From a well-shuffled deck of ordinary cards, four cards are turned over one at a time without replacement. What is the probability that the spades and red cards alternate? Is the solution [(52c13)(51c26)(50c12)(49c25)+(52c26)(51c13)(50c25)(49c12)]/(52c4) correct? Any help is appreciated.
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I believe the answer posted is incorrect. The numerator is correct, but the denominator should be 52 choose 4 which ends up being (525150*49)/24. Feel free to confirm that. – Sep 28 '15 at 23:07
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Well the total number of possibilities for the four ordered cards is $52\cdot51\cdot50\cdot49$.
It can either go spade, red, spade, red, or red, spade, red, spade.
In the first case, the number of ways to choose the spade is $13$, then the number of ways to choose the red is $26$, then $12$ and $25$. The second case is the same since multiplication is commutative. We get
$$\frac{13 \cdot 26 \cdot 12 \cdot 25 + 26 \cdot 13 \cdot 25 \cdot 12}{52 \cdot 51 \cdot 50 \cdot 49}=\frac{26}{833}$$
Zubin Mukerjee
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