Playing Odd-Even Cricket, is there a perfect strategy This is a simple two-player game. One if the people is picked to 'bat'.
Both players simultaneous choose a number from 1 to 6. (When playing against a person, you use your hands to show the number). If both players show the same number, the person 'batting' is out. Otherwise, the person batting gets points (runs) equal to the number he picked.
This process repeats until the person batting is out. His final score is the sum of all the points he scored.
The person batting and the other player switch roles. The same process occurs. Whoever scored more points wins.
Is there a perfect strategy for this game?
When playing against a computer, you stand exactly 1/2 chance of winning. You may sometimes be able to predict humans and play better. Assume you don't know who you are playing? Is the perfect strategy just to always pick random numbers, or to predict the opponent's future moves based on his current ones?
 A: This is also not a complete answer, but it maybe approximates one under the right circumstances.
The simplest reasonable heuristic that gives an answer in every situation is to compute the strategy that maximizes your expected value (i.e., instead of playing to win, you're playing for a penny per run). This should be nearly correct when either you need a lot of runs to win or there's a lot of uncertainty about how many you need.
Suppose the batsman adopts the strategy of choosing the number $k$ with probability $a_k$, and the bowler chooses $k$ with probability $b_k$. It's actually useful (because it makes things more symmetric) to write things in terms of the bowler's inverse probability $\overline{b}_k=1-b_k$. Then the expected number of runs $E$ that will be scored by the batsman is given by:
$$
E=\sum_{k=1}^6 (k+E)a_k\overline{b}_k
$$
(this is saying that if the batsman plays $k$ and the bowler doesn't, the batsman scores $k$ runs and then gets to start over). The problem also implies the constraints
$$
\sum_{k=1}^6 a_k = 1, \quad\sum_{k=1}^6 \overline{b}_k = 5,\\ \quad a_k, \overline{b}_k \in [0, 1] \text{ for all } k \in \{1,2,3,4,5,6\}
$$
This will yield a finite value for $E$ whenever $\sum a_k\overline{b}_k < 1$, which the bowler can guarantee by choosing $\overline{b}_k < 1$ for all $k$. So the game terminates with probability $1$ unless the bowler is extremely foolish, and it's reasonable to ask about optimal strategies.
These will be given by saddle points of the function $E(a_k,\overline{b}_k)$. To deal with the constraints, we write $a_1,\overline{b}_1$ in terms of the other variables, so
$$
E=\sum_{k=2}^6 (k+E)a_k\overline{b}_k + (1+E)\left(1-\sum_{k=2}^6 a_k\right)\left(5-\sum_{k=2}^6 \overline{b}_k\right)
$$
Then, differentiating, we have
$$
E_{a_k}=(k+E)\overline{b}_k +E_{a_k}a_k\overline{b}_k+\left[(1+E)(-1) + E_{a_k}\left(1-\sum_{k=2}^6 a_k\right)\right]\left(5-\sum_{k=2}^6 \overline{b}_k\right) \\
E_{\overline{b}_k}=(k+E)a_k+E_{\overline{b}_k} a_k \overline{b}_k + \left[(1+E)(-1) + E_{\overline{b}_k}\left(5-\sum_{k=2}^6 \overline{b}_k\right)\right]\left(1-\sum_{k=2}^6 a_k\right)
$$
At a critical point of $E$, we must have $E_{a_k}=E_{\overline{b}_k}=0$. So these equations reduce to
$$
0=(k+E)\overline{b}_k -(1+E)\overline{b}_1\\
0=(k+E)a_k - (1+E)a_1
$$
which yield the relations
$$
a_k=\frac{1+E}{k+E} a_1 \\
\overline{b}_k=\frac{1+E}{k+E} \overline{b}_1
$$
(strictly speaking, we've only derived these relations when $k>1$, but they clearly also hold when $k=1$).
Note that this means that $a_k$, $\overline{b}_k$ both decrease as $k$ increases. Since $a_k$ decreases as $k$ increases, the batsman wants to have a bias toward lower numbers. Why? Because $\overline{b}_k$ decreases as $k$ increases, which means $b_k$ increases as $k$ increases. That is, the bowler is preferentially defending the higher numbers, and so the batsman would rather stay away from them — and keep playing — than take the risk of getting out in exchange for a slightly higher score in the current round.
