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At state $r$, urn 1 has $r$ red and $n-r$ blue balls, while urn 2 has $n-r$ red and $r$ blue balls. Conditional on the state, the two choices are independent, each assigning probability $1/n$ to every ball. A red–blue exchange decreases $r$; a blue–red exchange increases it; equal colours leave it unchanged. Thus the conditional probability mass function isThe three numerators sum to $n^2$. At $r=0$ or $n$, the outward transition has probability zero. These are the transitions of the Bernoulli-Laplace diffusion model. The displayed formula is a transition kernel on every state; a literal conditional probability at a particular time requires the conditioning event to have positive probability.

Extinction selectivity is systematic variation in the probability or rate of extinction among lineages with different traits, rather than trait-independent loss. Traits may be biological, ecological or geographical: for example, survival may differ between narrow-ranging and widespread taxa within the same extinction interval.

A test needs an explicit denominator of taxa at risk, a defined taxonomic level and time window, and a specified trait measurement. Regional disappearance is not automatically global extinction. Nor is an association necessarily a causal mechanism: traits covary with habitat, abundance and preservation, and their effects can change between background intervals and mass extinctions.

Test homogeneity of the four category probabilities among the three samples. The null hypothesis is $p_{iA}=p_{iB}=p_{iC}$ for every category $i$; the alternative allows at least one difference. This is a Pearson chi-squared test of homogeneity, assuming independently sampled grains with consistent classification.

The row totals are $(195,52,64,16)$, the sample totals $(131,135,61)$ and the grand total $N=327$. Under the null hypothesis, $E_{ij}=O_{i+}{O}_{+j}/N$, giving the expected value tableFor example, $E_{\mathrm{undulatory},C}=52(61)/327=9.7003$, so that cell contributes $(4-9.7003{)}^2/9.7003=3.34974$. The complete cell contributions areso $X^2=6.87754$. There are $(4-1)(3-1)=6$ statistical degrees of freedom. The supplied $5\%$ critical value is $12.592$, and $6.87754<12.592$, so do not reject homogeneity at $\alpha =0.05$. The approximate upper-tail p-value is $0.3323$.

One of twelve expected cells is below five, with none below one; the asymptotic approximation is broadly defensible by a common expected-count guideline but is not exact. An exact/conditional simulation could check it, especially if grain counts are clustered. Do not pool categories solely to obtain a desired result. The conclusion is that these counts do not demonstrate different category proportions; they do not prove that the rocks share one geological formation. Different formations can share such proportions, and real grains within one rock can fail the independent-sampling assumption.

Patch clamp connects an electrode to a small area of cell membrane through a solution-filled patch pipette. A gigaohm seal around the rim makes tiny electric currents through ion channels distinguishable from seal leak current. The beautiful change of scale is from a population's average response to individual opening and closing events in single-channel recording.

Separate two questions. The configuration determines which cell membrane is accessible: an intact patch in cell-attached recording, an excised inside-out patch or outside-out patch, or intracellular access in whole-cell recording. The recording mode determines what is commanded and what is observed: voltage clamp commands an electric potential difference and measures electric current; current clamp commands electric current and measures membrane potential. Whole-cell recording supports both modes. A patch can contain several ion channels; a small recording area does not establish that only one is present.

The 1976 article Single-channel currents recorded from membrane of denervated frog muscle fibres describes an extracellular fire-polished pipette with a $3$–$5\mu \mathrm{m}$ opening and $2$–$5\mathrm{M\Omega }$ electrical resistance. It contained saline and a channel-activating compound. Two separate intracellular electrodes supplied local voltage clamp of the muscle fibre; the patch amplifier measured local electric current. Cooling to $6$–$8{}^{\circ }\mathrm{C}$ and hyperpolarization helped resolve elementary events. Figure 2 shows step-like pulses with a $200\mathrm{Hz}$ filter at $-120\mathrm{mV}$ and $8{}^{\circ }\mathrm{C}$. Discrete amplitudes and overlapping events supported individual ion channel openings rather than a smoothly varying population response. This was neither whole-cell recording through the patch pipette nor current clamp. The specialized conditions improved visibility but limit generalization to warmer preparations; filtering can conceal brief openings.^[ref]

