Number of rounds

[tuanl]

You play a game with a opponent. The game consists of the number of rounds, dcided before the game starts. In each round, your probability of winning is 0.45 and the opponent's probability of winning is 0.55. A player gets 1 point after winning each round. Who has more points afer finishing all the rounds is the winner of the entire game. If both players end with the same number of points, result is considered to draw. If you have a position of determining the number of rounds. Then how many rounds should you chose to maximize your chance of being winner?

PS: I tried using Wolfram Alpha to find the value of N to obtain the maximum value of \(sum_{n=floor{N/2}+1}^{N} (0.45)^n*(0.55)^{N-n}* \binom {N}{n}\), but Wolfram failed. By brute-forcing using Excel for N up to 16, I think N=1 gives the maximum chance, which surprises me.

 

[DieterD]

Seems pretty intuitive to me. Since you're the underdog, you want to play as few rounds as possible. Similar to say playing roulette, the more often you play, the more likely you are to lose.

 

[pruse]

For \(n\) odd, you either win the last round, and you've come out at least even in the previous \(n-1\) rounds, or you lose the last round but you've come out on top in the previous \(n-1\):
\( P_n = 0.55P_{n-1}+0.45 \large( P_{n-1}+\binom{n-1}{\frac{n-1}{2}}0.45^{\frac{n-1}{2}}0.55^{\frac{n-1}{2}}\large)=P_{n-1}+\binom{n-1}{\frac{n-1}{2}}0.45^{\frac{n+1}{2}}0.55^{\frac{n-1}{2}} \)

For \(n\) even, you either win the last round and you've also won the \(n-1\) rounds before, or you lose the last round and in the previous \(n-1\) rounds you're up by at least 2 rounds:

\( P_n = 0.45P_{n-1}+0.55\large( P_{n-1}-\binom{n-1}{\frac{n}{2}}0.45^{\frac{n}{2}}0.55^{\frac{n-2}{2}}\large)=P_{n-1}-\binom{n-1}{\frac{n}{2}}0.45^{\frac{n}{2}}0.55^{\frac{n}{2}} \)

So you only have to check if your decrease from odd to even is less than your increase from even to odd. This shouldn't be hard.

 

[tuanl]

So you only have to check if your decrease from odd to even is less than your increase from even to odd. This shouldn't be hard.

Nice recursion! Yes, it's not hard on this part Did you get 1 as the result too?

 

[pruse]

Ha, are you serious? You were looking for a rigorous solution, and I set you up for one; the rest of it should be right-quick...

We have to check the ratio of \(\binom{2k}{k}0.45^{k+1}0.55^{k}\) to \(\binom{2k-1}{k}0.45^k0.55^k\), which is easily computed as \(2\cdot 0.45 = 0.9<1\). This shows that your probability decreases from an odd number of rounds, \(2k-1\) to an even number of rounds \(2k\), more than it increases from \(2k\) to \(2k+1\), confirming your conclusion that 1 is the number of rounds that optimizes your probability.

The lesson? If you want to verify an inequality, checking increments between successive terms can sometimes be more tractable than dealing with formulas that don't have closed forms.