I found an online blackjack that has no minimum or maximum bet limit. The dealer stands on soft 17. Also, in the event that both dealer and player gets blackjack, it is a push. There is no insurance policy and 21 on both sides are always a tie.
I have 10,000 credits to work with.
With the Martingale betting system, I figured that I am more likely to gain a profit than to lose credits. But I'm not quite sure how to calculate the math behind this and how to calculate the probability for winning a single blackjack hand.
I figured with 10,000 credits, I can start my bet at 1, then go to 2, then 4, etc (doubling my bet everytime I lose). This allows me to lose up to 12 hands before I am bankrupt. On the other hand, if I win any of these 12 hands, I gain profit and just reset my bet to 1 again. The way I see it, I have over a 50% chance to win at least one of these 12 hands, thus always profiting.
Am I favored to gain profit in this specific type of blackjack? How can I calculate the probability for this?
$\endgroup$ 44 Answers
$\begingroup$Whether the game is in your favor is independent of the betting system. No system of betting can rescue a losing game. You are correct that with Martingale you are more likely to win than lose a given series, but if the basic bet is losing the losses will be large and probable enough to give a negative expectation.
For your example, if your chance of winning an individual hand is $0.49.0.50,0.51$, your chance of losing $12$ in a row is about $0.000310,0.000244,0.000191$. If you lose $12$ in a row, you lose $4095$, so the expectation of a series is $-0.268,0,0.215$
You can't calculate the chance of winning a hand until you specify the strategy you will follow. It is still too complicated to compute by hand, so people resort to computer modeling. You use a random number generator to deal a lot of hands and count how many you win.
$\endgroup$ $\begingroup$I was recently curious about this betting strategy as well. Instead of trying to figure out the math behind it, I wrote a simple computer program that runs 1000 trials at each permutation of start balance, base wager, and target. The model assumes a coin toss for each trial 50% chance of success (random trial number based on uniform distribution).
The betting strategy doubles your base wager each loss, and returns it to the base wager after each win. The table below shows the frequency of wins (win=reaching target balance without running out of money). To start, I tested identical proportions of parameters to compare results. So far I'm attributing the variance in results to statistical error.
I conclude that you can increase your odds of doubling your money by around 9%. It's somewhat obvious, but you maximize your gain by using the lowest base wager as percentage of start balance. The maximum avg run length was 15 trials, so I'd say this is definitely a feasible betting strategy.
If anyone out there knows how to mathematically model this, I would be happy to test out the mathematical theory and compare against the program.
I 've done the calculation in mathematica (as well as excel to double check). Under the rules infinite deck, dealer stands on soft 17, double down rights on any hand and no splitting (yet!). The total expectation is -0.0116769, ie 1.16769% in favor of the house.
I am now working on a Monte Carlo simulation to see the convergence between the theoretical and simulated expectation. I am looking forward to meeting others with the same calcs to exchange views and opinions. Anyone to prove/disprove me ?
$\endgroup$ $\begingroup$Blackjack is a solvable game and I 've solved it! Take a look at the results here, Using Mathematica.
$\endgroup$
