How to compute the expected number of sats to arrive in a probabilistic payment flow?

Originally posted by Rene Pickhardt in bitcoin.stackexchange.com/questions/117447/how-to-compute-the-expected-number-of-sats-to-arrive-in-a-probabilistic-payment

Look at the following network example:

Assume S wants to send 3 sats to R. You can further assume that S has enough liquidity in each of its local channels to send up to 3 sats. Also assume the liquidity in channels (A,R), (B,R) and (C,R) is uniformly distributed.

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 of 2/3*3/5 = 2/5 = 0.4 = 40%

The question:

How to compute the expected value of Satoshis to arrive at R if S sends 3?

Option A

(which I already know is wrong but I write it down because I suspect some people might have a similar first thought)

Initially I thought this would just be 3 sats * 2/5 = 6/5 sats = 1.2 sats which is what one gets from multiplying the amount to send with the probability of the flow. This seems strange as sending 2 sats along S-->B-->R has a probability of 3/5 and with the reasoning of above an expectation value of 2 sats * 3/5 = 6/5 sats = 1.2 sats. as the expected value for 1 sat along the S-->A-->B path is larger than 0 this would be a contradiction to the additivity of the expected value.

Option B

Starting from the above reasoning we add the expected 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

Option C

Of course the 2 satoshi path S-->B-->R does not have to be sent as one onion but could be sent as two onions with 1 sat each:

The first has a probability of 4/5 and the second has a conditional probability of 3/4 which is extensively explained at this issue. With the logic from option B one should be able to add those expected values. so we have the expected value for sending two sats in two seperate 1 sat onions along S--> B --> R would 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 would 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 in option B

Option D

To make things worse I am confused if the expected values of dissecting the 2 sat onion in option C into two 1 sat onions can just linearly added up as the second onion is conditioned on having 2 sats of liquidity in the channel. If the first onion has failed the second one will certainly fail. Thus one would have to compute expected value for sending two 1 sat onions like this:

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

this would result in a total expected value of:

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

Thoughts

just for comparison here are the results

Option A: 18/15
Option B: 28/15
Option C: 33/15
Option D: 31/15

While Option B seems certainly right it makes sense to further dissect the 2 sats onion. In simulations I did it seems that Option D is correct which is a bit surprising for me. Using the formalism of probability theory the difference for the 2 sat path is:

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 said the simulated setting indicates that Option D is the correct answer but that is highly surprising to me as I would expect the second term to be a conditional probabilty.