The Anti-Martingale Strategy

I'm sure this has been discussed before, probably in the thread right below this one in the math section. But I'm heading to Las Vegas tomorrow morning and can't find the information that I need right away. I know Martingale blows chunks for the player so what if you reverse it and make it work for you. For example, you double up your bet on a win until you hit the table limit and then walk away.

Let's say that a table game has a 3 percent house edge, a $5 minimum and a $1000 maximum bet. You would have to win 8 times in a row before you reach the table maximum. Supposing that I had the bankroll to do it, how many hands (spins, rolls, tosses, twirls, etc) would it take to reach a %75 probability of having an 8 game winning streak?

I will play that many hands and if I don't get my streak, then I'll assume that it just wasn't in the cards.

Quote: s2dbaker

Supposing that I had the bankroll to do it, how many hands (spins, rolls, tosses, twirls, etc) would it take to reach a %75 probability of having an 8 game winning streak?

For 75% it's 1093 hands. The 3% HA advantage hurts you pretty badly. It would be 704 if there was no HA.

Try something a little less ambitious. Lay a 6 or and 8 in craps, and try to win 6 times in a row. The HA is much lower, and the 75% probability will occur in 160 bets resolved. Use $10 as your initial bet instead of $5 if the jackpot of 32*$5 is too small for you to feel good.

BTW, anti-martingale is just as meaningless as martingale.

My RNG came up with 1094 but I noticed an interesting thing. I counted the number of trials and stored it in a table. The most frequent number of trials to win 8 in a row is 8 trials. The longest I went without 8 in a row is 8331 trials. I'll let this run all night and see if it normalizes.

Quote: s2dbaker

My RNG came up with 1094 but I noticed an interesting thing. I counted the number of trials and stored it in a table. The most frequent number of trials to win 8 in a row is 8 trials. The longest I went without 8 in a row is 8331 trials. I'll let this run all night and see if it normalizes.

Well the difference between 1093 and 1094 theoretically is 75.006% and 75.038% so I'm not surprised that the simulation would miss it by that much.

Your second observation is not an artifact. Ask your self the much simpler question of how many trials do you expect to play before you win? Assume a 50/50 game initially.
The answer is
half the time you expect to win on 1st trial;
1/4 the time you expect to win on 2nd trial;
1/8 the time you expect to win on 3nd trial;
...
On average you expect to win on the 1+1/2+1/4+1/8+...= 2nd trial
But most of the time you will in one trial. That last fact won't change for a house edge.

It's basically the same reasoning for a streak of 8 wins. It will take 1093 trials to get a 75% average, but the majority of times when it happens will be the first 8.

Quote: s2dbaker

My RNG came up with 1094 but I noticed an interesting thing. I counted the number of trials and stored it in a table. The most frequent number of trials to win 8 in a row is 8 trials. The longest I went without 8 in a row is 8331 trials. I'll let this run all night and see if it normalizes.

It will never normalize since the distribution is NOT normal.
More geometric.
I like photos.
Here is the histogram
p = 0.47
run = 8

How about the curve
BTW x-axis is log

added:
I also show the prob at 75% for the chance of at least 1 run of length 8 in 1093 trials
also show the prob at ~40% for the chance of at least 2 runs of length 8 in 1093 trials

Nice stuff! Thanks!

BTW, it ran all night and eventually settled at 1093. One unlucky run went 14,677.

I do believe the verdict is in, the Anti-Martingale is the suck!

Quote: s2dbaker

I do believe the verdict is in, the Anti-Martingale is the suck!

It doesn't so much suck as it doesn't matter. Casinos like people who have betting strategies because it keeps player playing despite losses, because they think the strategy will pay off in the end.

But at least you can gauge what to expect. If you want to start with a larger amount (SAY $50) and intend to play until you have a streak of 4 wins, now you know that is expected to take 30 trials if there was no house edge. If you add a small house edge like laying a 6 or and 8, the number of trials moves up to 31. But if you add a larger house edge, the expected number of trials is now 34.5 .

But the expected number of trials will to get this streak of 4 wins will vary widely. But HA advantage is going to matter a lot if you play repeatedly. They didn't invent that term grindhouse for nothing. The HA just grinds away at you.

Which is why I don't understand the Chinese fascination with Baccarat. If you are going to grind away all day, why not play craps with full odds. The probability of doubling your money before going bankrupt is much higher than Baccarat.

Quote: pacomartin

But the expected number of trials will to get this streak of 4 wins will vary widely. But HA advantage is going to matter a lot if you play repeatedly. They didn't invent that term grindhouse for nothing. The HA just grinds away at you.

Which is why I don't understand the Chinese fascination with Baccarat. If you are going to grind away all day, why not play craps with full odds. The probability of doubling your money before going bankrupt is much higher than Baccarat.

You say that a lot, and I agree with it. Last time I was in Vegas, I took $1,000 to the craps table at Casino Royale and my friend matched it. We played $5 craps with 100 times odds. We were either going to walk away rich or go broke very quickly. Guess what happened? (I'm still here :p).

you could try 2 double up 7 times in 250 trials, or why not 350 trials

Quote: edward

you could try 2 double up 7 times in 250 trials, or why not 350 trials

The lower the number of trials the lower the probability of a run of a specific length.
the mean wait time for run7 is ~371 trials(the median is always lower)
and the std dev is almost the same ~365

added: formulas:
mean wait time ((1/p^n)-1)/(1-p) [One should be familiar with this one]
variance: Var(N) = 1−(p^(1+2r)−qpr(1+2r)/q^2p^2r
Sequences, Patterns and Coincidences page7

It is that possibility of a long wait time that will kill a marty especially when p <= 0.5