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Edit. The error was noted in a subsequent Riddler entry with a reference to an earlier MSE post:

enter image description here

The most recent Riddler Express says:

enter image description here

But suppose you had this same question with only one suit (say, hearts): What is the probability that you get through the 13 hearts without what you say ever matching what you deal?

It seems to me that the method above would suggest computing $(12/13)^{13} = 0.3532\ldots$

This cannot be right, though; the modified question is essentially asking about derangements for a set of $13$ elements. Checking the derangement sequence at OEIS yields as its thirteenth term $2290792932$; among the $13!$ combinations, this would yield a probability of winning the modified game equal to $2290792932/13! = 0.3678\ldots$

The numbers are not very different, but they are certainly different.

Back to the main question:

Is the solution for the "Riddler Express" above incorrect? [I strongly believe it is...]

If so: What is the correct solution, and how does one show this?

I had thought it would require a modification of the derangement formula using the Inclusion-Exclusion Principle or some such thing (cf. this MO answer from Richard Stanley). In fact, I computed it as $0.01623\ldots$ Again, this is "about 1.6 percent" (as in the solution write-up) but the methods are very different and the actual numerical results are not the same.

Am I overthinking? Is the Riddler underthinking? What gives?

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Benjamin Dickman
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    You are correct; and I too am at little miffed at The Riddler for publishing this as the 'correct' answer (as opposed to a 'pretty good' answer). – Chas Brown Jan 08 '17 at 23:06
  • @ChasBrown Thanks for the confirmation; I spent a bit of time last week trying to find an "express" solution, but (as suggested in my question) I view the 538 puzzle as harder than enumerating derangements, which is already pretty tough stuff... – Benjamin Dickman Jan 08 '17 at 23:16
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    The exact value is given in this answer with citations to back it up. It is close to the Riddler answer, but not equal. – Ross Millikan Jan 08 '17 at 23:21
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    Have you (or anyone else) emailed Oliver Roeder about the error? It should be done, but ideally only once. – Mark S. Jan 12 '17 at 05:45
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    @MarkS. You can find the error acknowledged in the most recent Riddler entry. See the edit to this post for a direct link. (It mentions the earlier MSE 1891958.) – Benjamin Dickman Jan 16 '17 at 18:54

3 Answers3

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The value of $P(win) = \frac{\int_0^{\infty}L_4^{13}(x)e^{-x}dx}{52!/(4!)^{13}} \approx 1.6233\%$ is given here with citations.

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Ross Millikan
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Yes, the "Riddler Express" solution is not correct, but it is a reasonably good approximation.

Robert Israel
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You can see that it is not correct by looking at the simpler case of an eight-card deck with A2 in each suit. To get zero matches, you need precisely 2A2A2A2A; the chance of this is exactly

$$\frac12 \times \frac47 \times \frac12 \times \frac35 \times \frac12 \times \frac23 \times \frac12 =\frac{1}{70}$$

But according to the Riddler's flaky logic, it would be $$\left(\frac12\right)^8=\frac{1}{256}$$

TonyK
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  • This is a nice example; thanks. Thinking beyond your full suited 2-card deck to a full suited 3-card deck already indicates how difficult this problem is! – Benjamin Dickman Jan 09 '17 at 03:17