortedByEdge[i].n);
}
g_arrStakes = a_dRealKellyStakes;
return {arrNames: a_sParlayNames, arrStakes: a_dRealKellyStakes};
}```
Points Awarded:
 the_fredrik gave Ganchrow 10 SBR Point(s) for this post.

SBR Founder Join Date: 8/28/2005

19. ganch sighting! ganch sighting!

20. Thank you Ganchrow, excellent post, very educative. I never liked flat betting it always seemed like a road to losing in the long run.

21. Originally Posted by Ganchrow
If the sum of all the implied probabilities is greater than 1, then for each bet calculate the quotient (1 – the sum of actual probabilities) / (1- sum of implied probabilities).
Damn, you make it too easy.

22. Great article, good job Ganchrow

23. Thanks a lot Ganch!

Perhaps I am pushing my luck here, but the acutal problem I am trying to solve is a bit more complex variation of the problem I described earlier: I still have 1500 simultaneous bets. Each can win or lose like before. But there are actually 4 different types of wins. The first win is still mutually exclusive like before: Either one of the 1500 bets scores win #1, or none of them scores win #1.

In addition, there are also 3 other types of (smaller) wins though. And they are not mutually exclusive: None, one or more of win type #2-4 can happen for each of the 1500 bets.

Is it still possible to solve this stake problem?

Thanks a lot in advance to anyone that can describe a solution to this, approximative or not.

(win #1 is less likely than win #2 that in turn is less likey than win #3 that in turn is less likely than win #4)

And Ganch, don't be embarrased by code that works!

24. Ganchrow! Fama profectionis tuae sunt omnia maiora!!

25. good stuff of part II and useful, my level will be up 1 step. Waiting for the next part...

Originally Posted by Ganchrow
A quantitative introduction to the Kelly criterion

Part II -- Maximizing Expected Growth

In Part I of this series we introduced the concept of expected growth, where we discussed why a bettor might reasonably choose to gauge the relative attractiveness of a given bet by considering its expected growth. In Part II of the series we'll look at how a bettor might use the notion of expected growth to determine how large a bet to place on a given event. This is the very essence of the Kelly criterion.

There are two extremes when it comes to placing positive expectation bets. On the one hand you have people like my aunt, who’s so afraid of risk that I doubt she’d even bet the sun would rise tomorrow (“But what if it didn’t? I could lose a lot of money!”). On the other hand you have people like my old college buddy Will, whose gambling motto was “Get an advantage, and then push it.”

One Saturday night during the spring term of my sophomore year, Will decided he was going to run a craps game. He put the word out to a number of the bigger trust fund kids and associated hangers-on and let the dice fly. After maybe 4 or 5 hours, Will was up close to \$8,000, which was far from an insignificant amount for us at the time. One player, an uppity gap-toothed British guy named Dudley, whose own losses accounted for most of that \$8K, loudly proclaimed that he was sick of playing for small stakes and wanted some “real” action. He told Will he was looking to bet \$15,000 on one series of rolls. Will paused for a moment and then quickly agreed. He just couldn’t back down from the challenge. It didn’t matter that this represented all of Will’s spending money for the entire semester -- the odds were in his favor and he knew it and as far as he was concerned the choice was clear.

So what happened? Well to make a long story short, the guy picked up the dice and without a word silently rolled himself an 11. Will paid him the next Monday and wound up having to work at the campus bookstore for the rest of the semester. I remember a few weeks later I ran into Will at work and we got to talking while he moved boxes around trying to look busy. I asked him if he and Dudley and were still friends.

“Sure,” he said, “But the guy’s a moron. Didn’t he realize the odds were in my favor?”

So there you have it. Will was quick to label Dudley a moron because he made a negative EV bet. What Will failed to realize, however, was that this guy certainly had the means to make \$15,000 bets, and ultimately wouldn’t have been all that impacted by the result were he to have lost. Will on the other hand, had no business making a \$15,000 bet that he stood to lose close to half the time. It didn’t matter that if he made the same bet 10,000 times over he’d almost certainly have come out well ahead, it only took making the bet one time to bankrupt him for the semester and render him incapable of staking any more craps games at all.

Dudley might very well have been foolish for having offered to make the negative EV bet, but Will on the other hand was foolish for having risked such a large chunk of his bankroll on the positive EV bet in the first place. Never mind that losing the bet forced Will to work in the bookstore, never mind that losing the bet forced Will to switch from his Heineken bottles to Milwaukee’s Best cans, losing the bet had probably the worst effect possible on an advantage bettor – decimating his bankroll.

