[Ganchrow]

Which of these two bet would you consider "riskier"?

For the sake of discussion you can assume the player has a $10,000 bankroll.

[juuso]

$100 wager with a 5% edge at odds of +1,000 is "riskier" IMO.

-1000 bet has a lot lower variance and hits at much higher clip. 5% edge at -1000 also better than 5% edge at +1000

[HedgeHog]

I'd say the "risk" level was equal. The "comfort" level ,however, may be different as one grand may be a lot of money to some--perhaps more than they can stand to lose (even if the chance of loss is under 10%).

[Ganchrow]

Originally Posted by HedgeHog

The "comfort" level ,however, may be different as one grand may be a lot of money to some--perhaps more than they can stand to lose (even if the chance of loss is under 10%).

The issue of a "comfort level" is of course separate (although related to) risk.

For the purposes of determining a "comfort level" you can assume a $10,000 bankroll.

[HedgeHog]

Originally Posted by Ganchrow

The issue of a "comfort level" is of course separate (although related to) risk.

For the purposes of determining a "comfort level" you can assume a $10,000 bankroll.

Ok, I was assuming this was a one bet deal. If this is a series of bets then the -1000 bets are less risky--as defined by higher likelihood to show a profit over dozens (or more) of bets.

PS And even if it were just a 1 bet thing, the -1000 bettor has a greater than 90% of profitability compared to less than 10% chance for the big dog bettor.

[ShamsWoof10]

I would "assume" the -1000 is a greater risk... Just because someone is favored does not mean they will be the better team on that day....

The problem I have with sports probabilities is they are not accurate... There are too many variables that can not possibily be known before gametime... The data output is only as good as the data imput... Having "coached" and "played" as well as being a bettor I always use to wonder... "if we were a college or pro team there is NO WAY ...AAAAAAANY handicapper could factor in what goes on here behind closed doors that makes all the differance in the world during the game...

In cards it's differant.. If a certain number of cards are gone then what's left is a certainty... In sports you can have a bet that Steve Nash will make his next FT and even if he knows and will do it to help you out he still can miss... He's a 90some percent FT shooter so even if he knows and wants to make it to help you out his percentage can't go much higher to 100%... It actually might go down if he gets to nervous...

Probabilities are cute but that's about it...

[CrazyLou]

Option B seems too standard/obvious...

And because it is a Ganchrow question I know it has to be what makes less sense on the surface, and in some advanced mathematical way is actually theoretically correct. So my vote is option A.

[LT Profits]

Assuming you are a good enough handicapper that both 5% edges are accurate, I would say both risks are equal. You expect to gain 5% in either case in the long run.

[Ganchrow]

Originally Posted by LT Profits

I would say both risks are equal. You expect to gain 5% in either case in the long run.

Might it be relevant that you're risking ten times more on the -1,000 bet than on the +1,000 bet ($1,000 versus $100 respectively) ?

[20Four7]

It comes down to your definition of risk.

Risk amounts to the chances of a negative impact to an asset (bankroll). Therefore having 10% of your bankroll riding on a bet would appear to be riskier than having 1% of your bankroll riding on a bet.

If your looking at volitility of returns etc, than the dog would show more volitility in my opinion.
As a smaller percentage of your funds are at risk you should be able to sustain the volitility and therefore would be less "risky"

Now if ganch wants some mathematical "proof" this is the case I'll pass.

[HedgeHog]

Originally Posted by LT Profits

Assuming you are a good enough handicapper that both 5% edges are accurate, I would say both risks are equal. You expect to gain 5% in either case in the long run.

I agree that "profit expectation" is equal for the long run. But I would think that risk would be higher during the interim with a +1000 bettor due to the inevitable long losing streaks-- where 20-30 failures in a row would not be unusual.

PS 20Four7 beat me to the punch!

[HedgeHog]

Originally Posted by Ganchrow

Might it be relevant that you're risking ten times more on the -1,000 bet than on the +1,000 bet ($1,000 versus $100 respectively) ?

Might it also be relevant that the +1000 bettor is ten times more likely to lose each time? You're toying with us Ganch.

[Ganchrow]

Originally Posted by 20Four7

Risk amounts to the chances of a negative impact to an asset (bankroll). Therefore having 10% of your bankroll riding on a bet would appear to be riskier than having 1% of your bankroll riding on a bet.

