Dooglus "AMA"

[Phildo]

Just some thought without any actual math. You are making a series of bets with a hosue edge of 1% so the result can be summarized as a bet with a house edge of 1%.

Let's compare it to roulette where we can actually make the bets at the same time. If you put 10 on red you have a 47.73% chance of winning 10 bucks. That's a hosue edge of 5.26%. If you put 10 on 1-12 and 10 and 13-24 you now have a 63.16% chance of winning 10 bucks but still a house edge of 5.26%.

This isn't a perfect analogue because I increased the wager, but just by guessing (and not doing stats/probability for a while) I would guess the theory might be the same.

[Light]

When you bet your whole bankroll in a single bet, you expect to lose 1% of it.

When you split it up and bet the pieces in order from smallest to biggest, and stop when any bet wins you often don't end up betting the whole bankroll, and so you expect to lose 1% of less than the whole bankroll.

By splitting it up you reduce the amount you expect to bet, and so you reduce the amount you expect to lose.

Ah ok, that makes a lot more sense now. Didn't think of it like that - keep thinking of them as analogous cases which were giving me different results (ie mathematically impossible).

I cannot believe I'm saying this but I think you might be right.

I went ahead and played around with a case where the player starts at 1 wants to move to 2 by making two bets (of any size between 0-1 inclusive). Hence I went to go graph the function to see if it was true and I got this:

https://www.desmos.com/calculator/obkfifgbnl

Where y = probability of succeeding
and d = the value of the first initial bet

Notice how at both d = 0 and d= 1 the probability is 0.495 as expected (as you are either betting nothing then 1 or 1 then nothing and both are equivalent cases). And in between you get a probability higher than the 0.495 offered for the single bet.

I've tried a few set values for cases where you split your value up to more than two and you do get a better result. I can only theorise that this is because as your bet size approaches 0 with the number of bets approaching infinity your expectation approaches 1.

However, what I do not understand at the present is why this is so. I almost fell out of my chair when the numbers came out (I checked like 6 times), as it's inferring that you can get better than what the house 'technically' offers. The problem with this is that your expectation is better than just flat betting and logically that doesn't make sense. Both should have the same expectation.

I'm going to mull it over.

Can you explain why the probability of success increases as you make more smaller bets? Is it for the same reason as well, as you have a better chance of losing less (as your take a risk with only parts of your bankroll and for you to lose it all, all of the bets must fail) it results in an overall better chance of succeeding? 

[Mitchell]

The system I described doesn't double the bet, because it doesn't need to; it's not betting at 49.5%.  But it does increase the bet such that winning the 2nd bet will make up for the loss of the 1st bet and cause you to end up with exactly the same profit as if you had won the first bet.

Maybe it's only strictly martingale if you're playing for a 2x multiplier, and double your stake on a loss, I don't know the script definition.  But the system I'm proposing is very martingale-ish in that it has the "making up for losses" part.

Hmmm, well you are right, it is very martingale-ish, but since it doesn't match the description of a martingale betting system (can be found here) I wouldn't call it that way, even though it's quite similar. But oh well, it doesn't matter. The technique described by you is amazing and I really enjoy reading about it in this thread, even though I fail to see how this is possible.

[zimmah]

Say, my target is to double my initial deposit.
Is it really possible to use martingale (or whatever strategy) to achieve the goal with higher than 49.5% success rate (one single bet)?

Yes.

And you're the first person who responded in such an open manner.

Everyone else just tells me I'm wrong.  

The trick is to split up your bet (the amount you were going to risk in a single bet) into a series of amounts which sum to a the same, and which form a sequence such that you can bet the smallest amount, and if it wins, you make the same as if you bet the whole amount at 49.5% (so you'll be betting with a smaller chance, and higher payout multiplier).  And if it loses, you want betting the 2nd amount to cover the first loss and make the same net profit.  Etc.