Hopefully, this example helps illustrate a key concept that was touched on in the last article. Specifically, that expected value and expected growth are both key components of proper long-term wagering. Most bettors instinctively recognize the importance of expected value – most everyone realizes that betting 2-1 odds on a fair coin flip is “smart”, while betting 1-2 odds on a fair coin flip is not. But very few people consider as much as they should the expected growth of their bankroll due to their wagers they make. When a bettor places too much importance on the expected value and not enough on expected growth, he puts himself in danger of winding up in the same predicament as Will – pushing around boxes at the Brown Bookstore and trying to look busy, despite having made a indisputably “smart” bet when only considering EV alone.

But let’s go back to Will’s initial decision to make the \$15,000 bet. Certainly it’s pretty clear that making the bet was a mistake, but it should also be clear that because the bet had positive EV there was obviously a certain (lower) risk amount for which Will would have been making the right decision in accepting the wager. For a person with unlimited access to funds, the decision of how much to bet on a positive EV wager is easy – bet as much as possible. But for a person with a limited bankroll who wants to survive until the next day so he can continue staking craps games, the decision isn’t quite so obvious. That’s where Kelly comes in sports arbitrage

You’ll recall from Part I of this article the equation for expected growth:
E(G) = (1 + (O-1) * X)p * (1 - X)1-p - 1
Where X represents the percentage of bankroll wagered on the given bet and O the decimal odds.

For a player like Will, who has his basic necessities already paid for (food, shelter, clothing), his only real goal is to grow his bankroll as much and as quickly as possible. As such, Will’s objective would be to maximize the expected growth of his bankroll. The size of the bet (always given as a percentage of the player’s total bankroll) is known as the “Kelly Stake” and is a function of the bet’s payout odds and either win probability or edge1.

Mathematically , the formula for the Kelly stake is derived using calculus2. The actual mechanics are rather unimportant, but the result is that in order to maximize the growth of one’s bankroll when placing only one bet at a time, one should bet a percentage of bankroll equal to edge divided by decimal odds minus 1. (This is assuming the player has a positive edge. If he doesn’t his optimal bet is zero.) In other words:
Kelly Stake as percentage of bankroll = Edge / (Odds – 1) for Edge ≥ 0
Put in terms of win probability the equation becomes:3
Kelly Stake as percentage of bankroll = (Prob * Odds – 1) / (Odds – 1) for Probability * Odds ≥ 1
Let’s take a look at a few examples:
1. Given a bankroll of \$10,000 and an edge of 5%, then on a bet at odds of +100 one should wager 5% / (2-1) = 5% of bankroll, or \$500.
2. Given a bankroll of \$10,000 and a win probability of 55%, then on a bet at odds of -110, one should wager \$10,000 * (55% * 1.909091 - 1) / (1.909091-1) = 5.5% of bankroll, or \$550.
3. Given a bankroll of \$10,000 and a win probability of 25% then on a bet at odds of +350, one should wager \$10,000 * (25% * 4.5 - 1) / (4.5-1) ≈ 3.57% of bankroll, or about \$357.
4. Given a bankroll of \$10,000 and a win probability of 70% then on a bet at odds of -250, one should not wager anything because edge = win prob*odds = 70%*1.4 = 98% < 1.

Let’s look at all this a little more closely. Consider a bet at even odds (decimal: 2.0000) -- in this case, the bankroll growth maximizing Kelly equation simplifies to:
K(even odds) = Edge/(2-1) = Edge for Edge ≥ 0
In other words, when betting at even odds, the expected bankroll growth maximizing bet is equal to the percent edge on that bet. So if you have an edge of 5% on a bet at +100, then you should be wagering 5% of your bankroll. If your edge were only 2.5% then you should be wagering 2.5% of your bankroll. Now let’s consider a bet at -200, or decimal odds of 1.5:
K(-200 odds) = Edge/(1.5-1) = 2*Edge for Edge ≥ 0
So this means that for a bet at -200, the expected bankroll growth maximizing bet size would be twice the edge on the bet. Similarly, for a bet at -300, one should bet three times the edge, and for a bet at -1,000 one should bet ten times the edge.

This fits rather well with the manner in which many players size their relative bets on favorites. For a bet at a given edge if they were to bet \$100 at +100, they’d bet \$150 at -150, \$200 at -200, \$250 at -250, etc.

Now let’s consider bets on underdogs (that is, bets on money line underdogs -- bets paying greater than even odds). In the case of a bet at +200:
K(+200 odds) = Edge/(3-1) = ½*Edge for Edge ≥ 0
The optimal bet size is only half the edge. Similarly at a line of +300, the optimal bet size would be a third of edge, at +400 a quarter the edge, etc.

