Simultaneous game about conserving resources Two players start with $50$ stones each. Every round, each player chooses some ($>0$, unless they have no stones left) of their remaining stones, and holds them in their hand. Then they simultaneously show each other their choices, and the player with the most stones in their hand wins the round (if equal both lose). The stones in their hands are then discarded, and the next round begins. The first player who wins $3$ rounds in a row wins the game. If both run out of stones before this happens, the game is declared a draw. 
An example game:




P1 Stones Remaining
P1 Choice
P2 Stones Remaining
P2 Choice
Winner




40
10
35
15
P2


30
10
20
15
P2


10
20
0
20
-


9
1
0
0
P1


8
1
0
0
P1


7
1
0
0
P1




P1 wins the game.
It is simple to see that this game has no pure strategy Nash equilibria:
For any pure strategy we can construct one that beats it: If choosing $1$ stone would result in you having $3$ or more stones than your opponent, choose $1$, otherwise beat their choice by $1$. If you have $3$ or more stones than your opponent, just choose $1$ more than your opponent $3$ times.
However, I am unsure about strategies that employ randomness. Is there a mixed strategy equilibrium that is describable? 
It is worth mentioning that this game is almost equivalent to tennis.
 A: Long comment :
(potential misinterpretation : I replaced "The first player who wins 3 games in a row wins the game." by "the first player to win 3 rounds (not necessarily in a row) wins the overall game")
When enemy is dumb :
Suppose Alice knows Bob employs no strategy . Then in the $n$th round Alice and Bob pick $X_n \in \{1,...,50-\sum_{i=1}^{n-1} X_i\} , Y_n\sim unif(1,50 - \sum_{i=1}^{n-1} Y_i)$ stones respectively .
Alice acts as follow , for $n < 3$  always pick $1$ stone , for $n\ge 3$ always pick
$50 - \sum_{i=1}^{n-1} Y_i$ stones .
This yields $\sim 0.9936$ win rate (over 10000000 games with sd = 2.899172e-05) .
Rationale  :
Numerical results suggest  $E[Y_n] \approx  \frac{1}{2} E[Y_{n-1}] $ , so the number of stones Bob is expected to pick is largest when $n$ is small and then decreases , so we can expect later rounds to cost less resources to win .
Of cause this gives rise to many other strategies too . The above one is unlikely optimal .
When enemy isn't dumb ? I don't think both players could forecast each other using the same model if it requires real time data .
