Let's take a look at the "Sorokin" RoR formula. (Please don't get hung up on the who first
applied the formula to VP. It just doesn't matter here) In a future email I will take a good
look at the RoRBR formula.
How much confidence to we have that the RoR formula is correct? Do we really know what
RoR represents? To begin to answer these questions, I am going to consider an incredibly
simplified version of VP. Though it is simplified , the VP game I will use captures all the
"elements" (I like to call it the "physics") necessary to understand RoR.
···
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So here is the game: Instead of multiple outcomes, there are only 2 as follows:
Outcome 0: No return (bet is lost) with a Probability of P0
Outcome 1: A return of w units with a Probability of P1, W>0
With Po+ P1 = 1 and P0>0 and P1>0
Only 1 unit can bet at a time. The EV of the game is EV = 0 * P0 + W*P1 - 1 (I subtract
the 1 so that EV is either positive, negative or zero)
Please don't balk at the use of such a simple game. Instead, be patient, and see where it
leads us. We will add some more complexity to the game later.
For now, I am going to assume (without loss of generality) that we have a 1 unit bankroll
and that we only bet 1 unit at a time (I already said that).
----------------------------------- RoR Formula
----------------------------------------
If we apply the RoR formula to this simple game, we get
RoR = x = P0 + P1 * x ^w
For w=1, this takes on a particularly simple form
RoR(w=1) = P0/ (1-P1) = 1 since 1-P0 = 1
For w =2, the solution is also readily available, but a little more complex
RoR(w=2) = [1 +/- sqrt( 1 - 4*P1*P0)] / 2*P1
Noting that PO = 1 - P1, we get,
RoR(w=2) = [1+/- (2P1-1) ]/2P1
Which produces 2 values,
RoR(w==2) =1 or RoR(w=2) = (1-P1)/ P1 = P0/P1
Let's limit ourselves to these simple cases, where w=1 and w=2.
So what have we learned so far?
For w=1, RoR =1. That is, for this simple game, ruin is inevitable. That makes sense. The
game is NEGATIVE, so RoR =1. Good
For w=2. RoR can take on 2 values, RoR=1 or RoR = P0/P1. Since w=2, the game is
positive, so we expected 2 answers, and one that said RoR <1 and we got it
----------------------------------- Other approach
----------------------------------------
Let's look at the calculation of RoR from first principles.
After any number of hands there are only 2 possible "states": Either we are "still playing"
or we are "ruined"
So,
P(ruined,n) + P(still playing,n) = 1
where
P(ruined, n) is probability that we are ruined after hand n
P(still playing, n) is probability that we are still playing after hand n
This is a very important result that bears repeating: There are only 2 states to this "game",
and since we must always be in one of the state, the sum of their probabilities must equal
1.
Now, let look at the case when w =1.
At the first play, we can either "Ruin" with a probability of P0 or be still playing, with a
Probability of P1,
P(ruin,1) = P0
P(still playing) = P1
Adding up the two probabilities yields,
P(ruin,1) + P(still playing) = P0 +P1 =1
as it should.
After the first hand, if we are still playing, we again have the possibility of being ruined or
not. Interestingly, the contributions from each play to the probabliity of Ruin are additive,
while those for "still playing" are multiplicative. (Please don't take my word for this, work
it out for yourself). So,
P(ruin,2) = P(ruin,1)+ P0 * P1 = P0 + P0*P1
P(still playing) = P(still playing, 1) *P1 = P1^2
By extension, after the n-th hand, we have
P(ruin,n) = P0 + P0*P1 + P0*P1^2 + .... + P0*P1^(n-1) = P0 * [ 1 + P1 + P1^2 + .... +
P1^(n-1)]
P(still playing) = P1+ P1^2 + ....+ P1^n = P1* [ + P1 + P1^2 + .... + P1^(n-1)]
Both of these geometrical series have well known formulas, so,
P(ruin,n) = P0 * ( 1 -P1^n)/ (1-P1)
P(still playing,n) = P1 * ( 1 -P1^n)/ (1-P1)
We can check that the sum of the two probabilities is equal to 1,
P(ruin,n) +P(still playing,n) =? 1
P0 * ( 1 -P1^n)/ (1-P1) + P1 * ( 1 -P1^n)/ (1-P1) = (P0 + P1) * *( 1 -P1^n)/ (1-P1) = 1
as it should.
Let's simplify R(ruin,n),
P(ruin,n) = P0 * ( 1 -P1^n)/ (1-P1) = 1-P1^n since PO = 1 -P1
So we have arrived at our answer,
P(ruin,n) = 1 -P1^n
Earlier we found that, using a different approach,
RoR(w=1) =1
By noting that, in the limit of n->infinity,
P1^n -> 0 since P1< 1,
so,
P(ruin,n->infinity) = 1
So, at this point we can conclude, RoR indeed gives the P(ruin,n) for infinite n (for w=1, a
negative game).
Though this is a trivial result, it sure is reassuring.
END OF PART 1. In the next part, I hope to tackle P(ruin, w=2, n)