Continuing with our solution, we have
$$
\sum_{k=1}^6 \frac{1+E}{k+E} a_1 = 1, \\ \sum_{k=1}^6 \frac{1+E}{k+E} \overline{b}_1 = 5
$$
which yields $\overline{b}_1=5a_1$, and so $\overline{b}_k=5a_k$ for all $k$.
Plugging all of this back into the original equation for $E$, we have
$$
E = \sum_{k=1}^6 (k+E) a_k \overline{b}_k = \sum_{k=1}^6 \frac{(1+E)^2}{k+E}a_1\overline{b}_1=(1+E)^2a_1\overline{b}_1 \sum_{k=1}^6 \frac{1}{k+E} 
$$
which implies that
$$
E = (1+E)\overline{b}_1,\\
E = 5(1+E)a_1
$$
and so $a_1=\frac{E}{5(1+E)}$, $\overline{b}_1=\frac{E}{1+E}$. Finally, this gives us a $1$-variable equation for the optimal value of $E$:
$$
E=\frac{E^2}{5}\sum_{k=1}^6 \frac{1}{k+E}
$$
which, upon dividing through by $E$, reduces to a sextic with only one meaningful root: $E \approx 16.777$. Since this is the only critical point and $E$ can be either zero or infinite on the boundary, it must be a saddle point — thus it gives the minimax strategy we're looking for. Plugging this back into the relations for $a_k$, $\overline{b}_k$ and using the fact that $b_k=1-\overline{b}_k$, we get
$$
\begin{array}{cc}
a_1 \approx 0.18875 & b_1 \approx 0.05625\\
a_2 \approx 0.17870 & b_2 \approx 0.10651\\
a_3 \approx 0.16966 & b_3 \approx 0.15169\\
a_4 \approx 0.16150 & b_4 \approx 0.19252\\
a_5 \approx 0.15408 & b_5 \approx 0.22960 \\
a_6 \approx 0.14732 & b_6 \approx 0.26342
\end{array}
$$
So the batsman should actually choose pretty close to uniformly. The bowler, on the other hand, should work fairly hard at defending against the high-scoring plays.
Qualitatively, I think the right way to make sense of this is as follows. Since $6 \gg 2$, the batsman can score more in the long run by not getting out than by maximizing his scoring potential in the current round. The best way to not get out is to be as unpredictable as possible — that is, to keep his distribution near-uniform. If you grant that the batsman's distribution will be near-uniform, the bowler has very little control over how long the game goes — the probability of stopping in any given round will be near $\frac{1}{6}$ no matter what she does. So all she has left to play for is score in the current round, meaning she wants to mostly play large numbers (while playing small numbers just enough to keep the batsman honest). Finally, this means that whatever deviation the batsman makes from uniformity should be in the direction of playing smaller numbers — this will keep the game going for slightly longer.
A: (Too long for a comment)
When scores are tied, either player can pick a number a random. The batsman would then win at least 5/6 times, whatever the bowler did; and the bowler would win at least 1/6 times, whatever the batsman did.
With 2 runs to win, say the batsman chose $1$ with probability $1-5p$, and any other number with probability $p$ each.  If the bowler picks $1$, he wins with probability $1-5p$, and for any other number he wins with probability $p+(1/6)(1-5p)=(1/6)+(1/6)p$.  So the batsman should choose $p=5/31$, and the bowler can't do any better than $6/31$.
With 2 runs to lose, say the bowler chose $1$ with probability $1-5q$, and any other number with probability $q$ each.  Again, the batsman's chances of winning by picking $1$ are $(5/6)(5q)$, and any other choice, it is $1-q$.  So the bowler should set $25q/6=1-q$, $q=6/31$, and the batsman can do no better than $25/31$.
So each play has a strategy that guarantees at least $25/31$, or $6/31$; and that is the probability with 2 runs to win.  (I gave a tie as a win for the bowler; you could fix that.)