If you can find such a sequence (and you always can, though it can involve some hairy math depending on the length of the sequence you're looking for) then the amount you expect to risk is less than your whole amount (since there's a non-zero chance that you will win before the last bet, and stop at that point), and so the amount you expect to lose, being 1% of the amount you risk, is less than when you make the single bet.

Here's a very simple example:

you have 1 BTC and want to double it.

* you could bet it all at 49.5%, and succeed in doubling up with probability 0.495

* or you could bet 0.41421356 BTC at 28.99642866% with payout multiplier 3.41421356x, and if you lose, bet the rest at the same chance.  If you win either bet, you double up, else you lose.  Your chance of doubling up is 0.4958492857 - a little higher than the 0.495 you have with the single bet.

Cool, huh?

That's breaking the single bet up into a sequence of length 2.

If you break it up into more, smaller bets, then the probability of success increases further.

The more steps, the closer to 0.5 your probability of success gets.

You'll be limited by real-life barriers, like the invisibility indivisibility of the satoshi, and the limit of 4 decimal places on the chance at JD.  But in theory you can get arbitrarily close to 0.5.  I think.  

Interesting, are you sure that's correct? As you're basically saying there are ways to nullify the house edge. I doubt this could possibly be true.

[zimmah]

Yes.

And you're the first person who responded in such an open manner.

Everyone else just tells me I'm wrong.  

I cannot believe I'm saying this but I think you might be right.

I went ahead and played around with a case where the player starts at 1 wants to move to 2 by making two bets (of any size between 0-1 inclusive). Hence I went to go graph the function to see if it was true and I got this:

https://www.desmos.com/calculator/obkfifgbnl

Where y = probability of succeeding
and d = the value of the first initial bet

Notice how at both d = 0 and d= 1 the probability is 0.495 as expected (as you are either betting nothing then 1 or 1 then nothing and both are equivalent cases). And in between you get a probability higher than the 0.495 offered for the single bet.

I've tried a few set values for cases where you split your value up to more than two and you do get a better result. I can only theorise that this is because as your bet size approaches 0 with the number of bets approaching infinity your expectation approaches 1.

However, what I do not understand at the present is why this is so. I almost fell out of my chair when the numbers came out (I checked like 6 times), as it's inferring that you can get better than what the house 'technically' offers. The problem with this is that your expectation is better than just flat betting and logically that doesn't make sense. Both should have the same expectation.

I'm going to mull it over.

Yes, I'm intrigued as well, I feel this should not happen and it should be wrong.

Beating the house edge should not be possible, that's the whole point...

[Minnlo]

Yes.

And you're the first person who responded in such an open manner.

Everyone else just tells me I'm wrong.  

I cannot believe I'm saying this but I think you might be right.

I went ahead and played around with a case where the player starts at 1 wants to move to 2 by making two bets (of any size between 0-1 inclusive). Hence I went to go graph the function to see if it was true and I got this:

https://www.desmos.com/calculator/obkfifgbnl

Where y = probability of succeeding
and d = the value of the first initial bet

Notice how at both d = 0 and d= 1 the probability is 0.495 as expected (as you are either betting nothing then 1 or 1 then nothing and both are equivalent cases). And in between you get a probability higher than the 0.495 offered for the single bet.

I've tried a few set values for cases where you split your value up to more than two and you do get a better result. I can only theorise that this is because as your bet size approaches 0 with the number of bets approaching infinity your expectation approaches 1.

However, what I do not understand at the present is why this is so. I almost fell out of my chair when the numbers came out (I checked like 6 times), as it's inferring that you can get better than what the house 'technically' offers. The problem with this is that your expectation is better than just flat betting and logically that doesn't make sense. Both should have the same expectation.

I'm going to mull it over.

Yes, I'm intrigued as well, I feel this should not happen and it should be wrong.

Beating the house edge should not be possible, that's the whole point...

It is still not going to beat the house.
What doog's strategy does is to lower your expected wagered amount and so lower your expected loss (house edge % * wagered amount) and equivalently increase the successful rate of hitting the target.