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[Hfleer]

My point is that your return on a single bet is better than your return on multiple bets.  When considered as a probability.

For a given bankroll you will end up with more money by placing a single bet.

But this is an average over many possible outcomes.  An actual gambler making a single bet has one particular outcome, either wins or loses the entire bankroll.

Here is another way to state it.  If you took the average gain of all the people who make one bet only they would have a higher return than the average of all the people who make many bets.  And of course all of these people would on average have a negative return.

Now that I say it that way it doesn't sound correct.  This is why I avoid saying stuff.  I'll think about it later.

I can't find the post but about 20 pages before this one doog confirms that when you split a bet into two smaller bets, you end up with slightly higher EV

That's wrong, you would only have a better chance of winning at least 1 bet, obviously.   EV wouldn't change.  Also you can go for a long time with martingale, but the odds aren't any better.

[1715037008]

1715037008

[Hfleer]

I'm look for the post.

meanwhile, I found this idiot: https://bitcointalk.org/index.php?topic=238613.msg2732846#msg2732846

The original post was deleted but that is mirror^^

Well that post was good for amusement purposes.  Some people will try anything.

[HollowIP]

My point is that your return on a single bet is better than your return on multiple bets.  When considered as a probability.

For a given bankroll you will end up with more money by placing a single bet.

But this is an average over many possible outcomes.  An actual gambler making a single bet has one particular outcome, either wins or loses the entire bankroll.

Here is another way to state it.  If you took the average gain of all the people who make one bet only they would have a higher return than the average of all the people who make many bets.  And of course all of these people would on average have a negative return.

Now that I say it that way it doesn't sound correct.  This is why I avoid saying stuff.  I'll think about it later.

I can't find the post but about 20 pages before this one doog confirms that when you split a bet into two smaller bets, you end up with slightly higher EV

That's wrong, you would only have a better chance of winning at least 1 bet, obviously.   EV wouldn't change.  Also you can go for a long time with martingale, but the odds aren't any better.

You are correct, i believe. It's like pseudo odds. You still have a, in this case, 49.5% chance for every roll, so the odds really don't change. It's when you look at it as losing 10 49.5% rolls in a row that it becomes more improbable. Even though the odds are still the same for each roll, it's just something peculiar that you wouldn't always expect. It's totally possible to lose 1000 rolls in a row at 49.5% but the probability of that happening must be less than just losing 2 in a row right? Or perhaps that is another fallacy.

[dooglus]

No matter what complicated strategy you use, you're better off making a single straight-up bet for the amount you aim to win.

That's not actually true.

If you have 3 BTC and need 4 BTC, you're better off to bet 1 BTC at 2x and if you lose, bet the other 2 BTC at 2x than you are to place the whole 3 BTC on a single bet at (4/3)x.

You probably won't believe this.  But you will agree that both strategies turn 3 BTC into either 0 BTC or 4 BTC.

Then, if you work out the chance of each strategy succeeding, you'll find that the martingale is better!

When the edge is 1%, the chance of success is calculated as:

(chance_of_success = (100-edge) / bet_multiplier)

So the 2x game has a (100-1)/2 = 99/2 = 49.5% chance of winning.

[dooglus]

You can run the numbers and see this quite clearly, but there's also an intuitive way to think about it. JD takes, on average, 1% of the amount you bet. If you make a single bet, you have an expected 1% loss on the amount wagered. If you use a Martingale-style strategy, you bet the same money more than once.

Aha.  I should have kept reading.  Your intuitive reason can help you understand why Martingale is better, not worse - it's because you risk less:

In the 3 BTC -> 4 BTC case, when you do a single bet of 3 BTC, you bet 3 BTC and so expect to lose 0.03 BTC.

When you bet 1 BTC and then 2 BTC only if you lose, then you're risking either 1 BTC or 3 BTC.  You only risk 3 BTC if the first bet loses, which happens only 50.5% of the time.  So on average you only risk a little more than 2 BTC, and so expect to lose only a little over 0.02 BTC.

[dooglus]

I tried explaining this to him for a very long time..

It won't work. He's a persistent bugger, he's having fun though, so fair enough.

Do you still think Martingale is worse than the equivalent single bet?

I thought we had been through that before and that you finally agreed with me.

We were going to bet on it, until you realised I was right, no?

[dooglus]

This was a good explanation.  To simplify it: A single bet with a particular probability of winning has a higher expected value than multiple bets with the same combined probability.

Not true.

Someday I will try to write out the expected value of a martingale sequence.  Has somebody done this already?

You should do it.  I would in general suggest working out whether what you're saying is true before saying it, rather than after. 

Here: suppose you have 3 BTC, bet 1 BTC at 2x, and if you lose, bet the other 2 BTC at 2x.  2x bets have a 0.495 probability of winning.  You win 1 BTC total unless both bets lose.  Both bets lose with p=0.505*0.505 = 0.255025, and so you win with p=0.744975

The equivalent single bet would pay out 99/74.4975 = 1.32890365448505x and would return 3.98671096345515 BTC.

That's less than the 4 BTC returned by the martingale sequence with the same probability.

