Can you win money on a bad bet?

  • Thread starter techmologist
  • Start date
  • Tags
    Money
In summary, the Kelly betting system suggests that sometimes it is advantageous to bet a fraction of your money on a bet with negative expected value. This is demonstrated in an application where four horses with varying win percentages and odds are bet on using a fixed fraction and reinvesting winnings. The system may seem counterintuitive, but it is based on the idea of betting on mutually exclusive outcomes and taking into account the odds set by a bookmaker. While the past performance of horses may not always be a reliable indicator of future success, this system can still be useful in making strategic bets. Ultimately, it is important to also consider factors such as buyer sentiment and overall knowledge of the sport when making betting decisions.
  • #1
techmologist
306
12
I'm just now learning about the Kelly betting system, and I can't wrap my mind around the result that sometimes you should bet a fraction of your money on a bet with negative expected value. Here is an application of the system when you can bet a fixed fraction on any combination of mutually exclusive outcomes, reinvesting your winnings (if any) each time:

The same four horses run a long series of races on the same track. You know from tracking past results for a long time that the win percentages for the horses are

horse 1: 28%
horse 2: 55%
horse 3: 15%
horse 4: 2%

A bookmaker offers the following odds on the horses:

horse 1 at 3:1
horse 2 at 1:1
horse 3 at 5:1
horse 4 at 7:1

If past winning percentages can be taken as the probability that a horse wins, then bets on horses 1 and 2 have positive expectation, while bets on horses 3 and 4 have negative expectation. So you would think one would bet on horse1, or maybe horses 1 and 2, avoiding 3 and 4. But according to the Kelly strategy you do best in the long run if you bet 22% of your current wealth on horse 1, 43% on horse 2, and 11% on horse 3. The remaining 24% is not bet. I simulated this strategy and it works, beating strategies where you don't bet on horse 3. I understand the mathematical derivation of the formula, but I don't understand how it could be true. It is just too counterintuitive. Any thoughts?

Also, notice that there is a "take", in the sense that the odds add up to more than 100%. So there are no cancelling bets in which you get back exactly what you bet, no matter what the outcome.
 
Mathematics news on Phys.org
  • #2
techmologist said:
Here is an application of the system when you can bet a fixed fraction on any combination of mutually exclusive outcomes, reinvesting your winnings (if any) each time:

The key words here are "mutually exclusive". In other words, in anyone race, you can only win one of your bets, and if there are no other horses in the race you are guaranteed to win one of the 4 bets (but of course you might not win enough to cover the cost of making all four bets).

Your simple "avoid negative expectations" strategy ignores those facts, and allocates your money as if each of the 4 horses is running in a different race. In that situation, you could have any number of winnning bets from 0 to 4.
 
  • #3
Yeah, that's true. I guess it would be an example of hedging, right? I was familiar with the idea of betting on negatively correlated events to reduce the variance, but I was still thinking that both bets needed to have positive expectation for the result to be better than simply betting on the outcome with the greatest expectation by itself. This idea of betting on something with negative expectation still seems counterintuitive to me.
 
  • #4
techmologist said:
I'm just now learning about the Kelly betting system, and I can't wrap my mind around the result that sometimes you should bet a fraction of your money on a bet with negative expected value. Here is an application of the system when you can bet a fixed fraction on any combination of mutually exclusive outcomes, reinvesting your winnings (if any) each time:

The same four horses run a long series of races on the same track. You know from tracking past results for a long time that the win percentages for the horses are

horse 1: 28%
horse 2: 55%
horse 3: 15%
horse 4: 2%

I can't address the specifics of your abstract mathematical problem. But I can offer some practical advice.

The past performance of horses reveals much less than you'd think. It's common for owners and trainers to hold back a horse in a series of races to build up the price in some future big-stakes race. In other words one of the tricks of the trade in professional horse racing is to make horses look worse than they really are.

