EV for Martingaling the 6 & 8?

My friend just told me about an interesting craps system that involves Martingaling the 6 & 8 and praying the 7 doesn't hit 5 times without any 6 or 8.

The Martingale starts with $6 bets and then progresses to $18, $42, $90, and $204 bets.

He told me he's played 49 hours so far and is up about $1.8K without having to place any $204 bets yet.

Can anyone sim this system to calculate his risk of ruin for an 8 hour session and his hourly EV in the long run?

Quote: drjohnny

My friend just told me about an interesting craps system that involves Martingaling the 6 & 8 and praying the 7 doesn't hit 5 times without any 6 or 8.

The Martingale starts with $6 bets and then progresses to $18, $42, $90, and $204 bets.

He told me he's played 49 hours so far and is up about $1.8K without having to place any $204 bets yet.

Can anyone sim this system to calculate his risk of ruin for an 8 hour session and his hourly EV in the long run?

How is it that you get an hourly EV playing a -EV game? No matter what system crazy system you come up with you are still exposing money to the house edge. Math never lies, but sometimes(most of the time) friends who play table game systems do. You don't even have to be good at math, as long as you believe math always tells the truth. Just about every system you can come up with has been thought of, and the math always says it sucks.

Both my friend and I know craps is a -EV game and the house edge can never be defeated. He's not allowed to play BJ anymore at his local casino, so he started playing craps for fun. We're just curious about the math of his system and how bad his expected loss is in the long run.

Quote: AxelWolf

How is it that you get an hourly EV playing a -EV game?

Seems like our resident AP is unaware that EV can be a negative value! The OP is aware he will most likely lose, and wants to know to what expected level. He also asked a few other very reasonable questions. Someone here will likely do the math for him.

This system will lose 1.52% of money bet, on average.

In the short run, the Martingale can be deceiving, with many small wins. However, the big losses will more than wipe out those small wins, eventually.

Quote: drjohnny

My friend just told me about an interesting craps system that involves Martingaling the 6 & 8 and praying the 7 doesn't hit 5 times without any 6 or 8.

The Martingale starts with $6 bets and then progresses to $18, $42, $90, and $204 bets.

He told me he's played 49 hours so far and is up about $1.8K without having to place any $204 bets yet.

Can anyone sim this system to calculate his risk of ruin for an 8 hour session and his hourly EV in the long run?

You would first have to offer a starting bankroll in order for someone to calculate the risk of ruin for you. RoR basically refers to busting out.

For example, playing eight hour sessions on a buy-in of four million dollars using the exact system you described, I can pretty confidently say his risk of ruin is zero.

Anyway, how much he loses per hour even depends on how many rolls per hour are taking place at the table. The number of rolls per hour typically negatively correlates with the number of players (or, more to the point, the number of side bets to deal with) at the table. IOW, that's also going to be variable.

What I will do is calculate for you the expected loss of each individual run of the system and then you can do as you wish with that information.

Probability

Okay, so the first thing that we have to start with is the probability of the seven coming up compared to the other two numbers, as those are the only results that matter:

There are five ways that the roll can be six; there are five ways that a roll can be an eight and there are six ways that a roll can be seven:

(6/16) = Probability of a Seven

(10/16) = Probability of a Six or Eight

Expected Values and Possible Results

SUCCESS FIRST ATTEMPT: (10/16) * 7 = $4.375

SUCCESS SECOND ATTEMPT: (6/16 * 10/16) * (21-12) = $2.109375

SUCCESS THIRD ATTEMPT: (6/16 * 6/16 * 10/16) * (49 -36 - 12) = $0.087890625

SUCCESS FOURTH ATTEMPT: (6/16 * 6/16 * 6/16 * 10/16) * (105 - 84 - 36 - 12) = $-0.88989257812 (Why create a Martingale that loses money even if the fourth attempt succeeds?)

SUCCESS FIFTH ATTEMPT: (6/16 * 6/16 * 6/16 * 6/16 * 10/16) * (238 - 180 - 84 - 36 - 12) = $-0.9146118164

***WHAT!? It loses money even on success on the fourth or fifth attempt? The goal of a Martingale is usually to cover all previous losses and win some small amount, so this system is pretty weird. Of course, anything is just house edge * amount bet in the long run, but I don't know that I'd exactly call this a Martingale.

FAILURE: (6/16)^5 * (408 + 180 + 84 + 36 + 12) = -$5.33935546875

Okay, so let's figure out our results:

(4.375+2.109375+0.087890625) - (0.88989257812+0.9146118164+5.33935546875) = $-0.57159423827

Hey, do you want to know something fun? This expected loss per system attempt also tells us roughly how much we expect to bet, on average, each time we run the system...just divide by the House Edge:

0.57159423827/.0152 = $37.6048840967

On average, you will make a total of $37.605 in bets each time you run the system. We can prove that:

(10/16) * 12 = 7.50

(6/16 * 10/16) * 36 = 8.4375

(6/16 * 6/16 * 10/16) * 84 = 7.3828125

(6/16 * 6/16 * 6/16 * 10/16) * 180 = 5.9326171875

(6/16)^4 * 408 = 8.068359375

8.068359375+5.9326171875+7.3828125+8.4375+7.50 = 37.32128906

Differences due to rounding.

