What is the probability of winning the max amoun

Question

What is the probability of winning the max amount?

Answer

AssumptionTo rephrase my assumption in the answers heading, I'm assuming we can only play through the deck one, otherwise the trivial solution of a probability of 0 applies (you will pick up a red card when playing the game forever).SolutionThe key here is to note that events are not independent. Hence, the typical probability of intersection formula changes to:     P(A ∩ B) = P(B|A)*P(A)instead of     P(A ∩ B) = P(B)*P(A).For the first round (call it A if you like), winning probability is easily calculated as 26/52.Winning the second round provided the first was won (call it B|A), we note that we are calculating the probability based on exactly 1 less black card in our deck. This means, because of A, the deck has 25 black cards and hence 51 cards in total. This gives us a probability of 25/51 for winning the second round.By applying this logic recursively, we see a pattern emerging that on the i-th round, the probability of winning is      (26-i)/(52-i). So, the probability of winning the maximum amount is simple the probability of winning all the rounds which is this entire chain of 'product rules':P(Max won) = (26 x 25 x 24 x ... x 3 x 2 x 1)/(52 x 51 x ... x 29 x 28 x 27)                                                                                    NoteIf you are interested, this probability simplifies to 1/(52 combination 26). Thus, there is a much shorter argument based on combinatorics hiding in the details, which is a nice exercise for students struggling with combinatorics.

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What is the probability of winning the max amoun

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