More on multi-way matches -2

Let $X_i$ be the indicator random variable for the $i$ th round being a 4-way match. That is, $X_i = 1$ if the $i$ th round is a 4-way match and $X_i = 0$ otherwise.

Since the decks are shuffled, the $i$ th card in each deck is random. Also, the probability of any card from any of the three decks being a particular color is 1/4.

Therefore $\begin{aligned} \mathbb{E}[X_i] = \mathrm{Prob}(X_i =1) &= \sum_{\mbox{color } c} \mathrm{Prob}(\mbox{4-way match of color \$c\$ on round } i) \\ &= 4 \left(\frac{1}{4} \right)^{4} \\ &=\frac{1}{64} \end{aligned}$

Let $X$ be the total number of 4-way matches in the game. Then $X = \sum_{i=1}^{64} X_i$ . Also $Y = \sum_{i=1}^{32} X_i$ is the number of 4-way matches in the first half of the game.

By Linearity of Expectation we have $\mathbb{E}[X] = \sum_{i=1}^{64} \mathbb{E}[ X_i] = \sum_{i=1}^{64} \frac{1}{64} = 1$

By Markov's Inequality (see this wiki )

$\mathrm{Prob}( X \ge 2) \le \frac{\mathbb{E}[X] }{2} = \frac{1}{2}$

Moreover, since clearly there is a positive probability of 3 or more 4-way matches, we must have

$\mathrm{Prob}( X \ge 2) < \frac{1}{2}$

Note that this is the probability of event E3.

Similarly,

$\mathbb{E}[Y] = \sum_{i=1}^{32} \mathbb{E}[ X_i] = \sum_{i=1}^{32} \frac{1}{64} = \frac{1}{2}$

and

$\mathrm{Prob}( Y \ge 1) < \frac{1}{2}$

the strict inequality coming from the positive probability of two or more 4-way matches in the first half.

Note that this is the probability of event E1.

Finally, event E2 is the complement of event E3, so its probability is at least 1/2

Thus the most probable event is $\fbox{ E2}$ .

Note: The random variables $X_i$ are not independent, so it is fortunate that they do not need to be. Linearity of Expectation applies to any linear combination of random variables.