Why is my intuition wrong that one of two archers win in a tournament?

Take an example where the probability of Robin and Tuck hitting a target respectively are $0.9$ and $1.0$. Your method would give the probability of Robin winning being $\dfrac{0.9}{0.9+1.0}$ when the true probability is $0$.

Going back to your original probabilities of hitting of $0.45$ and $0.38$, you would do better saying the probability of Robin winning overall might be $\dfrac{0.45\times(1-0.38)}{ 0.45\times(1-0.38) + 0.38\times(1-0.45)}$ by looking at the decisive and mutually exclusive events of one hitting and the other not. This gives your $0.5717\ldots$


Whatever the probability distribution of the number of rounds, on any round Robin wins with probability $\propto r\bar t$ and loses with $\propto \bar rt$, hence the ratio

$$\frac{r\bar t}{r\bar t+\bar rt}=\frac{0.45\cdot0.62}{0.45\cdot0.62+0.55\cdot0.38}\approx0.572$$ as the game ends with probability $1$.


The easiest way to get there is to say the chance that exactly one of them hits the target is $0.45(1-0.38)+(1-0.45)0.38=.488$ The chance that Robin wins given that somebody did is $\frac {0.45(1-0.38)}{0.488}\approx 0.57172$ which matches your second calculation.

The error in your intuitive calculation is that Tuck loses more of his hits to ties because Robin is more accurate, so his chance of winning should be less than $1-0.542$

Tags:

Probability