For a new single-channel recording experiment, useful controls include a matched recording without the activating compound, controlled membrane potential, repeated seal tests, and an explicitly stated filter frequency response. These are proposed controls, not claims that every one was performed in that 1976 figure. Changing membrane potential tests the single-channel current-voltage relation; changing the recording filter tests whether a short pulse's apparent amplitude is an instrument artifact. A flat baseline establishes neither an absence of ion channels nor an absence of very brief events.

A payoff is the number assigned to a player's outcome at a strategy profile, with larger numbers preferred. In the usual expected utility model, expected payoffs are probability-weighted averages of these numbers. If a payoff is specified as money and a player has preferences involving risk, first apply the relevant utility function; maximizing expected money assumes risk neutrality. In a two-player zero-sum game, one player's payoff is the negative of the other's. General bimatrix games instead require a payoff matrix for each player.

Let $X_1,X_2$ be independent fair signs, each taking $-1$ and $1$ with equal probability. Set $N=1$ if $X_2=1$, and $N=2$ if $X_2=-1$. Deciding whether $N\le 1$ requires the second sign, so $N$ is not a stopping time for the natural filtration. The four equally likely possibilities giveBoth the increments and the count are bounded, but the identity fails because the second contribution is retained only when it is negative. This is selection using an unobserved outcome, rather than a legitimate sequential decision. Boundedness alone cannot replace the stopping time hypothesis.^[ref]

A percentage point is one hundredth on a probability or proportion scale. A change from $20\%$ to $25\%$ is $5$ percentage points, while the relative increase is $(25-20)/20=25\%$. These answer different questions and have different denominators.

For the rooted infinite model defining percolation probability, the percolation critical probability isBelow $p_c$, the root has probability zero of belonging to an infinite percolation cluster. Above $p_c$, that probability is positive, by the definition and the monotone coupling of Bernoulli percolation. The definition alone does not determine behavior at $p_c$, or the numerical value of $p_c$: these require results specific to the graph and the model.

A finite graph with $M$ edges has $2^M$ open/closed configurations. The probability of any connectivity event is a finite sum of terms $p^k(1-p{)}^{M-k}$, hence a polynomial in $p$. The square's $2p^2-p^4$ is an example. These finite probabilities vary smoothly, although an infinite-system limit can distinguish a whole interval with $\theta (p)=0$ from one with $\theta (p)>0$. A steep finite connection curve and a mathematically proved percolation critical probability are different kinds of information.

For independent bond percolation on an infinite graph with a specified root $o$, let $C_o$ be its percolation cluster. The percolation probability isIt concerns the root, rather than the existence of an infinite percolation cluster somewhere else. Even above the percolation critical probability of the square lattice, an isolated root has probability $(1-p{)}^4$, because its four incident edges can all be closed. Thus $\theta (p)\le 1-(1-p{)}^4<1$ when $p<1$. Positive percolation probability does not mean every vertex is connected to infinity.

On the square lattice, the events that $o$ is joined by an open path to the boundary of $[-n,n{]}^2$ decrease as $n$ increases. Their probabilities converge to $\theta (p)$ by continuity from above of a measure. This relates an infinite question to finite observations, without equating any one finite boundary connection with survival forever. The monotone coupling of Bernoulli percolation shows that $\theta$ is a monotone function of $p$.

Percolation theory asks how many local connections must be available before a large connected structure appears. Start with a graph, randomly retain some edges or vertices, and study the resulting percolation clusters. In independent bond percolation, every edge is open with probability $p$, independently of the others. In independent site percolation, the random decisions concern vertices instead. A percolation cluster is a connected component of a graph formed by the available connections; an open path joins two vertices exactly when they belong to the same percolation cluster.

