Say you have a 100$ starting bankroll and you get a series of 6 coinflips each, 5 of which are 50/50 to win, but if you lose 5 in a row, the last coinflip will be 100% guaranteed to win. That is, if you win the first flip a new series starts, the same goes if you win on the 2nd, 3rd, etc.

You can even bet 0, and wait to lose 5 (in a row, since only the 6th is guaranteed) and then take the double up on the 6th flip. If you want to double your roll the quickest way possible, is there a betting strategy that is better than just martingaling it until you reach 200?

At $1 bets, Martingale of 1, 2, 4, 8, 16, 32 guarantees $1 every 6 rolls or less, so on a very unlucky run it will take you 600 rolls to earn $100.

However, you could start your second Martingale on roll 2, regardless of the outcome of roll one, and your third on roll 3, etc.

The worst case scenario would be LLLLLW.

(Max capital outlay at any one time is 120 units so your unit would actually have to be 83c.)

Martingale 1: 1 2 4 8 16 32
Martingale 2: 1 2 4 8 16
Martingale 3: 1 2 4 8
Martingale 4: 1 2 4
Martingale 5: 1 2
Martingale 6: 1

On roll 6 you will bank 6 units profit which is $5.

This will reduce the maximum number of rolls required to make $100 profit to 120 rolls altogether.

Every time there is a winner you will land one unit profit per active Martingale, so in doesn’t actually matter when the wins occur. If you had the sequence LLW (start again with Mart 1) W (start again with Mart 1) L (Mart 2 joins in) WLLLW:
you would have staked 1 3 7 1 1 3 1 3 7 15
with running profit of -1 -4 +3 +4 +3 +6 +5 +2 -5 +10

So, basically, you win one unit every flip, guaranteed in the long run.

The number of rolls can be reduced considerably by ploughing the overall profits back in to boost the capitalisation of the next rolls. After any winner, you know that your maximum capital needed is 120 units and so your new capital divided by 120 is your new unit until the next winner. After 24 rolls ending on a winner, your capital would have been boosted to $120. Your units until the next winner would now be $1.

You can’t evaluate the precise number of rolls but you can certainly get it within 6. If you use the worst case scenario of LLLLLW repeated, it would not be so difficult to calculate. I'll leave you with that problem.

I am not sure I understand how you mean going to a second martingale regardless of the outcome of the first roll in your scenario? If you win the first one, the rolls reset, so how do we have enough to cover 5 consequent losing rolls beginning at 2x?

Yes, I was thinking starting with 1.58, and going up to 32x and then increasing the bet amount by a few cents, reinvesting every winning roll. But if it takes 120 rolls without reinvestment, or somewhere half of that with reinvestment, then is it not better to simply wait ~32 turns, until we lose 5 straight, and not wager anything, so as to not lose from your bankroll, and then simply going all in on the 6th guaranteed roll every time?

To cope with the worst case scenario of LLLLLW occurring:

(a) On the first flip you start your first M and bet 1 unit. (Total bet so far is 1 unit.)

If you win the first flip, the return is 2, so you have won 1 unit and you goto (a)

(b) If you lose the first flip, you continue your first M and bet 2; but you also start a new M and bet a further 1 - a total of 3 units on flip 2. (Total of bets so far is 4 units: 1 + 3)

If you win the second flip, the return is 6, so you have won 2 units and you goto (a)

(c) If you lose the second flip, you continue your first M and bet 4; you continue your second M and bet a further 2; but you also start a third M and bet a further 1 - a total of 7 units on flip 3. (Total of bets so far is 11 units: 1 + 3 + 7.)

If you win the third flip, the return is 14, so you have won 3 units and you goto (a)

Etc, until:

(f) If you lose the fifth flip, you continue your first M and bet 32 units; you continue your second M and bet 16 units; you continue your third M and bet 8 units; you continue your fourth M and bet 4 units; you continue your fifth M and bet 2 units; but you also start a sixth M and bet a further 1 - a total of 63 units on flip 3. (Total of bets so far is 120 units: 1 + 3 + 7 + 15 + 31 + 63.)

You must win the sixth "flip". The return is 126 units so you have won 6 units (6 x 83c = $5) and you goto (a).

IOW, you are guaranteed to double your BR in 120 flips.

If you wait for worst case scenarios, and they don't happen to come up, you could be waiting for a very long time before being able to make your winning AI bet. Of course, it is possible that you could double up on the sixth "flip".

Aha, I see, thought you had meant that the second martingale you are starting would be double that of the first and so on.

But still, even if there's significant variance on the waiting times to lose 5 in a row, on average it would take 32 turns to do so. Then wouldn't this be a quicker way to doubling up by default, as opposed to the guaranteed expectancy, even if we were to reinvest the winnings?

Let's say that every possible set of flips occurs and occurs only once, and the guaranteed W-set always occurs last. (Extraordinarily unlikely, but possible.)

