Roulette: a Naive, but Fundamental Question about Probability The other day, my friend and I were discussing a simple betting strategy in roulette that would tilt the odds in the bettor's favor. (Empirically, it did well for me in the casino, but I know that luck is luck, and Vegas wasn't built on winners.)
The strategy basically comes down to this: on each roll of the wheel, I have a 2/3 chance of winning a profit, and a 1/3 chance of losing my bet (this is ignoring the "0" and "00" outcomes).
So, we spoke about where I'd put the bets. The chance of my bet falling on a losing position twice in a row is $\frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9}$; in general, the chance of my bet falling on a losing position $n$ times in a row is $p=(\frac {1}{3} )^ n$. Thus, the probability of the wheel falling on a losing position 5 times is $p=(\frac {1}{3} )^ 5 = \frac {1}{243}$ which is roughly 0.41%. At least, this is what every undergraduate probably text (that I've read) says by using examples of a coin toss.
Now, suppose I actually go to the table, and make some bets. Placing a consistent bet, I lose the first roll; the lose the second, and then the third, and then the fourth. What are my chances losing the next hand? Since each roll is independent, the probability should be $\frac {1}{3}$- or should I be taking into consideration the conditional probability? Previously, we calculated that the probability I'd lose all 5 rolls is $\frac{1}{81}$ of $\frac {1}{3}$! 
You can probably now understand my frustration. Which is it? How do I reconcile these calculations together?
Pardon me if I'm missing something very obvious. 

The conditional probability of $A$ occuring given that $B$ occurs is defined as $P(A|B) = \frac{P(A\cap B)}{P(B)}$.
In your example, let $A$ represent the event that you lose all five games, and $B$ represent that you lost the first four (possibly winning fifth and possibly losing fifth).
Since for you to lose five in a row, you must have lost at least the first four in a row, you see that in this specific case $A\cap B = A$.  (note, this is not true in most general cases but happens to be true for this scenario)
So, $P(A|B) = \frac{P(A\cap B)}{P(B)} = \frac{P(A)}{P(B)} = \frac{1/3^5}{1/3^4} = \frac{1}{3}$
Your value of $\frac{1}{81}$ of $\frac{1}{3}$ was indeed the chance of losing all five, however that was with absolutely no restriction on the outcomes of the first four games.  Once you know what the outcomes were, you should use the definition of conditional probability as written above.