Another thing you might wonder is why these average bet values aren't the same, or at least closer. Why are some eights and one is below six? The reason why is the discrepancies in how much is being bet compared to previous bets, see:

18/6 = 3x Previous at Stage 2

42/18 = 2.33x Previous at Stage 3

90/42 = 2.14285714286x Previous at Stage 4 (This is the smallest difference in terms of multiplier of previous bet)

204/90 = 2.26666666667x Previous at Stage 5

Conclusion

Well, that was fun. As far as expected loss per hour is concerned, it depends on how many rolls per hour there are.

Do systems change the house edge? No, but you already know that.

Do I like the system? Not really. The reason why is because you can have the system, "Succeed," and still lose money. I'd do something more like:

$6---$12---$36---$108---$324

Why? Because each individual bet is the total amount that you have lost to that point, except the first one.

FIRST ATTEMPT SUCCESS: $7 Profit

SECOND ATTEMPT SUCCESS: $2 Profit

THIRD ATTEMPT SUCCESS: $6 Profit

FOURTH ATTEMPT SUCCESS: $18 Profit

FIFTH ATTEMPT SUCCESS: $54 Profit

If I were inclined to play this system at all, which I'm not.

Quote: drjohnny

My friend just told me about an interesting craps system that involves Martingaling the 6 & 8 and praying the 7 doesn't hit 5 times without any 6 or 8.

The Martingale starts with $6 bets and then progresses to $18, $42, $90, and $204 bets.

He told me he's played 49 hours so far and is up about $1.8K without having to place any $204 bets yet.

Can anyone sim this system to calculate his risk of ruin for an 8 hour session and his hourly EV in the long run?

Question: if a 6 is rolled before a 7 or an 8, when do you make your next $6 bet on the 6? Immediately? Wait until a 7 or 8 is rolled? Wait until an 8 is rolled, or you lose a $204 bet on the 8?

Another question: how many rolls per hour should I assume?

Quote: ThatDonGuy

Question: if a 6 is rolled before a 7 or an 8, when do you make your next $6 bet on the 6? Immediately? Wait until a 7 or 8 is rolled? Wait until an 8 is rolled, or you lose a $204 bet on the 8?

Another question: how many rolls per hour should I assume?

I assumed on a first attempt that both bets would just stay up and you'd take the winnings. For every subsequent attempt, I assumed you would pull the winnings and reduce both bets back to $6 apiece.

It's not how much he losses, it's how much he gets short paid.
To explain that, lets take that $4 million. He will never lose $4 million. He will be short paid his winnings times minus 1.2%
I'm sure someone can come up with a better series of bets.
Of course no one with $4 million to invest is going to make $6 bets!

If you're gonna lose $720 on a 5 step Martingale, maybe bet 6 & 8 to lose for the first two bets per shooter, then bet to win.

Quote: ChumpChange

If you're gonna lose $720 on a 5 step Martingale, maybe bet 6 & 8 to lose for the first two bets per shooter, then bet to win.

That's it! Bet two sets of negative payouts to get a win!

Assuming that, if a bet is won, it is not replaced until the other number is resolved, the exact expected loss per "run" (until either both 6 and 8 appear before a 7, or a 7 is rolled) is 297,590/161,051, or about $1.8478.

Can you do the numbers for a system that goes 8 attempts at just the 6 with doubling bet on loss?

Quote: Bsballedge23

Can you do the numbers for a system that goes 8 attempts at just the 6 with doubling bet on loss?
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Assuming the initial bet is 6, since a 6 pays 7-6, and you double it after each 7:
EV(win the first bet) = 5/11 x 7
EV(win the second bet) = 6/11 x 5/11 x (14 - 6)
EV(win the third bet) = (6/11)^2 x 5/11 x (28 - 18)
EV(win the fourth bet) = (6/11)^3 x 5/11 x (56 - 42)
EV(win the fifth bet) = (6/11)^4 x 5/11 x (112 - 90)
EV(win the sixth bet) = (6/11)^5 x 5/11 x (224 - 186)
EV(win the seventh bet) = (6/11)^6 x 5/11 x (448 - 378)
EV(win the eighth bet) = (6/11)^7 x 5/11 x (896 - 762)
EV(lose all eight) = (6/11)^8 x (-1530)

Total EV = -215,622,815 / 214,358,881 = about -1.0059 based on an initial bet of 6, or 16.765% of whatever your initial bet is

Quote: drjohnny

Both my friend and I know craps is a -EV game and the house edge can never be defeated. He's not allowed to play BJ anymore at his local casino, so he started playing craps for fun. We're just curious about the math of his system and how bad his expected loss is in the long run.
link to original post

Martingaling (or any progression) doesn't change the EV of the bet. Your expected loss is 1.6% (IIRC) of the total amount wagered.

Again, a Martingale does not change the EV. What it does do is trade a small probability of a huge loss for a high probability of a small win. It may win regularly for a period of time, but sooner or later, you will hit a losing streak that breaks you (or runs into your loss limit, or the table limit). In any case, you will frequently find your self risking a lot of money to win a little.

I've been maintaining for a while now that it actually increases the amount you are expected to lose, in the long run, because your total action exposed to the house edge increases [this assumes you wind up betting more than you normally would]

Quote: odiousgambit

I've been maintaining for a while now that it actually increases the amount you are expected to lose, in the long run, because your total action exposed to the house edge increases [this assumes you wind up betting more than you normally would]
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Good point:
Expected value expressed as a percentage wagered is not affected by Martingaling and will evaluate to the house edge and be a negative value for the player.
Expected value expressed in dollar terms will tend to be a larger number than the number of wagers x minimum wager because the wagers will be in growing amounts. This EV has a good chance of equalling the player's bankroll.