[Rannasha]

@dooglus: Yes, you're right with that example. My statement was a bit too general.

But what we're seeing more is people not doing a single Martingale-sequence, but repeating it over and over, with a very small minimum bet, until some level of profit is achieved (or all is lost).

If I set out with some amount of BTC with the intention of doubling it, a common approach is to run many Martingale sequences from a small starting bet until the target is reached. Lets take 1/1023 of the bankroll as starting bet. This starting bet runs my bankroll dry if I lose 9 bets in a row. The odds of this happening at x2 are 0.505^9 = 0.0021361. Consequently, the odds of winning an amount equal to the starting bet are 0.997864.

Now, to double my bankroll, I need to win 1023 of these sequences in a row. The probability of that is 0.997684^1023 = 0.112189 or a little over 11%. Or I could've taken a single spin at 49.5% odds to double up. Now there is a mitigating factor for the Martingale scenario, being that in this example you don't go broke after 9 consecutive losses (unless you hit it on the very first sequence) as you'll have your previous profits to fall back on, but recovering from this position will be difficult, you will have less than your original bankroll to play with, so while this factor increases the probability of the Martingale style to double the bankroll, I don't see it bridging the gap between 11% and 49.5%.

As your desired profit decreases and/or your starting bet increases (relative to your bankroll), the performance of the Martingale will approach and eventually pass (as seen in your example) that of the single bet.

But for most gamblers on JD, the aim seems to be small starting bets and aiming for "safe" strategies that, while have a negligible chance to lose a sequence, require so many bets that eventually the total amount wagered will easily pass the initial bankroll.

[ghibly79]

For reference, the chance of a 7 loss streak happening is 0.00004074% so there is a 99.99995926% chance of it not happening.

My interpretation of that is the 7th bet has a 99.99995926% chance to win.

Wrong: it's that chance over 7 rolls. Over 500k+ is lower (close to the chance to win a single bet, depending on how many and how long losing streaks already happened before. Cba doing the exact math xD).

[bitcoin44me]

The most important result is that in order to win you have to stop.  In all simulations given enough time the initial bankroll is wiped out.

And in all simulations after enough time the house takes it's edge of all money wagered.

The highest EV is for the house because of the positive edge.  The next best is to not bet at all with an EV = 1.0*B = 0.

What I don't know but suspect is that the next best is to bet your entire bankroll once and just accept the outcome.

More bets will erode your bankroll at the rate of e*B per bet.  Eventually your bankroll will be gone.

This assumes a finite bankroll, i.e. that you won't keep putting wages, loans etc. into the bankroll.

Yes, and if you you want more than 50% chance to win, you need to stop before you double up.

[organofcorti]

Now, to double my bankroll, I need to win 1023 of these sequences in a row. The probability of that is 0.997684^1023 = 0.112189 or a little over 11%.

The probability is far less than that. What happens if the last 8 rolls are all losses - or even just the last? Your bankroll won't be doubled. So you have to include the probability of not losing the last 8, 7, 6 ... 1 rolls.

[mechs]

This was a good explanation.  To simplify it: A single bet with a particular probability of winning has a higher expected value than multiple bets with the same combined probability.

Not true.

Someday I will try to write out the expected value of a martingale sequence.  Has somebody done this already?

You should do it.  I would in general suggest working out whether what you're saying is true before saying it, rather than after.  

Here: suppose you have 3 BTC, bet 1 BTC at 2x, and if you lose, bet the other 2 BTC at 2x.  2x bets have a 0.495 probability of winning.  You win 1 BTC total unless both bets lose.  Both bets lose with p=0.505*0.505 = 0.255025, and so you win with p=0.744975

The equivalent single bet would pay out 99/74.4975 = 1.32890365448505x and would return 3.98671096345515 BTC.

That's less than the 4 BTC returned by the martingale sequence with the same probability.

Your correct that my first statement was wrong.  But your simple example is not a martingale sequence.  To figure out the EV you need to average over all the possible outcomes.

Here is the math for a single bet:

Bet - B
house edge - e
probability of win - p
Expected value - EV = (1-e)B, on just-dice EV = 0.99 B

Of course for a single bet you either win (1-e)/p or lose B.

I believe that for a martingale sequence EV will be less than 0.99 B, where for the martingale case B is the starting bankroll.

Thankfully their is a max bet - even if very high it capped at 1% of Bankroll.  Or you could see some crazy bastard making a 5000 BTC YOLO roll at 2X, win and really destroy the bank.

[HollowIP]

Thankfully their is a max bet - even if very high it capped at 1% of Bankroll.  Or you could see some crazy bastard making a 5000 BTC YOLO roll at 2X, win and really destroy the bank.

Yeah- that is a good thing. haha. Upon extra contemplation, I think I am realizing that a person's bankroll is probably a larger factor than I've considered before.

If your bankroll is say 5000btc, and you are only trying to get an extra 0.01, then a martingale is definitely the route to take. but if you only had 0.01 and needed another 0.01 then it really seems as if going all-in would be the method most effective.

