Let there be five objects V, W, X, Y, Z. For each pair of objects, a fair coin is tossed. If the coin returns heads, the two objects become connected. If one object is connected to another object, it is connected to every other object the other object is already connected to. What is the probability that all five objects are connected to each other?

I've tried cases with 2, 3 and 4 objects but found no pattern.


Let's work our way up.

With only one vertex, there is no coin tossing at it is always connected; $p_1=1$.

With two vertices, the probability is clearly $p_2=\frac12$.

With three vertices, we need all three or any two edges, so $p_3=\frac18+\frac38=\frac12$.

Four vertices start to get a bit complicated. We may end up with a single cluster of $4$ connected vertices, or $3+1$, or $2+2$, or $2+1+1$ or $1+1+1+1$. The probabilities of the latter cases are clearly ${4\choose 3}p_3p_12^{-3\cdot 1}=\frac14$, $\frac12{4\choose 2}p_2^22^{-2\cdot 2}=\frac 3{64}$, ${4\choose 2}p_2p_1^22^{-2\cdot 1-2\cdot 1-1\cdot 1}=\frac3{32}$, and $p_1^42 {-6}=\frac1{64}$. We conclude that the connected case occurs with probability $p_4=1-\frac14-\frac3{64}-\frac3{32}-\frac1{64}=\frac{19}{32}$

Now finally check the problem of hand with five objects: We may end up with a cluster of $5$ connected vertices, or $4+1$, or $3+2$, or $3+1+1$, or $2+2+1$, or $2+1+1+1$, or $1+1+1+1+1$.

  • The probability for the $1+1+1+1+1$ case is clearly $2^{-10}$ as all coin tossed must be T.
  • The probability for $2+1+1+1$ is ${5\choose 2}2^{-10}$ because we must pick exactly one edge that must be H and the other nine edges must be T.
  • Similarly, the probability for $2+2+1$ is $\frac12{5\choose 4}{4\choose 2}2^{-10}$.
  • $3+1+1$ gives us ${5\choose 3}p_3p_1^22^{-3\cdot 1-3\cdot 1-1\cdot 1}= \frac5{128}$
  • Similarly $3+2$ gives the same $\frac5{128}$.
  • For $4+1$, we find ${5\choose 4}p_4p_12^{-4\cdot 1}$

Now gather all these numbers togerther and then the desired $p_5$ is "the rest".


Let $G(n,p)$ denote the random Erdös-Rényi graph where each of the $\binom{n}{2}$ edges is present with probability $p$, such that each edge occurs independently. Then define $$P_n = \mathbb{P}(G(n,p)\text{ is connected})$$ and note that $P_1= 1$. We now want to establish a recursive relation. For this we need to define the couple of concepts recursively. $$\mathcal{P}_n = \{(i,x_1,\ldots,x_k):i\in\{1,\ldots,n-1\},(x_1,\ldots,x_k)\in \mathcal{P}_{n-i}\}\cup\{(n)\}$$ and $\mathcal{P}_1=\{(1)\}$, such that $\mathcal{P}_5 = \{(1,1,1,1,1),(2,1,1,1),(2,2,1),(3,1,1),(3,2),(4,1),(5)\}$, in line with what Hagen von Eitzen was stating. In this way $\mathcal{P}_n$ contains all tuples that correspond to a partition of the $n$ nodes. Then $$1-P_n = \sum_{x\in\mathcal{P}_n\,:\,x\neq(n)} \left(M_x P_{x_1}\cdot \ldots\cdot P_{x_k} \prod_{i,j\,:\,i<j}(1-p)^{x_ix_j}\right)$$ where $$M_x = M_{(x_1,\ldots,x_k)} = \binom{\sum_{i=1}^k x_i}{x_1,\ldots,x_k}$$ is a multinomial coefficient.

  • $\begingroup$ For $p=\frac 12$,m you find $P_1=1$, $P_2=\frac12$, $P_3=\frac38$. But clearly $P_3=\frac12$ (connected iff the complement is not) $\endgroup$ – Hagen von Eitzen Oct 3 '17 at 12:02
  • $\begingroup$ Why is this true? If, say, $n=3$ and $p=\frac 12$ this looks false. $\endgroup$ – lulu Oct 3 '17 at 12:02
  • $\begingroup$ The problem is that there is no guarantee that, in a connected random graph, the complement of a point is connected. $\endgroup$ – lulu Oct 3 '17 at 12:03
  • $\begingroup$ You are right, I'll edit my answer and set up a correct recursive relation. $\endgroup$ – Yannik Oct 3 '17 at 13:10

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