The attraction is that a local rule can produce a global change. One open edge says little about travel across a large graph. Many alternative graph paths can nevertheless make distant connections likely. The model separates the geometry of possible connections from the probability distribution of which connections are usable. This was the central perspective of Broadbent and Hammersley's 1957 paper, which discussed porous solids and disease transmission as interpretations of the same mathematical structure: ^[ref]. Independent bond percolation supplies a simplified model of those systems; the physical interpretation still requires justification of its independence assumptions.

For a completely calculable example, consider the four edges around a square and ask whether opposite corners $A$ and $C$ are connected. There are two graph paths, each using two edges. Each open path has probability $p^2$. Their events are independent events because the paths share no edges, even though they share endpoints. The inclusion-exclusion principle therefore givesAt $p=1/2$, the connection probability is $7/16$, larger than the $1/4$ for either chosen graph path. Counting all sixteen equally likely edge configurations gives the same answer: two configurations with exactly two open edges, four with three, and one with four connect $A$ to $C$. Connectivity benefits from alternative routes; adding their probabilities without subtracting their overlap would count the all-open configuration twice.

This finite calculation is already useful for studying redundant connections. The deeper question is what happens as the graph becomes arbitrarily large: does connectivity remain confined to finite percolation clusters, or can an infinite percolation cluster appear? The percolation probability and percolation critical probability make that question precise. A finite picture can illustrate the geometry, but cannot establish the existence of an infinite percolation cluster.

The two-point percolation two-point connection probability ${\tau }_p(x,y)={\mathbb{P}}_p(x\leftrightarrow y)$ is the probability that two fixed graph vertices belong to the same percolation cluster. On a translation-invariant lattice it depends on their displacement. Restricting the permitted connecting graph paths to a subgraph gives the corresponding confined percolation two-point connection probability. The Harris-FKG inequality implies ${\tau }_p(x,z)\ge {\tau }_p(x,y){\tau }_p(y,z)$.

Let $X_1,\dots ,X_n$ be independent random variables, each choosing a label from $\{1,\dots ,n\}$. Put their probabilities in the matrix $P=(p_{ij})$, where $p_{ij}=\mathbb{P}(X_i=j)$. Each row sums to one. All choices are distinct precisely when their ordered list is one of the permutations of the $n$ labels. These are mutually exclusive events, and independence makes the probability of one list $\sigma$ equal to ${\prod }_i{p}_{i,\sigma (i)}$. Adding the list probabilities gives the permanent of a matrix:In particular, this permanent lies between zero and one.

For a two-person example,The distinct outcomes $(1,2)$ and $(2,1)$ have probabilities $3/5$ and $1/20$. ThereforeThe colliding outcomes $(1,1)$ and $(2,2)$ have probabilities $3/20$ and $1/5$, totaling $0.35$. All four outcomes are shown below; their probabilities sum to one.

If every choice has the uniform distribution on a finite set of $n$ labels, then $p_{ij}=1/n$ and the distinct-choice probability is $n!/n^n$. For three choices and three labels it is $6/27=2/9$. This counts a subset of all $n^n$ ordered outcomes, not independent pairwise collision events. Independence of the original choices is essential: without it, the individual rows of $P$ do not determine the joint probabilities.

Fix positive $\alpha$, $D\ge 2$ and $B>0$. Suppose a size band $[k,Dk)$ of an Achlioptas process contains at least $\alpha n$ vertices at time $m$. For every fixed $k$, at least $\beta n$ of those vertices remain in unchanged graph components throughout the next $T=\lceil Bn/k\rceil$ steps, with high probability, where one may take $\beta =\alpha e^{-16BD}/4$. This is a deliberately nonoptimal positive constant.

For independent uniform candidate edges, an initial graph component with $w<Dk$ vertices is untouched if none of the $2T$ candidate edges meets it. A uniform edge hits it with probability at most $3w/n$. For sufficiently large $n$, $(1-3w/n{)}^{2T}\ge e^{-16BD}$, so the expected value of the untouched vertex mass $Y$ is at least $\alpha e^{-16BD}n$. Replacing one row of two candidate edges can change $Y$ by at most $8Dk$, because only the initial graph components hit by an old or new endpoint can change status. The McDiarmid inequality therefore makes $Y\ge \alpha e^{-16BD}n/2$ except on an event of probability $e^{-cn/k}$, for a positive constant $c$ depending only on $\alpha ,B,D$. An untouched graph component stays unchanged whichever offered edge is selected.