Now this is quite different from the manner in which many players choose to structure their underdog bets. If they were to bet \$100 on a line of +100, they might also bet \$100 on a bet with the same edge at +400. For a player wanting to maximize his bankroll growth, this is inappropriate behavior because it attributes, relatively , excessively large amounts to underdog bets. Assuming constant EV an expected growth maximizing player should only bet half of his +100 bet size at +200, and only a quarter his +100 bet size at +4004.

So what we see in the case of any bet (be it on an underdog or a favorite) is that the player should bet an amount such that the percentage of his bankroll he stands to win is the same as his percent edge. In other words, a player betting at an edge of 2% should place a bet to win 2% of his bankroll. This means that at -200 he’d be risking 4% of his bankroll, while at +200 he’d only be risking 1% of his bankroll. The rationale behind this should be clear when you consider the following example:

For a player betting at an edge of 5% and odds of -200, the proper Kelly stake is 10%. Over 100 bets, he has an expected return of 64.7% with a 36.7% probability of not turning a profit and a 3.4% probability of losing two-thirds or more of his stake.

For a player betting at the same 5% edge but at odds of +400, were he to bet the 10% stake of the -200 player, while he’d have the identical 64.7% expectation, he’d have a 73.5% probability of no profit, while his probability of losing two-thirds or more of his stake would be 55.8%.

Generalizing, for two same-sized bets of equivalent (positive) EV repeatedly made over time, there’s a higher probability associated with losing a given amount of money when making the longer odds bet.

Once again, we keep returning to the same simple but often overlooked point – expected value isn’t everything. Due to the fact that longer odds (for a given edge) imply greater a probability of loss, the Kelly bettor will bet less on longer odds and more on shorter odds. Any time an advantage player loses money he’s giving up opportunity cost as that represents money he can’t wager on +EV propositions down the line. As such the Kelly player will (for a given edge) always seek to minimize his loss probability over time by selecting the shorter odds bet, even though that necessitates risking more to win the same amount.

Taking the logic a step further, a Kelly player should be willing to even accept lower edge in order to play at shorter odds. For example:
• At odds of -200 (decimal:1.500) and an edge of 4%, the win probability would be p = (1+4%)/1.5 ≈ 69.33%, and Kelly stake would be K = 4%/(1.5-1) = 8%. This represents expected bankroll growth of:
(1+(1.5-1)*8%)69.33%*(1-8%)1-69.33% -1 ≈ 0.1624%
• At odds of +400 (decimal: 5.0000) and an edge of 10%, the win probability would be p = (1+10%)/5 = 22%, and Kelly stake would be K = 10%/(5-1) = 2.5%. This represents expected bankroll growth of:
(1+(5-1)*2.5%)22%*(1-2.5%)1-22% -1 ≈ 0.1221%

So what this tells us is that a Kelly player would prefer (and by a decent margin) 4% edge at -200 to 10% edge at +400.

In this article we’ve introduced Kelly staking. This represents a methodology for sizing bets in order to maximize the expected future growth rate of a bankroll5. The bet sizes determined by Kelly will necessarily not maximize expected value, because doing so would require betting one’s entire bankroll on every positive EV wager that presented itself. This would eventually lead to bankruptcy and the inability to place further positive EV wagers.

We’ve seen that Kelly may also be utilized to gauge the relative attractiveness of several bets. What we see is that for a given edge, an expected growth maximizing bettor will prefer the bet with shorter odds (in other words, the bigger favorite). This result, derived entirely from first principles, may be surprising to some advantage players who’ve come to find wagers on underdogs generally more profitable than bets on favorites. While our conclusion in no way precludes the possibility that underdogs may in general provide superior return opportunities than favorites, the fact that for two bets of equal expected return the bet on the favorite will yield greater expected bankroll growth is indisputable and needs to be acknowledged by all those seeking to manage bankroll risk.

In Part III of this series we’ll discuss how one may generalize Kelly so it may be applied to a greater range of circumstances including multiple simultaneous bets, multi-way mutually exclusive outcomes, and hedging.