The other interesting thing you said is that the bets are placed with a bookmaker. That introduces another element.

At the track, the odds are set via pari-mutual betting. All the bettors' money goes into a pool, and the pool is split by those betters holding the winning tickets. Of course the pool's paid out after the state, the racing association, the track, and who knows who else gets their cut. I think around 85-90% of the pool gets paid out. So on the one hand, over the long term the odds are against you just based on the healthy chunk of cash raked off by the government and private entities.

But given that, this is a very fair betting system. The odds directly reflect buyer sentiment. It's literally a perfect market.

But when you go to a bookmaker, you're just dealing with some guy who a) makes up his own odds according to his whims and typically to his benefit; and b) is probably doing something illegal, unless you're at a sports book at a legal casino.

So if you're just doing a math problem, feel free to ignore my comments.

But if you are planning to spend any of your money on this, you'd be better off hanging around at the race track rather than studying probability theory. There's a lot more to horse racing than statistics.
 
Last edited:
  • #5
SteveL27 said:
...But if you are planning to spend any of your money on this, you'd be better off hanging around at the race track rather than studying probability theory. There's a lot more to horse racing than statistics.

I'd like to do both, really. Are you trying to tell me I can't have it both ways? Thank you for your input, though. The abstract mathematical problem that I put forward in the first post is very unrealistic, I agree. But I'm hoping that even a toy problem like that can increase my intuition about gambling/investing.

As someone who gambles, don't you think it is strange that there are situations where you should bet when according to the payout and the chances of winning, you are getting the worst of it?
 
  • #6
Maybe for gambling, but if you view investments as some sort of gamble, then it makes perfect sense. Kelly Criteria ends up being fairly important in investment theory, and is often a quick and nifty way to add diversity. Compared to a lot of mutual funds, you tend to have a fairly good chance of making a fairly decent return using Kelly's Criteria.
 
  • #7
MarneMath said:
Maybe for gambling, but if you view investments as some sort of gamble, then it makes perfect sense...

Can you explain a little how it makes perfect sense? I don't make any distinction between gambling and investing, so it doesn't make sense in either context to me. I only know that it works somehow. And I can follow the individual steps in the mathematical derivation. But I don't have the insight.

The only practical difference I see between gambling and investing is the size of the markets. For the Kelly criterion to apply, you need to be able to make arbitrarily large bets without completely destroying the favorable situation, or running into some kind of maximum bet limit.
 
  • #8
I just want to make something very clear, when I say make perfect sense, I don't intend to imply that you can come up with Kelly's Criteria from an investment perspective, or that Kelly's Criteria is the end of all investment strategies for everyone. When I say, it makes perfect sense, you have to look at how investors invest.

The idea is this. When you invest, you can different routes you can take, you can put all your money in a stock and hope for it to raise or you can put it all in a bond and know you'll receive a return(risk-free alternative.) More than likely you will choose a mix between these and various other options out there.

What Kelly Criteria allows you to do is quantity how to hedge your bets. Assuming a symmetric continuous probability distrbution, you can find the optimal way to disturb your funds. In fact, you can even use it to determine when a short sell and or long buy should happen.

Overall, it's a useful too, as I said before, to disverify yourself and gives you a reasonable way on how to diversify based on your history.

Is it perfect, nope. The market is not symmetric, there exist randomness and violatity, and it probably won't make you richer than your wildest dreams, but it serves its purpose to mitigate possible losses, while still producing growth.
 
