De Méré's Problem

The two games are equally good on average: the expected number of successes is $4\cdot\frac16=\frac23$ in Game 1 and $24\cdot\frac1{36}=\frac23$ in Game 2. That is the source of the tempting answer “equal”. The probability of at least one success is a different quantity, and it is easiest to find through the complement.

Write $p_1,p_2$ for the winning probabilities. The rolls are independent. You lose Game 1 exactly when none of the $4$ rolls shows a $6$. You lose Game 2 exactly when none of the $24$ rolls is a double-six, which has probability $\frac1{36}$ per roll. Hence

We show that $p_1>\tfrac12>p_2$.

Game 1. We have $\left(\tfrac56\right)^4=\tfrac{625}{1296}$. Since $2\cdot 625=1250<1296$, this is less than $\tfrac12$, so $p_1>\tfrac12$.

Game 2. We use $1+x<e^x$ for $x>0$. This gives

Now $\tfrac{24}{35}\approx 0.686<0.693\approx\ln 2$, so $e^{24/35}<2$. Therefore $\left(\tfrac{35}{36}\right)^{24}>\tfrac12$, and so $p_2<\tfrac12$.

Numerically, $p_1=\tfrac{671}{1296}\approx 0.518$ and $p_2\approx 0.491$.

Key idea. $\mathbb{P}(\text{at least one})=1-(1-p)^n$ is not proportional to $np$. When $np$ is held fixed, it is larger for fewer trials with larger $p$. This is because $(1-p)^n$ is smaller than $e^{-np}$, and it gets closer to $e^{-np}$ as $p\to 0$.