If the problem stated that the second flip was always a guaranteed win, the possible sets of two flips are:

WW
LW

As a set ends after a W, in this case, of the two sets of possible flips you would only actually see 3 flips:

W, LW

If the problem stated that the third flip was always a guaranteed win, the possible sets of three flips are:

WWW
WLW
LWW
LLW

As a set ends after a W, in this case, of the four sets of possible flips you would only actually see 7 flips:

W, W, LW, LLW

And so on:

if the problem stated that the fourth flip was always a guaranteed win, of the eight sets of possible flips you would only actually see 15 flips;

if the problem stated that the fifth flip was always a guaranteed win, of the sixteen sets of possible flips you would only actually see 31 flips;

as the problem states that the sixth flip is always a guaranteed win, of the thirty-two sets of possible flips you would only actually see 63 flips.

Of course, there is absolutely no guarantee that LLLLLW will occur in the thirty-two sets of possible flips, and you might see hundreds of flips before that required set did occur - although, again, of course, it could occur within the first 120 seen flips of the sets of flips "behind the scenes".

I would stick with my system to guarantee doubling in 120 flips. Once you try other systems you can only calculate their average expected wins, not precise ones.

Like, how about trying a trebling Martingale of 1 3 27 81 243 729 = 1084 units, with a unit being about 9c?

This gives variable rolling profits of 1 or 2 or 23 or 50 or 131 or 374 depending where the win falls in a set of 6. If you struck three LLLLLWs in a row you would be done in 18 rolls! But it could take one helluva lot longer if you kept striking a win early in a set of six.

e&oe

*

Edit: Doh.

Right, but then the two systems should have the same guaranteed expectancy over a large sample size, no (unless you reinvest earnings)? Except one would be guaranteed and variance free, while the other would be high variance.


Also, why did you go 27 instead of 9 on the third roll when trebling?

In that case, why not wait for only four losers then bet 25, 50? Half the time you win 25, half the time 50. Average expectation is 37.5 every 16 sets.

You might be sitting in your hypothetical chair for a lot less time than by waiting for five losers in a row. Maybe not, though, of course.

gg

Right, except 33/66 and you’d have a $49.5 expectancy every 16 sets.

But, yes with the 6 running martingales you are guaranteed to win a unit every time and thereby eliminate all variance. But waiting it out, or say trebling, increases variance and you can end up waiting longer or less. So it would be like an allinev graph that has huge deviations vs. the real as opposed to a linear graph.

But is any one strategy inherently better I mean? To me it seems they converge on similar results over the long run no? Or am I off base here?

I guess I should have asked expectancy, is there any one that guarantees, instead of a quicker double up, say a higher profit over an infinitely large sample (well the profit would be infinity then, but you get what I mean – a very large sample).

No. You'd have 33 expected value.

Ah, right.

Just realized you meant a 25/75 bet, thought 25/50 for the bets. But again, then isn’t 0/100 still better in terms of expectancy? Win/Lose 0 half the time, win 100 half the time for 50 ev.

Looks like it, when we introduce the hypothetical long term.

Okay, but then betting it all on the sure thing can't be beat in terms of expectancy, no? Since we'd have a 100 ev after 32 sets.

So then between the guaranteed martingale and waiting it out mostly comes down to the "luck" of losing 5 in a row. And then the average expectancy for the martingale coming from 32 sets, is the average of 1 to 5 units won per set, for a total of 79.68?

So then it's a question between a guaranteed 79.68 ev per 32 sets or a very high variance 100 ev per 32 sets right?

Yes, but theoretically you may never double except with an infinite number of rolls when you and the solar system would already have ceased to exist infinity years before. This would entail rather a long wait for $100.

Of course any possible waiting time can be assigned a probability by application of the binomial distribution.

The 1% tail is considered statistically highly significant and medical studies for new drugs, etc, rely on it to decide whether a drug is actually effective after a rigorous clinical trial. Basically, you compare what has happened with what might have occurred by chance. If the happening, (miracle cure for cancer or whatever), would only have had a 1% chance of occurring by fluke, then you statistically accept that it is a fitting drug. Like, you test 100 “terminally ill” people and 87 get better, so you have discovered a miracle cure for "statistically certain”, (even though it is not actually certain).

In your game you would have to imagine that there was a run of say 100 4-rolls without any LLLLs occurring and apply it to the binomial distribution tables to see if it is a statistically significant variation of at least the 5% level to decide whether to bother playing the zero/hundred game. You can never be sure but you can be pretty sure. It was not certain that the typhoon would strike Tokyo, but it was so statistically likely that they battened down the hatches and cancelled some rugby. The whole world runs on this and similar calculations. Meteorites may wipe us out before we agree anything, of course.

Too bad for that. I did think that the France moneyline or the handicap were worthwhile +ev bets.

Also, just to check that I am doing this right, the average ev for the treble martingale system would be

1 x 0.5 + 2 x 0.25 + 6 x 0.125 … 122 x 1/64 = 6.378 units (at 27c per unit) x 32 sets = ~55.1$ ev after 32 sets, correct?