  • #9
techmologist said:
I'm just now learning about the Kelly betting system, and I can't wrap my mind around the result that sometimes you should bet a fraction of your money on a bet with negative expected value. Here is an application of the system when you can bet a fixed fraction on any combination of mutually exclusive outcomes, reinvesting your winnings (if any) each time:

The same four horses run a long series of races on the same track. You know from tracking past results for a long time that the win percentages for the horses are

horse 1: 28%
horse 2: 55%
horse 3: 15%
horse 4: 2%

A bookmaker offers the following odds on the horses:

horse 1 at 3:1
horse 2 at 1:1
horse 3 at 5:1
horse 4 at 7:1

If past winning percentages can be taken as the probability that a horse wins, then bets on horses 1 and 2 have positive expectation, while bets on horses 3 and 4 have negative expectation. So you would think one would bet on horse1, or maybe horses 1 and 2, avoiding 3 and 4. But according to the Kelly strategy you do best in the long run if you bet 22% of your current wealth on horse 1, 43% on horse 2, and 11% on horse 3. The remaining 24% is not bet. I simulated this strategy and it works, beating strategies where you don't bet on horse 3. I understand the mathematical derivation of the formula, but I don't understand how it could be true. It is just too counterintuitive. Any thoughts?

Also, notice that there is a "take", in the sense that the odds add up to more than 100%. So there are no cancelling bets in which you get back exactly what you bet, no matter what the outcome.


It seems to be a sort of hedging, where the idea is to reduce the variance. Reduced variance is often called reduced "risk" Note that by this definition a bet that is a certain loser has no risk, because it has no variance.

In gambling in casinos and racetracks your expectation is always negative, and there is usually no system to beat it. If there were, the casino or racetrack would not be in business. To make things worse the casino or racetrack also has an incentive to cheat, and is unlikely to get caught. Recently a big online poker company got caught stealing millions from their customers, so yes, it surely happens.

It is possible to make money in games of skill like poker. I used to hang out (virtually) with people who had done this. They ALL agreed that it was a losing game. There were two scenarios. Either you went bust, or you caused other people to go bust and had wasted your life playing poker. So in the end you could not win no matter what. Some day someone you busted is going to go back to his hotel room and shoot himself dead.

If this sort of think interests you then investing in stocks seems like a better deal, as overall there is a positive expectation. But paradoxically, small players also tend to lose all their money even with a small positive expectation.

To sum it all up: in the long run you are CERTAIN to lose all your money if you persist in making bad bets. The appeal of such things has always been a mystery to me. I guess people like the excitement. Losing gives them an emotional kick, and they are willing to pay for that.
 
  • #10
Okay. Let's take stock of what has been learned:

1) I shouldn't gamble because for one person to win, someone else must lose.
2) I shouldn't study math because it's a waste of time.

and more to the point

3) The Kelly system can be useful for investing because it helps you diversify.

I don't want to argue against any of these observations. Each is undeniably true and admirable in its own way. But none of them answers my question: It is downright surprising to me that when you have several bets to choose from, some of which tend to gain money in the long run and some of which tend to lose money, that sometimes you do best by betting on the "good bets" and one of the "bad bets". Better than you would do if you spread your money out among the good bets (which are themselves negatively correlated) and left the bad bet alone. This is really so obvious to everyone else? John Kelly, who was an early advocate of the Kelly system, thought it was noteworthy enough to remark in his http://www.bjmath.com/bjmath/kelly/kelly.pdf:

Kelly said:
It should be noted that...some bets might be made for which...the expected gain is negative. This violates the criterion of the classical gambler who never bets on such an event.


AlephZero actually read my question and tried to answer it...

AlephZero said:
The key words here are "mutually exclusive". In other words, in anyone race, you can only win one of your bets, and if there are no other horses in the race you are guaranteed to win one of the 4 bets (but of course you might not win enough to cover the cost of making all four bets).

But that last part in parentheses is the whole point. My (evidently wrong) intuition says that the consolation of knowing that you will win at least one of your bets does not make up for the fact that you are losing money on a losing bet. Yes, when all the "winning bets" fail to pay off, your "losing bet" comes through. But I still find it hard to see how that could ever compensate for all the money you have to throw away on the losing bet most of the time. But somehow, something similar to that happens in the example I gave above (you bet on 3 horses, not all 4). It works, and I don't get it.
 
