Four cards are selected from a pack of 52 cards. Calculate the probability that no two cards have the same value.
My attempt: Paying cards have 13 possible values $2,3,4 … , 9, 10, J, Q, K, Ace.$ We first choose 4 values then we choose one card for each of the values.
Hence, probability equals $\frac{{13 \choose 4} 4^4}{{52 \choose 4}}= 0.676$.
Is this correct?
Best Answer
It is simpler to think that the order of drawing cards matters (it does affect the final probability). Then the total number of ways to draw $4$ cards is
$$52\cdot 51\cdot 50\cdot 49$$
and the number of ways to draw 4 cards of different value is
$$52 \cdot 48 \cdot 44 \cdot 40$$
so the probability is
$$\frac{48 \cdot 44 \cdot 40}{51\cdot 50\cdot 49}\approx 0.676$$