# Assessing Probability – Common Myths and Mistakes

Let us have a look at common myths and mistakes that players of casino games make when assessing probability and let us also find out why it is necessary to beware of rash or seemingly "clear" conclusions.

## Myth: There Must Be a Change After a Repetition

The following case is a common mistake that players make. Imagine that you play Roulette and the red color has shown up 5times in a row. Now which color would you bet on and why? The answer is: It does not matter, which color you pick, since the chance to win is still the same. How is this possible? Do you think that the black color gives better chances? Unfortunately not.

It is necessary to realize that each spin in Roulette represents a brand new event with the same probability `18/37`

(in French Roulette) for either red or black color to come up. Roulette has no memory! Despite that many players would bet on the black color, but it is absolutely unjustified. There is also a famous Roulette system called Martingale, which is based on a similar principle (the stakes are doubled after a lost spin).

However in the long-term point of view (this is certainly not a night spent in a casino) there is a tendency of the outcomes to average. For instance, the more times you roll a dice, the greater is the chance that the numbers 1–6 would be distributed in a roughly equal way (assuming we have a perfectly symmetric dice) – we talk about the law of probability and the concept of expected value.

## Comparison: Simple vs Multiple Events

The similar laws are also valid for small and big, even and odd numbers in Roulette, when rolling a dice or a coin and essentially for all other casino or gambling games. We talked about a simple / individual / new event.

There can also be events (or wagers) that follow each other – we bet on a series of outcomes. Imagine that you bet with your friend that you would be able to roll the same side of a coin in three successive attempts. What is the probability of your success?

We will use the combinatorial rule of product as the probabilities of multiple events have to be multiplied. The probability to roll either side of the coin is` 1/2 = 0.5`

. The probability to roll the same side of the coin 3 times in a row is then `0.5 × 0.5 × 0.5 = 0.125`

(or 12.5% if you like).

That is the difference between e and a multiple event (or series of events). There is another strategy that uses multiple events – the Anti-Martingale Roulette system.

## Example: Four Children

Thirdly we can demonstrate the assessment of probability on this particular example. Imagine parents who would like to have four children. What is the most probable distribution of boys and girls? Do you think that two boys and two girls is the most probable outcome? Unfortunately it is not the correct answer.

If we disregard some biological factors, we can assume that both boys and girls can be born with the same probability 0.5. Note: In fact there is 0.515 probability that a boy would be born, since the long-term statistics have shown that there is about 515 boys out of 1000 children born in average; this is how the nature works, on the other side a man's life is shorter. Let us write out all possible combinations of the children – from boys (B) only to girls only (G).

1. BBBB 2. BBBG 3. BBGB 4. BGBB 5. GBBB 6. BBGG 7. BGBG 8. BGGB 9. GBBG 10. GBGB 11. GGBB 12. GBBB 13. GBGG 14. GGBG 15. GGGB 16. GGGG

We can see that there are 16 combinations or distributions that can occur. As an illustration we can calculate all probabilities.

The probability that only boys would be born is `1/16`

(only 1 case out of 16). We can also demonstrate the combinatorial rule of sum – for example in this case: what is the probability that either boys only or girls only would be born? There is the only one possibility that four boys would be born and as well as the only possibility that four girls would be born. The probability that the children would be of the same gender is then `2/16 = 1/8 = 0.125`

(or 12.5% if you like).

Now let us go back to our initial question. The probability that two boys and two girls would be born (distribution 6.–11., see the illustration above) is then `6/16 = 0.375`

.

There are 8 possibilities left, which mean that either 1 girl or 1 boy would be born and the rest of the children would be of a different gender. Therefore this distribution with the probability `8/16 = 0.5 > 0.375`

is the most probable outcome. Thus the answer, which seemed to be the most likely "at first sight", is not correct.