Click to hide footnotes

 Technically, because odds, edge, and win probability are linked by way of the equality Odds * Prob = 1 + Edge, any two of these variables could be used to determine the Kelly stake.The calculus is rather simple. We need to maximize E(G) = (1 + (O-1) * X)p * (1 - X)1-p - 1 with respect to X, subject to X lying on the unit interval [0,1]. To simplify the analysis, however, we can take the natural log of both sides of the equality and seek to maximize the log of expected growth. This is equivalent because the log function is monotonically increasing. So our problem becomes:Maximize wrt X: log(Growth) = p*log(1 + (O-1) * X) + (1-p)*log(1 - X) s.t. 0 ≤ X ≤ 1 which gives us: dlog(G)/dX = p*(O-1)/(1 + (O-1) * X) - (1-p)/(1 - X) setting to zero and solving yields: X = (Op-1)/(O-1) with d2log(G)/dX2 ≤ 0 for all feasible 0 ≤ X < 1This may also be extended to include bets that include a third push outcome where the at-risk amount is returned to the bettor in full (such as in the case of an integer spread or total). In order to generalize this article to include bets with ternary outcomes, one need only consider the "probability of winning conditioned on not pushing" instead of pure "win probability". In general, given a win probability of PW, a loss probability of PL, and a push probability of PT (where PW + PL + PT = 1), then the probability of winning conditioned on not pushing would be:P*W = PW / (1 - PT)and the probability of losing conditioned on not pushing would be:P*L = PL / (1 - PT)So assuming decimal odds of O, Edge would be:Edge = O × PW / (1 - PT) - 1     -or- Edge = O × PW - (1 - PT)which in either case is just the same as:Edge = O × P*W - 1And the Kelly stake would remain unchanged as:Kelly Stake as percentage of bankroll = Edge / (Odds – 1) for Edge ≥ 0So why do so few players do this? It’s my opinion that the only explanation for this inconsistent behavior (risking the same amount on all underdogs while betting to win the same amount on favorites) is the manner in which US-style odds are quoted. Odds of -200 imply one would need to bet \$200 to win \$100 so it would seem to make sense to bet in increments of that \$200. Odds of +200 imply one would need to bet \$100 to win \$200, and so it would seem to make sense to bet in increments of that \$100. What if, however, US odds on under dogs were also quotes as negative numbers? What if a +200 underdog were written as a -50 underdog (meaning a player would need to risk \$50 to win \$100) and a +400 dog as a -25 dog? The two methods for expressing odds are obviously identical, but it’s my belief that if odds were quoted in this manner you’d have far fewer bettors undertaking the questionable practice of betting an equivalent dollar amount on all underdogs.An equivalent way of looking at this is that Kelly maximizes both the bettor’s median and modal future bankroll over a large number of bets. In other words, applying expected bankroll growth to the current bankroll yields both most likely bankroll outcome (the mode) and the outcome which has an equal likelihood of being outperformed and underperformed.

26. I feel so lost in this thread.

27. I think I'm just going to use that calculator, lol.

28. bump

29. I miss ganchrow

30. I wonder what he does now.

\$20
Angelman
donation 02/18/2019

31. Originally Posted by Enkhbat
I wonder what he does now.
Risk mgmt at a book in CR

32. Originally Posted by nanyin2017
I miss ganchrow
He'll be back.

\$20
Angelman
donation 02/18/2019

33. Props to Ganch. At least the guy was trying.

Thought-provoking stuff. One example I'll present:

* Betting a -200 Fav w/ 70% Win Probability. Compare to +200 Dog w/ 35% Win Probability.

Which is a better risk?

Well, each bet is a +5% ROI. So, we're even on Expectation. The -200 Fav is a MUCH better risk.

One vantage point is to view how your Bankroll will oscillate over 100 similar plays. You can chart this on a Binomial Distribution. On the string of -200 Favs, I'm very unlikely to show a Net Loss on 100 plays.

On 100 +200 Dogs, I could easily show a Net Loss, just based on bad variance.

\$20
Angelman
donation 02/18/2019

34. Originally Posted by ChuckyTheGoat
Props to Ganch. At least the guy was trying.

Thought-provoking stuff. One example I'll present:

* Betting a -200 Fav w/ 70% Win Probability. Compare to +200 Dog w/ 35% Win Probability.

Which is a better risk?

Well, each bet is a +5% ROI. So, we're even on Expectation. The -200 Fav is a MUCH better risk.

One vantage point is to view how your Bankroll will oscillate over 100 similar plays. You can chart this on a Binomial Distribution. On the string of -200 Favs, I'm very unlikely to show a Net Loss on 100 plays.

On 100 +200 Dogs, I could easily show a Net Loss, just based on bad variance.
Scott has learned and forgotten more about stats and gaming theory than all of the SBR collective. To say he was trying is somewhat accurate, he was like a college math professor trying to teach kindergarteners Differential Equations or Calculus of Variations

First 123 Last
Top