Scientific Gambling – “House Advantage”

In previous post we looked at how Betting Shops, Casinos etc make money, fundamentally by ‘salting’ the odds just a tiny bit in their favor.

Let’s use two very simple games to illustrate how this works, tossing coins and throwing dice.

Let’s start with a game of coin toss, assuming a fair coin, that is, the probability of getting heads or tails is fifty-fifty, that is, 0.5. The corresponding ‘straight’ odds for this game are thus 2 (decimal format), or 1:1 (fractional format). By ‘straight’, I mean odds that are directly given by the outcome probabilities. However, no professional betting shop can issue straight odds, if they would do so, they would very soon go out of business, simply because playing with straight odds is a zero sum game, that is, in the long run, the expected value of the game is zero, for both the player and for the house.

How to see that…? Let’s pretend we play coin toss repeatedly, very many times. In the long run, we should expect winning in half (50%) of the games, losing in the other 50% of the games. As an example, let’s say we play 100 times, the stake per game is 1$, and we have fair odds, that is 2 (inverse of 50% probability). So, these 100 games, where each game costs 1$, will cost us 100$. Statistically, we should win 50 times, lose 50 times. To make this a fair game, i.e. a zero sum game, we need to win back the cost of all the 100 games, that is 100$ within the 50 winning games. Dividing 100$ by 50 wins means that each win should return 2$, what is what will happen if we have straight odds set, i.e odds of 2 for this game.

So, in order to make money, the betting shops do not use straight odds, but instead ‘salt’ the odds a tiny bit in their favor, to ensure a healthy margin.  Let’s markup the probabilities by 5%, i.e. applying a markup of 1.05 to the odds. Now,  and instead of odds 2, we have odds 1.90 for both outcomes. Still, assuming a fair coin, we can expect winning 50% of the time. So, again, with 100 games, we should expect to win 50. The cost of the 100 games is still 100$, but the returns now are not 50 * 2 == 100, but 50 * 1.90 == 95$. That is, in the long run, we should expect to lose 5% of our stake. That markup of  of the odds, here 5% , thus generates a house advantage of 5%, which is the reason for being for any professional betting shop or casino.

For the game of throwing dice the same thing applies: in dice throwing, the probability of a fair dice to land on any of the 6 sides is 1/6, which gives that fair betting odds are 6 (decimal), or 5:1 (fractional). In order to make a profit, the betting shop must markup the odds, exactly as for the coin tossing example above.

Below two graphs shows simulations of 1.000.000 games of coin tossing and dice throwing, where the red lines show the cumulative results (total win/loss) after the 1.000.000 games. As can be seen, with fair odds both games are very close to zero sum, while with a 5% markup, the house makes a healthy profit of 5%, i.e. about 50.000 $.


About swdevperestroika

High tech industry veteran, avid hacker reluctantly transformed to mgmt consultant.
This entry was posted in Gambling, Math, Probability, Statistics and tagged , , , . Bookmark the permalink.

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