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The backward algorithm computes suffix likelihoods in a Hidden Markov model. With ${\beta }_t(i)=P(o_{t+1},\dots ,o_T\mid Z_t=i)$, its recurrence is ${\beta }_t(i)={\sum }_j{a}_{ij}{b}_j(o_{t+1}){\beta }_{t+1}(j)$. Together with the forward algorithm, it gives posterior state probabilities proportional to ${\alpha }_t(i){\beta }_t(i)$. The terminal condition must include any explicit End-state transition.

The forward algorithm computes sequence likelihoods by summing over all hidden paths through a Hidden Markov model. For emitting states, ${\alpha }_t(i)=P(o_1,\dots ,o_t,Z_t=i)$ obeys ${\alpha }_{t+1}(j)=b_j(o_{t+1}){\sum }_i{\alpha }_t(i)a_{ij}$. It reuses partial sums rather than enumerating exponentially many state sequences. Silent states require additional transitions without consuming an observation. Scaling or log-sum-exp prevents floating-point underflow.

The sequence has five characters, while the four match states account for four compulsory emissions. Consequently exactly one insertion state emission is needed. An insertion at $I_3$ would have to emit T, which has probability zero there. The only nonzero hidden path is thereforeMultiply its transition probabilities and emission probabilities, including the probability of exiting $I_2$ immediately and bypassing $I_3$:Normally the forward algorithm sums over all compatible paths. Here all other paths have probability zero, so this single path gives the full sequence probability.

Use a uniform independent DNA background, so each nucleotide has probability $1/4$. Conditional on length five, its sequence probability is $4^{-5}=1/1024$. The requested score, often called log-odds in sequence analysis, is a log-likelihood ratio against this background:Using natural logarithms instead gives $\ln (27/4)\approx 1.90954$. Identically distributed independent positions do not, by themselves, require equal base frequencies. If the intended background instead has specified frequencies $q_A,q_T,q_G,q_C$, the general answer isThe numerical value above uses the conventional equal-frequency random DNA model.

Taking a logarithm only after multiplying probabilities does not prevent floating-point underflow: the product may already have rounded to zero, and its logarithm is then $-\infty$. Accumulate logarithms before forming tiny products; where alternative paths must be added, use log-sum-exp in the forward algorithm or rescale its intermediate probabilities.

Let $Z_t$ be the hidden state. With the usual convention ${\alpha }_t(i)=P(o_1,\dots ,o_t,Z_t=i)$ and ${\beta }_t(i)=P(o_{t+1},\dots ,o_T\mid Z_t=i)$, the Markov property and the conditional independence of emissions factor the joint probability asThe requested state-path probability after observing the complete data is the corresponding conditional probability:Here the forward algorithm and backward algorithm give the same normalizer at any interior time,There is no extra factor $b_i(o_t)$: it is already included in ${\alpha }_t(i)$. If an unconditioned joint probability with the observations is intended, it is the numerator alone. Summing the boxed expression over all $i,j,k$ gives one.