  • #11
techmologist said:
But that last part in parentheses is the whole point. My (evidently wrong) intuition says that the consolation of knowing that you will win at least one of your bets does not make up for the fact that you are losing money on a losing bet. Yes, when all the "winning bets" fail to pay off, your "losing bet" comes through. But I still find it hard to see how that could ever compensate for all the money you have to throw away on the losing bet most of the time. But somehow, something similar to that happens in the example I gave above (you bet on 3 horses, not all 4). It works, and I don't get it.

This reminds me of the investor and writer Nassim Nicholas Taleb, who as an investor, made his living betting on "Black Swan" events ... events that are extremely unlikely, but that pay off hugely when they come in.

As an investor, Taleb would come to work every day, place his trades, (which invariably lost a small amount of money every day), and spend the rest of the day working on his mathematical theory of making money from extremely unlikely events.

Every so often, one of those events would happen. All the conventional investors would get killed, and Taleb would collect.

http://en.wikipedia.org/wiki/Nassim_Nicholas_Taleb

You might take a look at his books. One is called Fooled by Randomness: The Hidden Role of Chance in Life and in the Markets.

https://www.amazon.com/dp/0812975219/?tag=pfamazon01-20

and the other is The Black Swan: The Impact of the Highly Improbable.

https://www.amazon.com/dp/081297381X/?tag=pfamazon01-20

If I'm understanding you correctly, this is the phenomenon you're talking about. I'm not sure if Taleb's Black Swan events have a negative expectation. But typically they're regarded as so unlikely that nobody bothers to calculate them at all.

Also what you're saying reminds me of hedging. Investors buy stocks and then buy options against the stocks. If the stocks go up, the options are like unused insurance. If the stocks go down, the options cover the loss.

So buy making the "bad bets" you are in effect trading some of your potential winnings for a protection against an adverse effect.

Actually just think of car insurance. It's a terrible bet for the buyer. The insurance companies make sure of that! But you still buy car insurance. You need to insure against a rare outcome.
 
Last edited by a moderator:
  • #12
SteveL27 said:
This reminds me of the investor and writer Nassim Nicholas Taleb, who as an investor, made his living betting on "Black Swan" events ... events that are extremely unlikely, but that pay off hugely when they come in.

As an investor, Taleb would come to work every day, place his trades, (which invariably lost a small amount of money every day), and spend the rest of the day working on his mathematical theory of making money from extremely unlikely events.

Every so often, one of those events would happen. All the conventional investors would get killed, and Taleb would collect.

http://en.wikipedia.org/wiki/Nassim_Nicholas_Taleb

You might take a look at his books. One is called Fooled by Randomness: The Hidden Role of Chance in Life and in the Markets.

https://www.amazon.com/dp/0812975219/?tag=pfamazon01-20

and the other is The Black Swan: The Impact of the Highly Improbable.

https://www.amazon.com/dp/081297381X/?tag=pfamazon01-20

If I'm understanding you correctly, this is the phenomenon you're talking about. I'm not sure if Taleb's Black Swan events have a negative expectation. But typically they're regarded as so unlikely that nobody bothers to calculate them at all.

Also what you're saying reminds me of hedging. Investors buy stocks and then buy options against the stocks. If the stocks go up, the options are like unused insurance. If the stocks go down, the options cover the loss.

So buy making the "bad bets" you are in effect trading some of your potential winnings for a protection against an adverse effect.

Actually just think of car insurance. It's a terrible bet for the buyer. The insurance companies make sure of that! But you still buy car insurance. You need to insure against a rare outcome.


Now we are getting somewhere. I'm glad you brought up Taleb. I read The Black Swan a couple of years ago and would benefit from reading it several more times. Haven't gotten around to Fooled By Randomness yet.

