I’m working on my own hare-brained scheme to Break Vegas…or at least not lose all my money so fast. I’m been playing free blackjack on the net here and there and using my latest betting scheme, I always come out ahead. It’s a slow progression but I always end up in the black. I just have a few questions before I take it to Atlantic City.
Assuming I’m on a standard Blackjack table where the dealer hits on a soft 17 and BJ pats 3-2, and using the basic BJ strategy, is it possible to figure out my odds of losing seven consecutive hands?
Secondly, is there a standard maximum for every minimum? For instance, is every maximum on a $5 table the same and every $10 table the same? And what would that be exactly? Are the mins - maxs on video BJ the same as the table games? I’d prefer the video because I can play faster.
Well, I don’t want to lose more that 6 consecutive hands. You will always come out ahead unless you lose 7 straight. This idea of mine also hinges on what the maximum bets are also…
Well, a quick Monte Carlo (no pun intended) analysis where your odds are exactly 50%, you’ll get 7 losses in a row after about 256 hands, on average. Since house rules will never give you 50% odds, you should expect to see 7 in a row sooner.
Whoa…I knew I couldn’t have been the first person to think of this, but I couldn’t find it anywhere on the internet. I like where it explains why the casinos encourage this type of play. You have to have discipline…
I got involved in on-line gambling for the first (and last) time a while back trying this. The initial deposit was $50, and a couple hours later I had a balance over $500. I wanted to cash out, and that’s when I learned just how difficult these sites are to deal with.
Pretty much no place in Vegas offers 3:2 payouts, either. They’ve all swtiched to 6:5 and hit a soft 17, which really beefs up the house edge. In my observation, the max at a table seems to be 100X the minimum most of the time. But I wouldn’t swear to it.
Well, the expected value per hand will be less than 100% of the bet, including the splits/double-downs/etc, so you should expect to lose even more often, but make it up with some larger wins.
With this scheme you are accepting a small risk of catastrophic loss in order to achieve a high likelihood of small gains.
Anyone who has spent any time thinking about gambling has thought of (or heard of) the martingale scheme. If it worked, casinos would not be profitable and wouldn’t exist.
If you want to make money by gambling, play poker and take it from the other players. You’re not going to beat the house (inventive card counting schemes possibly excepted).
I haven’t been to Vegas in a year or two but when I was there, the 6:5 payout was usually just for single deck games. Has this become the standard now? The dealer hitting a soft 17 was about 50/50. Those that did hit it usually offered an early surrender.
King Friday, casinos encourage every kind of betting system except for cheating.
There is one very important thing to remember about gambling schemes: If a single game (or hand, or roll, or whatever) has a negative expectation, then any combination of bets on any combination of games will also have a negative expectation, which will be at least as bad as any single game. In other words, if it’s a sucker bet, it’s always a sucker bet.
Now, there are actually a very small number of cases where you can get a positive expectation bet. Most of them involve betting against other suckers, so the house doesn’t care. If you count cards in blackjack, you can just barely beat the odds playing against the house, but it’s so much hard work for so little profit to pull it off well, that it just isn’t worth it on that score.
The real way to win in Vegas is to enjoy the game you’re playing (whatever it is), and not mind the fact that you’re losing money to play it.
I was just there a couple of weeks ago, and all of the multiple deck games (6 decks) were 3:2 payouts. All of the single deck and double deck games were 6:5.
Not really discipline. I mean, seriously, how much discipline does it take to stick to Martingale? It’s a very simple system and, as has been said, most gamblers happen on it at some point. You need deep pockets and no limits on the table to implement the system ideally.
To answer your question, the odds of you losing 7 hands in a row is going to be a little better than 1 in 128 (that number assumes 50-50 odds on all hands.) So, as has been stated before, the vast majority of the time, you will win a little bit of money with this system. But when you lose, you will lose big. If you play long enough, you should expect to win (or, rather, lose) the same amount as you would playing single hands with fixed bets.
#!/usr/bin/perl -w
my $trials = 100000;
my $total = 0.0;
my $square = 0.0;
my $failures = 7;
for(my $i = 0; $i < $trials; ++$i)
{
my $j = 0;
my $l = 0;
while($l < $failures)
{
$j++;
if (rand() < 0.5)
{
$l++;
}
else
{
$l = 0;
}
}
$total += $j;
$square += $j * $j;
}
my $avg = $total / $trials;
my $stdDev = sqrt($square/$trials - $avg * $avg);
print "$avg hands to fail $failures times.
Standard deviation: $stdDev
";
Sample run:
254.32722 hands to fail 7 times.
Standard deviation: 249.492980997606
So running it once I got a number really close to 256, but from running it more times and checking for different number of repeated failures, it’s not just some simple power of 2 thing. For example:
6.00834 hands to fail 2 times.
Standard deviation: 4.70525987001781
Okay, looking at the progression of expected hands, the monte carlo simulation gives number very close to the series 2, 6, 14, 30… which looks a whole lot like 2 * (2^f - 1). That would make the expected number of hands for f = 7 equal to 254, not 256.
Interesting. I wonder what I’m missing. I guess it’s a slightly different way of phrasing the answer. What are the odds of getting two heads in a row? 1 in 4. On average, how many times do you have to flip a coin before you see two heads in a row? I guess the answer is 6, according to your simulation.