A coke hand in bridge is one where none of the thirteen cards is an an ace or is higher than a 9. What is the probability of being dealt such a hand?
Attempt: Suppose the thirteens cards are amoung the: 2,3,4,5,6,7,8,9, but not 10,A,K,Q or J.
Then since there are 4 different suits for each card, there is a total of 8*4 = 32 possible coke hand cards. Thus there is a 32_C_13 ways of selecting. Now the P(probability of being dealt such a hand) = 32_C_13/ 52_C_13.
Is this correct? Please can someone please help me? thank you.
Best Answer
$\checkmark$ Yes.
The total number of ways to deal a hand of 13 cards from a deck of 52 is: $^{52}\mathsf C_{13}$
The number of ways to deal 13 cards selected from the $8\times 4$ favoured cards is: $^{32}\mathsf C_{13}$
The probability of the favoured event is: $\dfrac{^{32}\mathsf C_{13}}{^{52}\mathsf C_{13}}$