Yes, that does seem similar to what I'm talking about. But I assumed when I was reading Taleb that his bets actually had a positive expectation simply because they are so profitable when they finally pay off. At that point I had never read Kelly's paper (linked to in my previous post), so it was "obvious" to me that any bet that you can consistently make money with must have a positive expectation. I am surprised that there are exceptions to that rule, however unlikely such a situation may be in the real world.

I like the insurance example. I have car insurance, but I'm not hoping I'll get in a wreck just so I'll get my money's worth :)
 
Last edited by a moderator:
  • #13
And as for the hedging, I have been accustomed to thinking of hedging as something you do to sleep better at night. You give up some of the profitability to reduce risk, because people usually don't like to be exposed to huge losses even when the expected value is high. But in my example above, the "hedge" actually maximizes the long-term growth rate. This is the part that is surprising to me. Maybe I am just unsophisticated. Compared to this, the Monty Hall problem seems easy to grasp. However, I was exposed to that problem at a younger age (as homework), so I may have missed out on the opportunity to be wowed by it.
 
  • #14
So let me get this straight, if you have a market with;

1/1
3/1
5/1
7/1

and you bet on horses 1+2 then you get a 0.75 overround and a 55+28 = 83% strike rate. Throw in horse 4 and now you have a 0.875 overround and an 85% strike rate which makes it unprofitable.


Ok.

1. Your market is crap. The overround for it is 104% and a typical bookies will have it set at around 114%

1/1
2/1
4/1
8/1

(114%) This is a better book, use this.


2. Horse strike rates are generally much lower than the odds suggest! There is a website (google it, i forgot the name) which tracks data from all of the horse races in the uk. I recall seeing that horses with an SP of even money have around a 45% strike rate. Your strike rate figures may be overestimated.


3. If you are right with those strike rate figures then the 1/1 horse will be that price (despite winning 55% of the time) because other people don't fancy the horse! There may be a number of factors you fail to consider and thus a situation where you believe you have positive expectation actually ends up in you having significantly negative expectation.

4. You still have to get over that overround figure! 114% is a large edge for the bookies...

Using my new book above the overround for horses 1+2 is 0.8333 and their strike rate is 83%, so by simply making a more realistic book I have made it pretty much break even. DON'T FORGET YOUR RATHER GENEROUS STRIKE RATE FIGURES! I just read those figures more as probability of winning rather than strike rate in previous races. Even so, getting 55% for a 1/1 horse is buying money and should be encouraged wherever possible!

5. Even if you think you can read the stats, don't forget that all the bookies are in regular contact with each other when it comes to setting prices. They also employ people who sit there trawling through even stat possibly just to make as tight a book as possible. Horses are usually subject to guide prices which use stats, but admittedly the punters have more control in the prices. Suffice to say, if a horse with a 55% chance of winning is running at 1/1 then you take it! Realistically this horse would be priced at around 1/2 or 8/13



I mean ok if you can find situations like this then bloody well go for it! With a book of 104% you could probably make a bit of money on betfair or any other betting exchange with ease. They just don't exist :p:
 
  • #15
techmologist said:
Okay. Let's take stock of what has been learned:

1) I shouldn't gamble because for one person to win, someone else must lose.
2) I shouldn't study math because it's a waste of time.

and more to the point

3) The Kelly system can be useful for investing because it helps you diversify.

I don't want to argue against any of these observations. Each is undeniably true and admirable in its own way. But none of them answers my question: It is downright surprising to me that when you have several bets to choose from, some of which tend to gain money in the long run and some of which tend to lose money, that sometimes you do best by betting on the "good bets" and one of the "bad bets". Better than you would do if you spread your money out among the good bets (which are themselves negatively correlated) and left the bad bet alone. This is really so obvious to everyone else? John Kelly, who was an early advocate of the Kelly system, thought it was noteworthy enough to remark in his http://www.bjmath.com/bjmath/kelly/kelly.pdf:




AlephZero actually read my question and tried to answer it...



