Posted by: mbuckley56 | December 7, 2008

Gambling Strategies

Seeing as Casino Night is coming up, I decided to make my post about gambling strategies. The first thing I am going to do is analyze the situation for the night. Basically, lets say everybody has equal odds going in to the night and everyone has a similar skill level for all of the games we will be playing.  Also, the games that are being played are blackjack, craps, and roulette. All of these are played against the “house” as opposed to poker, where players play against eachother. Basically, since the games are against the house, in the long run everyone is going to lose (gamblers fallacy). In one night, most people will lose, although due to random luck some people will come out on top. My theory is you need to maximize your probability that you will be one of the people who can take advantage of random luck. According to the Atlantic City rulebook, the probability of winning a black jack hand is 43.31%( If you play 100 hands of blackjack, you will likely lose about 57 and win about 43. Maybe you’ll get lucky and win 57 and lose 44. But the more games you play, the more likely you will be part of the “long run” losers segment(law of large numbers). So, my strategy is to play a very few number of games. For instance, bet half of your money on one blackjack hand twice in a row. Yes you might lose very quickly, but this is the same outcome that would most likely occur if you played 100 games. If you happen to get lucky and win the first two hands, (lets say .43*.43 = .18). Then you end the game. Chances are most other people in the room wont be using a similar strategy, and will have lost all their money to the house by the end of the night. Even the lucky winners will probably only come out with a very small amount of money. Thus, you basically have given yourself a .18 probability to be the top winner of the night, whereas if you used the same strategy as everyone else your chances of being on top would be 1/n or roughly .04 for our class.  Another good strategy would be to not bet any money, but you would most likely lose out to the individuals who get lucky and win for the night. I actually wrote this blog post last week because I knew I had a busy week this week. So I don’t really know if I will use my strategy or how it would fare or how lucky people will get.

The only game I really know about from the three at casino night is blackjack. 

Other than counting cards, the only real strategy in black jack involves memorizing the table that tells you what to do in every situation. Although most people don’t realize, constructing the table is very mathematical and deals with the odds that one would have in winning each of the situations.  A different strategy that I thought of once is very simple. Basically, you bet $1. If you lose, you bet $2. If you lose, you bet 4$. if you lose, you bet $8, etc. etc. If you win a bet, than you start back at $1. Eventually, you will win 2 bets in a row, thus capturing that $1 profit. This seems to be a perfect game. The only problem is, casinos set table limits on bets. So, if you get unlucky enough, and you lose like 10 hands in a row (Very unlikely with a .47 chance of losing a hand). You could switch tables to get a higher limit, but eventually there might not be a table that takes large enough bets. I guess you could switch casinos, but then its getting ridiculous. The other problem is that you could run out of money. People with inifinite money usually don’t play blackjack, I would imagine. After writing this little blurb, I decided to research on wikipedia. Apparently this strategy is called the Martingale betting system. Wikipedia has a mathematical breakdown of the reasons I gave for it not being a good strategy. I have copied that math here:

     Let one round be defined as a sequence of consecutive losses followed by a win, or consecutive losses resulting in bankruptcy of the gambler.       After a win, the gambler “resets” and is considered to have started a new round. A continuous sequence of martingale bets can thus be            partitioned into a sequence of independent rounds. We will analyze the expected value of one round.

Let q be the probability of losing (e.g. for roulette it is 20/38). Let y be the amount of the commencing bet. Let x be the finite number of bets you can afford to lose.

The probability that you lose all x bets is qx. When you lose all your bets, the amount of money you lose is

\sum_{i=1}^x y \cdot 2^{i-1} = y (2^x - 1)

The probability that you do not lose all x bets is 1 − qx. If you do not lose all x bets, you win y amount of money (the initial bet amount). So the expected profit per round is

(1-q^x) \cdot y - q^x \cdot y (2^x - 1) = y (1 - (2q)^x)

Whenever q > 1/2, the expression 1 − (2q)x < 0 for all x > 0. That means for any game where it is more likely to lose than to win (e.g. all chance gambling games), you are expected to lose money on average per round. Furthermore, the more times you are able to afford to bet, the more you will lose.




  1. At this time I am ready to do my breakfast, when having my breakfast coming yet
    again to read further news.

  2. Hi there! Quick question that’s entirely off topic.
    Do you know how to make your site mobile friendly? My blog looks weird when viewing from my apple iphone.

    I’m trying to find a theme or plugin that might be able to correct this problem.
    If you have any recommendations, please share. Thank you!

  3. I hardly leave comments, however i did some searching and wound
    up here Gambling Strategies | The Math 152 Weblog. And I do
    have a couple of questions for you if it’s allright. Is it just me or does it look as if like a
    few of the responses look like left by brain
    dead people? 😛 And, if you are writing at other sites, I would like to keep up with
    everything fresh you have to post. Would you list of every one of all your social sites like your Facebook page, twitter feed,
    or linkedin profile?

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