Or since BTC is divisible down to the satoshi, lets say your bankroll is 50 satoshis, and you need 100. You only need 7 losses to bust you, but you need 50 winning streaks to double. Getting 7 losses in a row out of a string of 100 bets (if everything happened perfectly and you lost roughly ever other bet) would most likely happen.

[HollowIP]

Here is something of interest. The longer you continue a 50% series, the larger the 'spread' becomes.

In short, if you flip 10 coins, than the heads to tails ratio would most likely be more like the ideal 50% than if you actually flipped 100 coins, and 100 coins would most likely be closer to 50% than if you flipped 1000 coins. Martingale works, but as time progresses, or number of bets increase, this becomes more and more ineffective, making actually doubling your bankroll quite hard. obtaining maybe 1/4th your bankroll is better suited for the martingale strategy indeed (unless you have my luck)

[infested999]

Here is something of interest. The longer you continue a 50% series, the larger the 'spread' becomes.

In short, if you flip 10 coins, than the heads to tails ratio would most likely be more like the ideal 50% than if you actually flipped 100 coins, and 100 coins would most likely be closer to 50% than if you flipped 1000 coins. Martingale works, but as time progresses, or number of bets increase, this becomes more and more ineffective, making actually doubling your bankroll quite hard. obtaining maybe 1/4th your bankroll is better suited for the martingale strategy indeed (unless you have my luck)

It checks out http://www.youtube.com/watch?v=vpTs0Zag2pw

[organofcorti]

Here is something of interest. The longer you continue a 50% series, the larger the 'spread' becomes.

In short, if you flip 10 coins, than the heads to tails ratio would most likely be more like the ideal 50% than if you actually flipped 100 coins, and 100 coins would most likely be closer to 50% than if you flipped 1000 coins. Martingale works, but as time progresses, or number of bets increase, this becomes more and more ineffective, making actually doubling your bankroll quite hard. obtaining maybe 1/4th your bankroll is better suited for the martingale strategy indeed (unless you have my luck)

Well done- you've rediscovered the Binomial distribution!

[HollowIP]

Well done- you've rediscovered the Binomial distribution!

Haha, You can tell I was great in math class! Well, at least I've learned it in my life! Better to discover it on your own then being told so anywho.

[organofcorti]

Well done- you've rediscovered the Binomial distribution!

Haha, You can tell I was great in math class! Well, at least I've learned it in my life! Better to discover it on your own then being told so anywho.

Exactly right.

Here's something to look into next - you can discover far more about the expected values of a martingale sequence if you approach it as a geometrically distributed random variable instead.

[Dabs]

Dabs, you pay back 100 BTC for the 92 used, however my question is, from reading one of your "updates" is that you keep some of the profit made. How much does the 92 BTC turn into before you pay people out?

It should turn into maybe 112, or maybe a little bit more. That's 8 that goes to the "new investors" and 12 that goes to "old investors" or those who previously sent coins. What's left is proportionally distributed according to their initial contribution.

In a way, I am using new people's new money to pay old people's old money, but both new and old people get paid out, so this does not become close to any sort of pyramid or ponzi scheme. I tell you what it is though, it is a gamble.

As for everyone else with their real math and real probabilities, ... hey, what did I start? And dooglus and ooc has answered some of the more difficult questions.

My only question was, and now still is, who wants to bet?

According to another formula on probability that I discovered: where p = 0.877779

* The average number of loss runs until the next 7 losses in a row: ((1 - p)^-n)/p =  2,796,365.8140
* The average number of loss runs until the next greater than 6 run of losses in a row: (1 - p)^-(n+1) = 2,454,591.1878

You pick which one you like, it still comes out to about 1 in 2 million.

The key word here is average. This means this number can be higher or lower. It's vague on purpose, because it can not be exact.

Add to the mix my magic beanstalks, and it boils down to "Do you feel lucky?"

Will Dabs lose the 7th bet in his sequence now? Will he have lost it or will he have won it if we didn't know the most recent 6 bets that lost?

[dooglus]

Your correct that my first statement was wrong.  But your simple example is not a martingale sequence.

Are you telling me that betting 1, stopping if I win, and doubling to 2 if I lose isn't a martingale sequence?

I would say it's a 2-step martingale sequence.  I can show the same effect with longer sequences if the 2-step one doesn't convince you.  The effect (martingale is better than single bet) becomes more pronounced the more steps in the sequence.

I believe that for a martingale sequence EV will be less than 0.99 B, where for the martingale case B is the starting bankroll.

I know you do, but unfortunately it's not true.  It always feels rude to me to tell people they're wrong.  But it doesn't seem right to leave them believing falsehoods either, so I'm kind of stuck not knowing what to do other than to say "you're wrong" and try to explain why.  But I already explained why ("the equivalent martingale sequence expects to bets less and so expects to lose less") so notw I'm really stuck.

If the martingale is risking the same amount, and potentially winning the same amount, then its EV is greater than 0.99 B.  That's what the post you're replying to demonstrated.

And now I feel like a meanie.

Edit: and am also a little concerned that the peopleperson most aligned with my views according to recent posts is Dabs!