But that last part in parentheses is the whole point. My (evidently wrong) intuition says that the consolation of knowing that you will win at least one of your bets does not make up for the fact that you are losing money on a losing bet. Yes, when all the "winning bets" fail to pay off, your "losing bet" comes through. But I still find it hard to see how that could ever compensate for all the money you have to throw away on the losing bet most of the time. But somehow, something similar to that happens in the example I gave above (you bet on 3 horses, not all 4). It works, and I don't get it.

Your intuition is entirely correct. This example makes no sense. So let me mutate it into a situation where it DOES make sense.

Suppose you have a thousand dollars, and are flipping coins with some guy who has a million dollars. You flip coins and make one dollar bets until one or the other goes bust. The expectation in both cases is zero, but the distribution is different. Joe Sixpack as one chance in a thousand of making a million dollars, and Daddy Warbucks 99.9% chance of making a thousand dollars.

As an even more extreme example, consider that you have one dollar, and are flipping coins with God, who has infinite dollars. Every time you win, you double your bet. Your expectation from this game is infinite! Really! But you have zero chance of winning that infinite amount of money. The point of this is that a bet with infinite positive expectation can actually be a sure loser, so it is always a good idea to use a bit of common sense in applied mathematics.

As for the horse race where you are betting a trivial amount then the hedging strategy makes no sense at all. Hedging comes in when large amounts of one's net worth are bet. Someone with a million dollars who is betting the entireity of that amount has a very different attitude. The most important thing is to not lose the entire thing and become destitute. So one will pay to protect oneself against such things.

In modern portfolio management one may be "investing" billions of other people's money. Most people don't like to see wild fluctuations in the price of their assets. So such fluctuations are often called "risk." It is simply assumed that all investments have a positive expectation, then there are small investments with negative expectation called "insurance." Supposedly this allows higher return with minimal risk. This fallacy was perhaps the primary cause of the 2008 financial meltdown, which was a combination of fraud and misapplication of basic statistics by economists.

SO, if you are betting a sum of money that would hurt to lose, then this hedging strategy makes sense. If you were betting your entire net worth, then you would certainly bet on that 2% dark horse as well.

----

Often in economics one reads about risk-return. The idea is that investments with higher variance ("risk") have higher expectation ("return.") This is not true. There are plenty of just plain bad investments in this world.

As for this Black Swan's thing, the reason that he could make money was not that the investments were high risk. He could make money because the investments were undervalued. Now that the news is out the price will very likely rise and his return drop.
 
  • #16
techmologist said:
And as for the hedging, I have been accustomed to thinking of hedging as something you do to sleep better at night. You give up some of the profitability to reduce risk, because people usually don't like to be exposed to huge losses even when the expected value is high. But in my example above, the "hedge" actually maximizes the long-term growth rate. This is the part that is surprising to me. Maybe I am just unsophisticated. Compared to this, the Monty Hall problem seems easy to grasp. However, I was exposed to that problem at a younger age (as homework), so I may have missed out on the opportunity to be wowed by it.

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.
 
  • #17
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.
 
  • #18
SteveL27 said:
This reminds me of the investor and writer Nassim Nicholas Taleb, who as an investor, made his living betting on "Black Swan" events ... events that are extremely unlikely, but that pay off hugely when they come in.

As an investor, Taleb would come to work every day, place his trades, (which invariably lost a small amount of money every day), and spend the rest of the day working on his mathematical theory of making money from extremely unlikely events.

Every so often, one of those events would happen. All the conventional investors would get killed, and Taleb would collect.

http://en.wikipedia.org/wiki/Nassim_Nicholas_Taleb

You might take a look at his books. One is called Fooled by Randomness: The Hidden Role of Chance in Life and in the Markets.

https://www.amazon.com/dp/0812975219/?tag=pfamazon01-20

and the other is The Black Swan: The Impact of the Highly Improbable.

https://www.amazon.com/dp/081297381X/?tag=pfamazon01-20

If I'm understanding you correctly, this is the phenomenon you're talking about. I'm not sure if Taleb's Black Swan events have a negative expectation. But typically they're regarded as so unlikely that nobody bothers to calculate them at all.

