Probability problem from CMU contest prep course Your sports team is playing a best-of-15 series against a single opponent. Against this opponent, your
team wins with probability 40%, unless it is behind in the series (with strictly fewer wins than losses
so far in these 15-game finals), at which point your team wins with probability 60%. What is the
probability that your team wins?
I'm not great at math so I wrote a dynamic programming algorithm to solve it. If my algorithm is correct, the answer would be 40%. I found that if the number of matches is odd, the probability that you win is always 40%, for any number of matches.
Can anyone verify this and provide a probabilistic explanation of the solution?
Thanks!

Note: What's crucial here is that $0.4 + 0.6 = 1$. Otherwise, this will not work.
Let the series be denoted by an element of $ S =  \{ L, W \} ^{15}$.
Hint: Create a bijection $f: S \rightarrow S$ between matches that you won, and matches that you lost

 - If you won the series, consider all of the matches (you won more games)
 - Take that series, and 1) Swap each of the W into a L and vice versa, and 2) Reverse the game order
 - This then becomes a series that you lost (since you lost more games)

Claim: P(You win) / 40% = P(You lost ) / 60%.

 Proof: Say you won with $s$.
 For the last match that you won, you won with probability 40%.
 Now consider $f(s)$.
 For the first match that you lost, you lost with probability 60%.
 Ignoring that match in both series (which correspond to the same match), verify that the probability of the sequence happening is the same. (There's a small step here that you should verify the probabilities match up, which has to do with how the games are defined).

Corollary: P(You win) = 40%.

As an explicit example, consider a round of 5 games where $ s = WLLWW$ where you won.
Then $f(s) = LLWWL$ where you lost.
$P(s) = 0.4 \times 0.6 \times 0.6 \times 0.6 \times 0.4$.
Ignoring the last win, we get $P(s) / 0.4 = 0.4 \times 0.6 \times 0.6 \times 0.6 $.
$P(f(s) ) = 0.6 \times 0.4 \times 0.6 \times 0.6 \times 0.6 $
Ignoring the first loss, we get $ P (f(s) ) / 0.6 = 0.4 \times 0.6 \times 0.6 \times 0.6$.
Hence, the probabilities match up.
