Take a look at the next community instance:

Assume S desires to ship 3 sats to R. You’ll be able to additional assume that S has sufficient liquidity in every of its native channels to ship as much as 3 sats. Additionally assume the liquidity in channels (A,R), (B,R) and (C,R) is uniformly distributed.

one optimally dependable cost movement on this diagram appears to be like like this:

1 sat: S --> A --> R   chance: 2/3
2 sats: S --> B --> R  chance: 3/5

This movement has a complete chance of 2/3*3/5 = 2/5 = 0.4 = 40%

The query:

Tips on how to compute the anticipated worth of Satoshis to reach at R if S sends 3?

Possibility A

(which I already know is unsuitable however I write it down as a result of I think some individuals may need an analogous first thought)

Initially I believed this could simply be 3 sats * 2/5 = 6/5 sats = 1.2 sats which is what one will get from multiplying the quantity to ship with the chance of the movement. This appears unusual as sending 2 sats alongside S-->B-->R has a chance of 3/5 and with the reasoning of above an expectation worth of 2 sats * 3/5 = 6/5 sats = 1.2 sats. because the anticipated worth for 1 sat alongside the S-->A-->B path is bigger than 0 this could be a contradiction to the additivity of the anticipated worth.

Possibility B

Ranging from the above reasoning we add the anticipated values for the disjoint paths so:

E[3 sats] = 1 sat * 2/3 + 2 sat * 3/5 = 10/15 sats + 18/15 sats = 28/15 sats

Possibility C

In fact the two satoshi path S-->B-->R doesn’t should be despatched as one onion however might be despatched as two onions with 1 sat every:

The primary has a chance of 4/5 and the second has a conditional chance of 3/4 which is extensively defined at this challenge. With the logic from choice B one ought to have the ability to add these anticipated values.
so we have now the anticipated worth for sending two sats in two seperate 1 sat onions alongside S--> B --> R can be computed as:

E[2 sats] = 1 sat * 4/5 + 1 sat * 3/4 = 31/20 sats 

If we add the 1 sat onion from the S-->A-->R which was 2/3 sats

we might count on to have

E[3 sats] = 31/20 sats + 2/3 sats = 93/60 sats + 40/60 sats = 132/60 sats = 33/15 sats

That is 5/15 sats = 1/3 sats greater than the reply in choice B

Possibility D

To make issues worse I’m confused if the anticipated values of dissecting the two sat onion in choice C into two 1 sat onions can simply linearly added up because the second onion is conditioned on having 2 sats of liquidity within the channel. If the primary onion has failed the second will definitely fail. Thus one must compute anticipated worth for sending two 1 sat onions like this:

E[2 sats] = 1 sat * 4/5 + 1 sat * 3/5 = 7/5 sats

this could lead to a complete anticipated worth of:

E[3 sats] = 2/3 sats + 7/5 sats = 10/30 sats + 21/15 sats = 31/15 sats

Ideas

only for comparability listed below are the outcomes

• Possibility A: 18/15
• Possibility B: 28/15
• Possibility C: 33/15
• Possibility D: 31/15

Whereas Possibility B appears definitely proper it is sensible to additional dissect the two sats onion. In simulations I did plainly Possibility D is right which is a bit stunning for me. Utilizing the formalism of chance principle the distinction for the two sat path is:

• Possibility C: E[2 sats] = 1 sat * P(X>=1) + 1 sat * P(X>=2 | X >= 1)
• Possibility D: E[2 sats] = 1 sat * P(X>=1) + 1 sat * P(X>=2)

As mentioned the simulated setting signifies that Possibility D is the right reply however that’s extremely stunning to me as I might count on the second time period to be a conditional probabilty.