Also what you're saying reminds me of hedging. Investors buy stocks and then buy options against the stocks. If the stocks go up, the options are like unused insurance. If the stocks go down, the options cover the loss.

So buy making the "bad bets" you are in effect trading some of your potential winnings for a protection against an adverse effect.

Actually just think of car insurance. It's a terrible bet for the buyer. The insurance companies make sure of that! But you still buy car insurance. You need to insure against a rare outcome.

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?
 
Last edited by a moderator:
  • #19
techmologist said:
My (evidently wrong) intuition says that the consolation of knowing that you will win at least one of your bets does not make up for the fact that you are losing money on a losing bet.
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%.
 
  • #20
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.

f1+ f2+...+ fn + 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+...+ fn + 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.
 
  • #21
techmologist said:
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.
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.
 
  • #22
mfb said:
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%.

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.

ImaLooser said:
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.

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.
-----------------------------------------------------------------------------------------


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.
techmologist said:
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+...+ fn + 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_n f_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{\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]

which tends to

[tex] G = p(1)\ln(b+\alpha_1f_1)+p(2)\ln(b+\alpha_2f_2)+...+p(n)\ln(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+...+ fn + b = 1, each term being nonnegative. Using the Lagrange multiplier method, this is the same as maximizing


[tex]p(1)\ln(b+\alpha_1f_1)+p(2)\ln(b+\alpha_2f_2)+...+p(n)\ln(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 will cover the procedure for determining which horses to bet on in a later post.
 
Last edited:
  • #23
techmologist said:
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.
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 fj 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?

ImaLooser said:
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.

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.
 
  • #24
Norwegian said:
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 fj 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?

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.
 
  • #25
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 fj > 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 fj = 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 bt 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, bt 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 bt 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 bt 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 bt. 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).
 
Last edited:
  • #26
techmologist said:
God bless anyone who actually reads this.

Like I said earlier ... the math is impressive; but it doesn't have much to do with horse racing.
 
  • #27
SteveL27 said:
Like I said earlier ... the math is impressive; but it doesn't have much to do with horse racing.
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.
 
  • #28
mfb said:
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.

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.

https://www.physicsforums.com/showpost.php?p=4081729&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.
 
  • #29
SteveL27 said:
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.

https://www.physicsforums.com/showpost.php?p=4081729&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.

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.
 
Last edited:

1. Can you really win money on a bad bet?

Yes, it is possible to win money on a bad bet. However, it is not very likely and should not be relied upon as a consistent strategy for winning money.

2. How is it possible to win money on a bad bet?

In some cases, a bet may be considered 'bad' due to low odds or a high probability of losing. However, if the bet does end up winning, the payout can be significantly higher than the initial wager, resulting in a profit.

3. Is it worth taking a chance on a bad bet?

It ultimately depends on the individual's risk tolerance and their understanding of the odds. Taking a chance on a bad bet can result in a big payout, but it is important to understand the potential risks and not rely on it as a consistent source of income.

4. Are there any strategies for winning on a bad bet?

There is no guaranteed strategy for winning on a bad bet. However, some people may use techniques such as hedging or betting on multiple outcomes to increase their chances of winning.

5. Can you turn a bad bet into a good one?

No, a bad bet does not become a good one. However, if a bad bet does end up winning, it can be viewed as a fortunate outcome but it does not change the initial assessment of the bet being a bad one.

Similar threads

Replies
11
Views
2K
  • General Math
Replies
4
Views
733
Replies
9
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
2
Views
4K
  • General Math
Replies
28
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
931
Replies
3
Views
2K
  • Programming and Computer Science
Replies
9
Views
3K
Replies
1
Views
2K
Replies
14
Views
6K
Back
Top