If you were to measure your risk in such a manner ("the chances of a negative impact to an asset (bankroll)") then wouldn't the -1,000 bet be less risky in that you'd only lose that bet approximately 4.545% of the time versus 90.455% of the time with the +1,000 bet?

[Justin7]

You can precisely define risk. But another way to look at it:

Which wager is closer to the "optimal" wager?

[Ganchrow]

Originally Posted by HedgeHog

Might it also be relevant that the +1000 bettor is ten times more likely to lose each time? You're toying with us Ganch.

Do those two factors (risking 10x as much on a bet winning about 9.48x more frequently) cancel each other when it comes to figuring a dollar expected return?

Think about it this way ... assuming no chance of bankruptcy, would you expect to have more money betting betting 10 units on the favorite many, many times over or 1 unit on the underdog the same many, many times over?

[Ganchrow]

Originally Posted by HedgeHog

You're toying with us Ganch.

And my intention is certainly not to "toy" with anyone here, but rather to stimulate discussion and critical thinking on a topic of great import to successful sports betting.

Everyone talks about the necessity of proper bankroll management, but many people appear to have misconceptions about what that really entails.

[donjuan]

The $100 bet is far riskier. You are betting Kelly*2 whereas with the $1000 bet you are betting Kelly/5. With the $1000 bet, your expected growth per bet is $59, whereas with the $100 bet your expected growth is $0.

[donjuan]

Originally Posted by ShamsWoof10

I would "assume" the -1000 is a greater risk... Just because someone is favored does not mean they will be the better team on that day....

The problem I have with sports probabilities is they are not accurate... There are too many variables that can not possibily be known before gametime... The data output is only as good as the data imput... Having "coached" and "played" as well as being a bettor I always use to wonder... "if we were a college or pro team there is NO WAY ...AAAAAAANY handicapper could factor in what goes on here behind closed doors that makes all the differance in the world during the game...

In cards it's differant.. If a certain number of cards are gone then what's left is a certainty... In sports you can have a bet that Steve Nash will make his next FT and even if he knows and will do it to help you out he still can miss... He's a 90some percent FT shooter so even if he knows and wants to make it to help you out his percentage can't go much higher to 100%... It actually might go down if he gets to nervous...

Probabilities are cute but that's about it...

Are you seriously trying to make a run for dumbest post of the year?

[donjuan]

Originally Posted by LT Profits

Assuming you are a good enough handicapper that both 5% edges are accurate, I would say both risks are equal. You expect to gain 5% in either case in the long run.

You are confusing expected value with expected growth.

[Ganchrow]

Originally Posted by donjuan

The $100 bet is far riskier. You are betting Kelly*2 whereas with the $1000 bet you are betting Kelly/5. With the $1000 bet, your expected growth per bet is $59, whereas with the $100 bet your expected growth is $0.

You appear to be basing your answer on the presumption that a double-Kelly bet will always be riskier than a fifth-Kelly bet.

Is this valid?

[durito]

Don Juan is correct.

The higher the odds, the more variability in your finishing bankroll. Therefore the underdog here is a more risky bet.

I believe this situation comes up a lot, as you see a lot of people using a money management system where they bet to win the same amount (assuming the same edge) regardless of the odds. Which, please correct me if I'm wrong Ganch, is not the optimal (kelly) way to grow your bankroll.

[Ganchrow]

Originally Posted by durito

I believe this situation comes up a lot, as you see a lot of people using a money management system where they bet to win the same amount (assuming the same edge) regardless of the odds. Which, please correct me if I'm wrong Ganch, is not the optimal (kelly) way to grow your bankroll.

Assuming identical edges, then betting to win the same amount regardless of odds is in keeping with Kelly. Risking the same amount regardless of odds is not.

[donjuan]

Originally Posted by Ganchrow

You appear to be basing your answer on the presumption that a double-Kelly bet will always be riskier than a fifth-Kelly bet.

Is this valid?

IIRC, double Kelly will always be riskier than fifth Kelly because double Kelly has zero or negative expected bankroll growth while fifth Kelly would always have positive expected growth. However, my answer was not quite correct because given two bets at full Kelly with the same edge, you would prefer the one that has a better chance of winning because it will have higher expected growth.

[durito]

OK, I ****ed that up and that makes sense now that I think about it.

The $100 bet is the riskier play here -- as if he were betting to win the same amount, he'd only risk $10 on the +1000 play.

[ShamsWoof10]

Originally Posted by donjuan

Are you seriously trying to make a run for dumbest post of the year?