For candidates restricted to absent edges, generate each by an independent uniform proposal stream and reject already present edges. During any interval of $O(n)$ steps with $O(n)$ present edges, each proposal has rejection probability $O(1/n)$ conditional on the past. The number of rejections is at most $O(\log n)$ except on an event of probability smaller than any inverse power of $n$: if there were $r$ rejections among $2T+r$ proposals, a union bound over their positions bounds this by $\left(\genfrac{}{}{0ex}{}{2T+r}{r}\right)(C/n{)}^r$. The first $2T$ proposals are independent and give the previous untouched-mass estimate. Extra proposals can touch at most $2r$ additional initial graph components, losing only $O(Dk\log n)=o(n)$ vertices for fixed $k$. This leaves the stated $\beta n$ bound. A union bound makes these estimates simultaneous over starting times $m\le 3n$. They hold through the entire interval, so deterministic or random stopping times within it are allowed.

An optical element that adds a controlled wave phase to one path. On two path states, an ideal lossless phase shifter has the displayed unitary matrix. It changes their relative phase while leaving individual path probabilities unchanged; subsequent interference can turn that phase into an output-intensity change.

Choose the following original phenomenological mathematical model solely to calculate a paired-pulse readout. It is not experimental data, a fit to the teaching experiment, or a simulation of the Hodgkin-Huxley model. The conditioning action potential defines time zero. Test pulses have fixed width $0.2\mathrm{ms}$; $I$ denotes their inward electric current amplitude. Impose an absolute neuronal refractory period $t_{\mathrm{abs}}=2\mathrm{ms}$. No finite test amplitude generates an event for $0<\Delta <t_{\mathrm{abs}}$. After that hold, stipulate a required amplitudewith $I_{\infty }=0.2\mathrm{nA}$, dimensionless $A=2$ and recovery time constant ${\tau }_r=3\mathrm{ms}$. Emit a test event exactly when $I\ge I_{\mathrm{req}}(\Delta )$, including equality. For this ideal event readout, assign events to pulse onset and assume the conditioning event is successfully evoked. There is no calculated membrane potential, spike waveform, response latency or stochastic success probability.

The exponential decay represents a combined recovery deficit; it is neither the Hodgkin-Huxley model's $h$ nor a measured potassium electrical conductance. One such curve cannot separately identify sodium recovery and persistent potassium electrical conductance. A constant pulse-width amplitude criterion also does not establish an equivalent rule for other pulse widths or voltage trajectories.

At the hold boundary, $I_{\mathrm{req}}(2\mathrm{ms})=0.2(1+2)=0.6\mathrm{nA}$. At $\Delta =4\mathrm{ms}$,at $\Delta =6\mathrm{ms}$,Thus a $0.35\mathrm{nA}$ test fails at $1\mathrm{ms}$ because of the imposed hold, fails at $4\mathrm{ms}$ because recovery is insufficient for that amplitude, and succeeds at $6\mathrm{ms}$. At $4\mathrm{ms}$, a $0.5\mathrm{nA}$ pulse would instead succeed: failure of the weaker pulse was relative refractoriness.

For $I>I_{\infty }$, solve $I\ge I_{\mathrm{req}}(\Delta )$ using the logarithm. The earliest allowed delay isFor $I=0.35\mathrm{nA}$ this is $2+3\ln (8/3)\approx 4.942488\mathrm{ms}$; for $I=0.5\mathrm{nA}$ it is $2+3\ln (4/3)\approx 2.863046\mathrm{ms}$. If $I\ge 0.6\mathrm{nA}$, the imposed $2\mathrm{ms}$ hold is the only delay restriction. If $I\le I_{\infty }$, no finite delay passes the rule: the required amplitude stays strictly above its asymptotic baseline. A finite experimental “fully recovered” time therefore also needs a tolerance and a readout definition.

