Double Up Strategy

[grimReaper]

Double up strategy is a losing one.

[FlipCoach]

Double up strategy is a losing one.

Yeah worked that one out mathematically in my dorm 7 years ago.. on a long enough timeline.. its effed.

[irishpkr]

Martingale betting strategy is how I beat blackjack Its a 110% proven winning strategy

[irishpkr]

NOTE: This link is NOT a real winning program in case some idiots decide 'Oh this look solid!'

Pretty funny, all his vids are in a similar vein: [ame="http://www.youtube.com/watch?v=9jakG8pYUQE"]YouTube- Roulette sniper review - I make $15000 per week[/ame]

Why is it always the Irish? Putting me to shame [ame="http://www.youtube.com/watch?v=hVd3GHr42FY"]YouTube- Roulette Sniper paid for my New Lexus[/ame]

[grimReaper]

Yeah worked that one out mathematically in my dorm 7 years ago.. on a long enough timeline.. its effed.

I wrote a computer simulation, definitely EV-. Btw, there a proof on wikipedia. As long as the probability of winning on trial is less than 50%, it's EV-, and according to my Stochastic Processes textbook (which I don't have on me right now), probability of winning one game of craps is something like 49.x%. (x=7?).

What's sad is there's probably people that use this strategy to make a living, then go broke, get depressed forever.

[travz21]

Whenever a bet is -EV, it doesn't matter what strat you use. People don't seem to understand this.

"Oh, I'll just double my bets and will have to win one of them!"

You're still losing a % of every bet, dude.

The people that use this strat to make a living playing table games are just hot runners. They have no specific skills that other intelligent people don't possess. All of them will eventually go broke. Lucky for them, "eventually" takes longer than some of their lives.

[FlipCoach]

"Oh, I'll just double my bets and will have to win one of them!"

All of them will eventually go broke. Lucky for them, "eventually" takes longer than some of their lives.

Yeah I mean theoretically this could go on for over a lifetime because if the double up strategy is to just double bets when you lose, you have to look at the entire string.

So if the house edge is 51% and you're willing to double your bet 14 times (from $5 doubling up to $40,960) - assuming table stakes allow it - the odds you'll lose all 14 are:

(.51*.51)^14 which is LIKE .0000805346

so the odds of losing the entire string are 1/.0000805346 = 1 in 12,417 strings of 14.

Therefore if you do this 3x per day, 5x per week and quit every time you get ahead, then you would expect to go broke only once every 15.9 years, and seeing how all this guys are older than time, they should definitely die first before going broke.

[grimReaper]

So if the house edge is 51% and you're willing to double your bet 14 times (from $5 doubling up to $40,960) - assuming table stakes allow it - the odds you'll lose all 14 are:

(.51*.51)^14 which is LIKE .0000805346

so the odds of losing the entire string are 1/.0000805346 = 1 in 12,417 strings of 14.

If you can afford 14 doubled up losses, total amount you can afford to lose is:

5+10+20+40+... = 5(1+2+4+8+...) = 5 Sum(i=0, i=13)2^i = 5*(2^14 - 1) = $81915

0.0000805346 is correct but not sure what (.51*.51)^14 is.

Therefore if you do this 3x per day, 5x per week and quit every time you get ahead, then you would expect to go broke only once every 15.9 years, and seeing how all this guys are older than time, they should definitely die first before going broke.

Bringing $82k to the table to win $5. br management at its finest.

[grimReaper]

I analyzed the variance:

Let X be our return using the martingale strategy and let q be the probability of losing one round, and A is the initial bet. Let n be the number of trials we can afford to lose before we go busto. Then,


X = A w/ probability 1 - q^n
X = -A*(2^n - 1) w/ probability q^n

E[X] = A(1-(2q)^n)
E[X^2] = A^2[1 + (2^2n)(q^n) - (2^(n+1))q^n]

Var[X] = E[X^2] - (E[X])^2 = (A^2)(2^2n)q^n[1-q^n]
SD[X] = sqrt(Var[X])

I plotted the SD[X] as a function of n:



Generally SD is interpreted as riskiness, and initially counter-intuitively, as we increase the # trials we can afford to lose, the riskiness of playing the martingale strategy increases exponentially! Though if you think about it, this makes sense as we double our bet each time we lose, and doubling a double double double bet can get very big...

So if we come in w/ $5,242,875 and made our initial bet $5, we can afford to lose 20 trials in a row. The SD of this is $6241.247, as you can see in the graph.

Now if we played the normal way where we don't double up etc, the standard deviation of that is 2A*sqrt(q*(1-q)), or 10*sqrt(0.2499) = 4.999999 = $5 per game. And if we played n regular games, the standard deviation of the overall profit is 5*sqrt(n).

Setting 5*sqrt(n) = 6241.247 and solving for n, n = 1558750.065, or 1,558,750. In other words, playing the game ONCE w/ the martingale strategy where we allow ourselves to lose 20 times is just as risky as playing the game 1,558,750 times regularly.

[travz21]

Econ started this thread just hoping we'd take the bait so he could maths all over this place.

And all tables have max bets, which would make the double up strat even worse. And roulette has 1 or 2 house numbers, depending on the casino, so you'd get really owned with double up strat there. Not really sure how to play craps, but there's gotta be some way you can get owned. Like if you totally miss the table with your throw you lose all your chips plus your hotel room.

[frob23]

We should also not forget that most places have a maximum bet amount. So we're not going to be able to double our bet 10-20 times in an attempt to win back our losses plus the $5. This would also decrease the amount of time until we ended up having a session where we lost it all.

The last casino cruise ship I went on I stood by the roulette table for the free booze. They had a min-bet of $1 and a max of $100 (on the "even" money bets). That's 7 losses before you reach an amount where you can't bet enough to cover your previous losses.

I am too lazy to do the exact math right now but that's a better than 1% chance of losing $127 in an attempt to win $1. Granted, it's not a huge chance but it seems pretty dumb anyway. And those times you win, you're up $1... nothing life changing.

[purplechip1]

http://www.rakeback.com/poker-forum/...ambler-3-6222/

[bigslick1]

mathy makey my brainy hurty

[grimReaper]

So as I said, the more bets you can afford to lose, the more riskier it is; but if we look at relative riskiness, i.e SD/br, where br is $ you can lose before busto, then that tells a different story:


(The y axis is the "absolute riskiness"/br. Let's call it relative riskiness)

Our br is $5,242,875, the relative riskiness when we can afford 20 losses in a row is 0.12%.

If we compare a 20-round game to a normal regular bet:
0.0012 = 6241.247/5,242,875 = 5*sqrt(n)/5,242,875

n=1558126 as before, obv, so playing that many games normally is just a risky (relative or absolute, doesn't matter) as playing the game once under matingale affording to lose 20 bets in a row.

[RiverstarsVictim]

Fkn maths...if you're gonna be that serious, run the casino!

[grimReaper]

Ah, RiverStarsVictim, nice to see you back.