Hey Don Juan or anyone else out there... Why don't you prove it...? Start a thread and post a pick in that thread over so many bets just to see what the results are.. If you are not willing to prove it then I don't want to hear it...

If I have to go out and get a 20 oz. bottle of Coke Classic and put a nail in it to prove what I believe (I'll be posting the results of that on Sat.) then F*CK YEAH you or someone has to prove this probable bullsh*t!!!!

[donjuan]

Shamswolf,

There are a couple of things here:

1. Each bet has a different edge.
2. I would have to post thousands of bets to "prove" this to you.
3. Posting my bets is -ev for me, so I won't do it.

[ShamsWoof10]

Originally Posted by donjuan

2. I would have to post thousands of bets to "prove" this to you.

THOUSANDS OF BETS JUST TO PROVE IT..??? Since it seems like you know... how many THOUSANDS OF BETS will it take it prove it...?

Originally Posted by donjuan

3. Posting my bets is -ev for me, so I won't do it.

Then don't f*ckin' quote me like that and keep your mouth to yourself...

Is anyone else willing to PROVE IT or are we all mouths here..?

[donjuan]

Originally Posted by ShamsWoof10

THOUSANDS OF BETS JUST TO PROVE IT..??? Since it seems like you know... how many THOUSANDS OF BETS will it take it prove it...?

Ever heard of variance?

Then don't f*ckin' quote me like that and keep your mouth to yourself...

Is anyone else willing to PROVE IT or are we all mouths here..?

I pointed out that it was an incredibly idiotic statement, which it was. You don't understand that math can be applied to sports. We get it. Now please go back to hitting 75,000% on your halftime and quarters bets while the rest of us learn something from Ganchrow.

[ShamsWoof10]

Do you understand ..

PROVE IT..???????????

and...by the way I am trying to learn something here but I want to see it proven... I don't believe it because his calculator says so... If it's valid all I ask is for it to be proven..

[Ganchrow]

Originally Posted by donjuan

IIRC, double Kelly will always be riskier than fifth Kelly because double Kelly has zero or negative expected bankroll growth while fifth Kelly would always have positive expected growth.

Does higher expected growth always imply lower risk? Isn't one advantage of logarithmic (full-Kelly) utility over quadratic (Markowitz) utility that the former takes into account higher-order moments? As such isn't it possible that due to (let's say) a sufficiently kurtotic outcome distribution the variance of a zero-growth +EV bet might nevertheless be lower than that of a higher growth bet with equivalent EV?

Originally Posted by donjuan

given two bets at full Kelly with the same edge, you would prefer the one that has a better chance of winning because it will have higher expected growth.

Compare (odds=+100, p=52.5%) with (odd=+1,000, p=13.636%) under full-Kelly.

[Ganchrow]

Originally Posted by ShamsWoof10

PROVE IT..???????????

and...by the way I am trying to learn something here but I want to see it proven... I don't believe it because his calculator says so... If it's valid all I ask is for it to be proven..

I don't understand what you're asking be proven.

[ShamsWoof10]

I would like to know or have proven which is more risker or if there is any differance at all.... Maybe as I said a thread posting only -/+1000 ML's to see what happens over time..

What is wrong with that..?

[HedgeHog]

Define "risk". And do the # of trials matter, say 1 vs 1000. Question may be ambiguous.

[Ganchrow]

Originally Posted by ShamsWoof10

I would like to know or have proven which is more risker or if there is any differance at all.... Maybe as I said a thread posting only -/+1000 ML's to see what happens over time..

What is wrong with that..?

There's nothing wrong with that, it's just that it's much easier to deduce an answer from first principles.

One could ask the question whether it's more likely for a fair roulette spin to land on red than it is for it to land on first third. Now while you and I could go take measurements at a casino, it would be considerably easier just to utilize the tools of probability to determine an answer without even looking at a wheel.

[ShamsWoof10]

Originally Posted by Ganchrow

There's nothing wrong with that, it's just that it's much easier to deduce an answer from first principles.

One could ask the question whether it's more likely for a fair roulette spin to land on red than it is for it to land on first third. Now while you and I could go take measurements at a casino, it would be considerably easier just to utilize the tools of probability to determine an answer without even looking at a wheel.

That may be but, as in class, they just don't go by the lecture although I wish they would because it's easier... They have the nerve to make you prove it in a lab and that is harder...

I would like to see things proven is all... You might see or notice something you normally would not just by going the easy way...