Ideal conditional application of a creation operator, followed by normalization. It reweights the input number distribution as well as raising each occupied number by one. A heralded physical implementation has a success probability; the algebraic creation operator alone is not a trace-preserving deterministic quantum operation.

Independent loss alone leaves normalized factorial moments invariant: if every input photon is retained with probability $\eta$, binomial thinning conditional on a total gives $⟨K⟩=\eta \mu$ and $⟨K(K-1)⟩={\eta }^2⟨N(N-1)⟩$. Both powers cancel from $g^{(2)}(0)$. Fewer coincidences still mean poorer statistical precision.

Now let each channel contain a signal rate $s_i$ and an independent Poisson process background rate $b_i$, independent also of the signal. Put ${\rho }_i=s_i/(s_i+b_i)$. The pair density is $s_1{s}_2{g}_{\mathrm{sig}}^{(2)}+s_1{b}_2+b_1{s}_2+b_1{b}_2$. Dividing by $(s_1+b_1)(s_2+b_2)$ proves the displayed formula, for the same bin and timing response. For chosen equal signal fractions ${\rho }_1={\rho }_2=4/5$, ideal signal values $0,1,2$ become $9/25,1,41/25$. Background pulls both dips and peaks toward one.

This calculation assumes additive, mutually independent background counts. Scattered coherent light that interferes with the signal, correlated electronic noise, or recovery effects require a different model. Correcting a dip without independently estimating the backgrounds cannot establish photon antibunching.

In 1998, Jian-Wei Pan, Dik Bouwmeester, Harald Weinfurter and Anton Zeilinger created two photon pairs by two passages of a pulsed pump through a crystal using spontaneous parametric down-conversion. Interfering the inner photons at a nonpolarizing beam splitter implemented a partial Bell-basis measurement. Narrow optical filters and spatial filtering removed distinguishing information; changing the outer polarizer bases checked the conditional correlations. Fourfold detections selected the entanglement swapping events. The reported interference supported conditional quantum entanglement, but the paper reserved a statistically significant Bell inequality violation for improved visibility; it was not a demonstration closing the detection loophole.^[ref]

The ideal calculation makes the distinction between conditional and unconditional quantum states explicit. Start with spin singlet states $|{\psi }^{-}{⟩}_{12}|{\psi }^{-}{⟩}_{34}$, where $|{\psi }^{-}⟩=(|HV⟩-|VH⟩)/\sqrt{2}$ and $H,V$ denote orthogonal photon polarizations. Taking the inner product with $⟨{\psi }^{-}{|}_{23}$ gives $-|{\psi }^{-}{⟩}_{14}/2$: expanding the four terms leaves precisely the two antisymmetric outer terms. That Bell-basis measurement outcome has probability $1/4$ and leaves the outer pair in a spin singlet state. Ignoring the inner outcome instead leaves the outer density matrix $I_4/4$, a maximally mixed state. A conditional preparation flag is useful for later Bell tests only with the timing and setting conditions in setting-independent heralding preserves Bell locality.

An explicit original ideal meter model explains how photon number survives while field quantum coherence changes. Let $\hat{N}$ be the number operator and prepare an atomic qubit in $|+⟩=(|e⟩+|g⟩)/\sqrt{2}$. The dispersive unitary operatorcommutes with $\hat{N}\otimes I$. For a field quantum superposition ${\sum }_n{c}_n|n⟩$, it produces ${\sum }_n{c}_n|n⟩|a_n⟩$, where $|a_n⟩=(e^{-in\chi /2}|e⟩+e^{in\chi /2}|g⟩)/\sqrt{2}$. The atomic meter states encode $n$ through relative quantum phase and can become entangled with the field. The second Ramsey interferometry pulse and detector turn those correlations into the outcome probabilities above when $C=1$.

