Questions
I was watching a Penn & Teller Fool Us clip, and Paul Gertner has a trick where he starts with a deck that is supposedly shuffled. I noticed, however, that there is no instance where there are more than two reds or more than two blacks consecutively. In fact, the cards appear in a consistent pattern of $…,1,2,1,2,2,1,2,1,2,2,…$ where 2 is a block of two cards of the same color.
You can see the sequence of cards here , if you like.
Assume you begin with a fairly and completely shuffled deck, then what are the chances of:
- No color appearing more than twice in a row?
- The pattern $…,1,2,1,2,2,1,2,1,2,2,…$ appearing through the entire deck?
Edit: Extra info for anyone that loves puzzles, math, and/or magic
Here is the sequence of cards in the trick.
Notice first that the sequence follows the questions posed above. Namely, that the entire sequence is of the form $…brrbrrbbrbbrbbrr…$ where $b$ is a black card and $r$ is a red card.
This may also be expressed as $…,1,2,1,2,2,1,2,1,2,2,…$ where $1 \in \{b, r\}$ and $2 \in \{bb, rr\}$.
Then consider, in addition:
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In each block of 2 cards, the suit is the same.
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For each adjacent block of 1 or 2 cards, the suits are always different.
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In fact, treating each 2-block as one card, then the pattern of suits is $…,c, d, s, h, c, d, s, h,…$ where $c$ is clubs, $d$ is diamonds, $s$ is spades, and $h$ is hearts.
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In all single card blocks, the card's value is $\in \{6, 7, 8\}$ (in the section shown in this clip).
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Where $x$ and $y$ are the numeric values of two different cards with $x, y \in \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13\}$, and an ace has a value of 1, a jack of 11, a queen of 12, and a king of 13, and the difference between the values of a two card block is defined as:
$$D(x,y) = {| x-y |}$$
Notice that $D(x,y) = 8$ for all two card blocks.
There are some tough and interesting counting/probability problems in there. It would be a very complex counting problem indeed to determine the chances of all of those conditions being true. There may be even more patterns I haven't noticed, as well. I don't know why the deck is arranged that way for the purposes of the trick, but it's an interesting puzzle!
Best Answer
Rough estimate:
Considering only the first three cards, the probability of three equal colours is approximately $\frac14$; the precise value $\frac{52\cdot 25\cdot 24}{52\cdot 51\cdot 50}=\frac 4{17}$ differs a bit from this because the colours are not independent. Fortunately, expected values are additive even in case of dependency. Therefore, we expect $50\cdot\frac 4{17}\approx 11.8$ instances of three consecutive cards of same colour.
Going by rule of thumb, we may estimate that this number is distributed with mean and variance both $=11.8$. So zero instances is about $3.4$ standard deviations too low - and that is very unlikely.
Less hand-wavy:
Let $f(n,k)$ be the number of ways to break a line $n$ same-colour cards into $k$ pieces of length $1$ or $2$. We have the recursion $$f(n,k)=f(n-1,k-1)+f(n-2,k-1) $$ and clearly, $f(n,k)=0$ if $k>n$ or $2k<n$. The recursion reminds of binomial coefficients, and indeed we find that $$ f(n,k)={k\choose n-k}.$$ If our deck has $k$ blocks of red cards (each of length $1$ or $2$), then it can have
Hence the total way to have a 52 card deck without blocks longer than $2$ is $$\begin{align} \sum_k f(26,k)\cdot& \left(f(26,k-1)+f(26,k+1)+2f(26,k)\right) \\&=\sum_k {k\choose 26-k}\cdot \left({k-1\choose 27-k}+{k-1\choose 25-k}+2{k\choose 26-k}\right)\end{align}$$ where we need only $k=13, \ldots, 26$. Evaluating this and dividing by $\frac{52!}{26!^2}$, we arrive at a probability of $$ \frac{23246263956\cdot 26!^2}{52!}=\frac{1937188663}{41326544412342}\approx 0.000047.$$
Fixed prescribed red/black pattern
For a fixed prescribed red/black pattern the answer is much simpler: If we prescribe which cards are red and which are black, there are $26!^2$ "good" sequences out of $52!$ in total. We arrive at a probability of $\frac{26!^2}{52!}\approx 2\cdot 10^{-15}$.