Dice game with an opponent I and my opponent play a game of dices. I win if my opponent didn't throw a dice in our game with a higher number than my highest number.
Example :
I have 2 dices, he's got 3 . 
Our throws : (1,4,2,4,3) , I threw 1,4 he 2,3,4. I won in this case because he didn't get any higher number than my 4. (he should've gotten a 5 or 6) The dices are also distinguishable.
Give the number of possible combinations in which I can win if:
1) I have 2 dices, he's got 1.
2) I have 3 dices and my opponent 2.
My incomplete solution:(let $i$ be my throws and $o$ for opponent)
1) (i1, i2, o1) here are possible $6^3$ total combinations .
$wins=6^3 - |o1>max(i1,i2)|$ (but I don't know how to count it)
2) no idea
Thank you.
 A: In general, if you have $\max(Z_1, Z_2, Z_3, ..., Z_n) = m$ for dice $Z_i$ and max outcome $m$, then you have $M_{m,n} = \sum_{k=1}^{n} \binom{n}{k} (m-1)^{n-k}$ ways to get that outcome.
You choose $k$ of the dice to equal $m$, and then the remaining $n-k$ dice can take on one of $m-1$ remaining values each. This simplifies to $M_{m,n} = m^n - (m-1)^n$
So for each max we can achieve, $m_1$, we wish to compute the number of ways we can achieve that max, multiplied by the number of ways the opponent achieves a max $m_2 \leq m_1$.
Let $m_1, a$ be our max and number of dice, and let $m_2, b$ be the opponent's max and number of dice. Let $F(a, b)$ be the number of ways we can win given $a, b$ dice.
$$F(a, b) = \sum_{m_1=1}^{6} \sum_{m_2=1}^{m_1} (m_1^{a} - (m_1-1)^{a})(m_2^{b} - (m_2-1)^{b})$$
This can be simplified to a closed-form expression:
$$F(a, b) = 1 - 2^{b} + 2^{a + b} - 2^{a} 3^{b} + 3^{a + b} - 3^{a} 4^{b} + 4^{a + b} - 4^{a} 5^{b} + 5^{a + b} - 5^{a} 6^{b} + 6^{a + b}$$
See that $F(2, 1) = 161$ and $F(3, 2) = 5593$.
A: $(1).~161$ 
$(2).~1071$ 
combination are possible to win the game.
