Probability of not dropping two games in row A team has a $40$ game season, and the probability of winning each game is $65\%$.  How could I determine the probability that the team would not drop consecutive games during the $40$ game season?
 A: If they lose $k$ games, there are aexactly $41-k\choose k$ possible win-lose sequences that avoid two consecutive losses. We have to weight these with the respective probabilities and sum over $k$ (up to a maximum of $20$ as beyond that consecutive losses are unavoidable), that is
$$ \sum_{k=0}^{20}{41-k\choose k}\cdot 0.35^k\cdot 0.65^{40-k}.$$
I compute $\approx 0.0177$ - so even an over-achieving team is verrry likely to have a bad streak.
A: I get the same answer as Hagen von Eitzen. I want to explain the formula that comes out from the combinatoric considerations: this is the way I did it using high school methods but I'm sure there are many more to choose from, some of them more concise! I included my favorite "sneaky" method as an addendum (not much harder to understand, but in my experience, people find it harder to "spot").
Clearly losing games must be separated from each other somehow. Several possible necessary and sufficient conditions would ensure this, e.g.:


*

*Each loss is followed by a win, except perhaps the final loss (it may be the 40th game)

*Each loss is preceded by a win, except perhaps the first loss (it may be the 1st game)


I'll take the first option here, but the second works similarly. The $k-1$ non-final losses can be paired up with the following win, and treated as a single unit. Treating the other individual games as single units, I have:


*

*$k-1$ two-game units (LW) containing the earlier losses and their subsequent wins (making $2k-2$ games in total),

*1 single-game unit (L) containing the final loss, 

*$40-(2k-2)-1=41-2k$ single-game units (W) containing the remaining (unpaired) wins.


But $41-2k\ge 0$ only if $k \le 20.5$ and since the number of losses is integer, there can be at most twenty losses before consecutive losses become inevitable. That matches our intuition.
I now have a total of $(k-1)+1+(41-2k)=41-k$ units, of three types, that can appear in any order except for the restriction that the final loss must occur after all the LW units. This seems complicated - I have three types of unit, and a restriction on the order they can appear in! But thinking about it, I don't need to distinguish the two types of loss-containing units. Whatever the arrangement of my $k$ loss-containing units and my $41-2k$ unpaired wins, the final loss-containing unit is necessarily "L" and the preceding ones are "LW". I'm now back in the easy territory of two types of unit, that can appear in any order.
I can choose the $k$ positions to be filled by the loss-containing units out of $41-k$ available, which gives $41-k\choose k$ possible win/loss sequences. Or I could choose the $41-2k$ positions to be filled by the unpaired wins, giving $41-k\choose 41-2k$. Which is, of course, exactly the same!
The final application of binomial probability gives $$ \sum_{k=0}^{20}{41-k\choose k}\cdot 0.35^k\cdot 0.65^{40-k}=0.0176547...$$ to 6 significant figures.
ADDENDUM: There's a fast but sneaky method I realized would work, but wanted to stick to the "pairing" method used in introductory textbooks (at least on the British high school curriculum) and therefore what I teach to my students. This alternative was taught in my first term at college. 
The $k$ losses split the season into $k+1$ winning streaks. The first and last winning streaks might be zero games long (as the first and/or last game of the season may be defeats) but the other winning streaks must be of positive length (to avoid consecutive losses). Annoyingly this breaks the symmetry, so I'm going to add two fake games to the sequence - a pre-season win, and a post-season win. Warned you it was sneaky! Now all my winning streaks are guaranteed positive length.
After those additions I have $42-k$ wins, and want to know how to subdivide them into $k+1$ winning streaks. In other words, how many ways are there to add $k+1$ natural numbers and reach the sum $42-k$? (If this sounds vaguely familiar you may be interested in this question.) I can imagine my $42-k$ wins having dividers between them: $$W | W | W | W | .... | W$$ How many such dividers are there? The fencepost problem tells me it's one less then the number of wins. So there are $41-k$ dividers, of which $k$ are going to be occupied by streak-breaking losses. The number of ways this can be achieved is, of course, $41-k \choose k$.
If you don't like the two fake wins, you can instead visualize a run of $40-k$ wins with $k$ losses able to occupy dividing positions between wins, or before the first win, or after the last: |W|W|...|W|. Fencepost considerations show there are $(40-k)+1=41-k$ possible slots for the $k$ losses to be chosen from. This is less contrived (no fake games!), but counting the length of winning streaks (perhaps one of the more "obvious" ways to think about this problem) doesn't work so nicely as we lose the nice analogy to the well-known problem of "how many ways can $a$ natural numbers be chosen that sum to $b$".
