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.

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: 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.

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.

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.

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

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.

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?

Dear Friends,

Watching that so far discussion about martingale stuff reminded me about an interesting text I once read.

The topic is well known (progressive systems do not work the way we all want), but the math is strong in order to explain that.

It's named: "Mathematical Proof that Progressions cannot overcome Expectation."

I leave here the direct link to it:

http://www.bjmath.com/bjmath/progress/unfair.htm

And there is still more about progressive systems here:

http://www.bjmath.com/bjmath/progress/progress.htm

Enjoy it!