What kind of bigram probability smoothing is this?

Question

I hope it isn't off topic but I need to understand this example. Given the corpus 12 1 13 12 15 234 2526 and smoothing factor of k=1. The example does the following operations:

Considers OOV(out of vocabulary) words and assigns them a zero times value, after that k=1 is added to the times every words appears, to avoid zero probabilities. So the result of smoothing the bigrams probability will be:

$P(1|12)=(1+k)/(2+2+6*k)=0.2$
$P(15|12)=(1+k)/(2+2+6*k)=0.2$
$P(13|1)=(1+k)/(2+6*k)=0.25$
$P(12|13)=(1+k)/(2+6*k)=0.25$
$P(234|15)=(1+k)/(2+6*k)=0.25$
$P(2526|234)=(1+k)/(2+6*k)=0.25$

My question is, What kind of smoothing is this? shouldn't be for example like this?; $P(1|12)=(1+k)/(2+6*k)=0.25$
Besides it also says "If OOV words appear, you need to use smoothing to return a value; $P(234|12)=1/((2/7)*6+6)=0.1296$"

PS: I take this example from a small section of the translated version of this chinese webpage, it is just explaining a code implementation.

Answer 1

This looks like it might be implementing Laplace smoothing (though I can't explain the extra +2 in the first two calculations).

There are many smoothing methods, e.g., Lidstone, Good-Turing, Kneser-Tey, Witten-Bell. You can also use interpolation and backoff to handle rare or low-probability events when computing conditional probabilities like this.