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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.

user20877
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2 Answers2

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$\checkmark$ 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: $\dfrac{^{32}\mathsf C_{13}}{^{52}\mathsf C_{13}}$

Graham Kemp
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Remember, the probability of event A and event B occuring is the product of the two probabilities. So, getting a heads immediatly followed by a tails is the product of the two probabilities. This same theory can be applied to your problem, except it gets a bit more complicated.

What you would want to do is find the probability of likely outcomes over total outcomes.

Probability of getting 1 coke card = 32/52 (favorable over total outcomes)

2 coke cards = product of 1 coke card and previous probability (31/51) because there are 31 coke cards left.

3 coke cards = probability of two coke cards times (30/50).

So on and so forth.

Repeat 13 times :)

However, your initial answer is correct, as notated by the other answer. It sometimes just helps if you can do it intuitively too.

louie mcconnell
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