Tracing out the atom multiplies a field density matrix entry ${\rho }_{nm}$ by $⟨a_m|a_n⟩=\cos ((n-m)\chi /2)$. Hence the nonselective quantum measurement preserves every photon number population ${\rho }_{nn}$ while changing off-diagonal quantum coherence. For $\chi =\pi /2$, coherence between $|0⟩$ and $|1⟩$ is multiplied by $1/\sqrt{2}$ per fresh meter, even with no photons absorbed. Field quantum phase information can therefore be disturbed by an energy-preserving probe. No universal self-adjoint phase operator or naive canonical number–phase commutator is needed for this conclusion.

One possible finite-contrast quantum instrument makes the distinction sharper. Define diagonal Kraus operators $K_e=\cos ((\varphi +\chi \hat{N})/2)$ and $K_g=\sin ((\varphi +\chi \hat{N})/2)$. After their ideal readout, flip the reported label with probability $r=(1-C)/2$. The unnormalized selective maps areTheir traces reproduce $L_e,L_g$ on a Fock state, and their sum reproduces the displayed dephasing quantum channel, independently of $r$. At $C=0$ the recorded labels carry no information although the physical meter interaction can still cause measurement backaction. This is an illustrative instrument, not a reconstruction of all laboratory imperfections; equal likelihoods alone do not determine the disturbance.

A photon-recoil lag correlation asks how often a photon record occurs at a specified time displacement from an electron record. Let $r_e,r_{\gamma }$ be observed separate-channel rates, $S$ the rate of detected genuine pairs before a timing cut, and $f(\ell )$ their normalized timing probability density function. For an original dilute, stationary model with observation time $T$, neglecting end effects and detector dead time, the expected pair density isThis counts all pairs of records, rather than at most one detector trigger per electron. It uses independent unrelated detections for the flat term and an added same-event contribution. A normal distribution for timing spread is a chosen approximation, not a claim about the historical instrument. Intrinsic emission delay, counter response and readout uncertainty can all affect the observed probability density function.

Define the accepted lag interval by $|\ell |<w/2$: $w$ is its full width. Integrating the flat baseline gives the existing accidental coincidence expectation $T{r}_e{r}_{\gamma }w$. The additional area is $TS{\int }_{-w/2}^{w/2}f(\ell )d\ell$. Moving a window of the same full width away from the peak tests the baseline under the model's stationarity assumption. It does not remove correlated instrumental backgrounds or changing rates automatically.

For the lag model, choose $T=1000\mathrm{s}$, $r_e=2{\mathrm{s}}^{-1}$, $r_{\gamma }=0.5{\mathrm{s}}^{-1}$, $S=0.1{\mathrm{s}}^{-1}$ and timing standard deviation $\sigma =1\mathrm{ms}$. A full width $w=2\mathrm{ms}$ gives $2$ expected accidental coincidences and $100\mathrm{erf}(1/\sqrt{2})\simeq 68.27$ accepted genuine pairs, using the error function. The total is approximately $70.27$, whereas the displaced interval from $4$ to $6\mathrm{ms}$ has approximately $2.003$ expected pairs. A finite sample fluctuates around these expectations; these values are not measured counts or a significance calculation.

The right panel isolates the information lost by retaining singles totals alone. In a separate idealized model, stipulate $M=10^6$ independent $1\mathrm{ms}$ bins, with Bernoulli distribution indicators $X,Y$ for electron and photon detection in a bin. Set $p_e=\mathrm{Pr}(X=1)=0.002$ and $p_{\gamma }=\mathrm{Pr}(Y=1)=0.0005$. Independence predicts $\mathrm{Pr}(X=1,Y=1)=p_e{p}_{\gamma }=10^{-6}$, hence one expected double detection. Alternatively chooseThese nonnegative probabilities sum to one and preserve both marginal probabilities, yet predict $101$ double detections in expectation. Their covariance is $0.000101-p_e{p}_{\gamma }=0.000100$. This is an exact joint probability distribution illustrating pairing; it does not model the timing spread in the left panel or reconstruct either historical hypothesis quantitatively. Its lesson is why two singles totals cannot decide whether the observed events share a cause.