How do I calculate the number of expected SATs and arrive at a probabilistic payment flow?

Take a look at the following network example:

Assume S I want to send it 3 Sit down R. You can assume that further S Each local channel has sufficient liquidity 3 soil. It also assumes channel fluidity (A,R), (B,R) and (C,R) It is distributed evenly.

One optimally reliable payment flow in this diagram looks like this:

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

This flow has a total probability 2/3*3/5 = 2/5 = 0.4 = 40%

question:

How to calculate the expected value of the arrival of Satoshu R if S send 3?

Option a

(I already know I’m wrong, but I think some people have similar initial thoughts, so I’ll write it down)

At first I thought this was just right 3 sats * 2/5 = 6/5 sats = 1.2 sats This is obtained by multiplying the amount of transmission with the probability of flow. This seems strange when sending two SATs S-->B-->R There is a chance of 3/5 And with the above reasoning 2 sats * 3/5 = 6/5 sats = 1.2 sats. The expected value of 1 is sitting along S-->A-->B The path is bigger 0 This is inconsistent with the expected additive.

Option b

Starting with the above reasoning, we add the expected value to the broken path.

E(3 sats) = 1 sat * 2/3 + 2 sat * 3/5 = 10/15 sats + 18/15 sats = 28/15 sats

Option c

Of course, two Satoshi Passes S-->B-->R You don’t need to send it as a single onion as an onion, but you can send it as two onions each with one soil.

The first is the probability 4/5 The second is a conditional probability 3/4 This is widely explained in this issue. You should be able to add these expectations using the logic in option B. Therefore S--> B --> R It is calculated as follows:

E(2 sats) = 1 sat * 4/5 + 1 sat * 3/4 = 31/20 sats 

Add one Saturday onion S-->A-->R That was 2/3 soil

We expect to have

E(3 sats) = 31/20 sats + 2/3 sats = 93/60 sats + 40/60 sats = 132/60 sats = 33/15 sats

This is 5/15 sats = 1/3 sats More than the answer to option b

Option d

Worse, I’m confused if there’s a chance that the expectation of analyzing two SAT onions in option C into two SAT onions could be linearly added, as the second onion is conditional to have two SAT onions in the channel. If the first onion fails, the second onion certainly fails. Therefore, in order to send such two Saturday onions, you need to calculate the expected value.

E(2 sats) = 1 sat * 4/5 + 1 sat * 3/5 = 7/5 sats

This gives you the following total expected value:

E(3 sats) = 2/3 sats + 7/5 sats = 10/30 sats + 21/15 sats = 31/15 sats

thought

There are results just for comparison

• Option a: 18/15
• Option b: 28/15
• Option c: 33/15
• Option d: 31/15

Option B certainly seems correct, but it makes sense to further analyze the two SATS onions. In the simulation, option D appears to be correct. This is a bit surprising to me. Using the formalism of probability theory, the difference between the 2 SAT paths is as follows:

• Option c: E(2 sats) = 1 sat * P(X>=1) + 1 sat * P(X>=2 | X >= 1)
• Option d: E(2 sats) = 1 sat * P(X>=1) + 1 sat * P(X>=2)

As mentioned earlier, the simulated setting indicates that option D is correct, but that is very surprising to me as I expect the second term to be a conditional probability.