Dirty Money Spinner
Corey Plover
Cumpston Sarjeant, Melbourne, VIC 3000
Corey_Plover@cumsar.com.au
1.
Introduction
Page 23 of Actuary Australia, Volume 111 [1] issued a challenge to readers involving a
casino game where the probability of spinning red is p and the probability of spinning black is
q = 1 – p. It asks for the probability that, given an infinite number of spins, the number of red
spins ever equals the number of black spins.
To begin, we recast the problem in the context of a random walk. Define the process Xt to
represent the cumulative number of reds minus the cumulative number of blacks after t spins.
Therefore after each spin, Xt either increases by 1 with probability p or decreases by 1 with
probability q.
For now, we assume that the process begins at X0 = 1 and we introduce absorbing barriers at
Xt = 0 and Xt = M. Define ψm to be the probability that Xt hits zero before it hits M, given that
X0 = m. By taking limits as M → ∞ and examining ψ1, and by mirror arguments ψ-1, we can
obtain the answer to the original question.
2.
Method 1: Recurrence relations
By definition ψ0 = 1 and ψM = 0. By conditioning on the result of next game we also have:
 m  p  m1  q  m1, m  1, 2,
, M 1
Since p + q = 1 we can rewrite this as
 m1  m 
q
 m  m1 
p
From this we can easily see that
 m1  m
q
 
 p
m
 1  0 
We now use telescoping sums to obtain an expression for ψM – ψ0 (which by definition equals
-1) and then for ψm – ψ0
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 M  0  1

M 1
  k 1  k 
k 0
M 1
q
  p
k 0  
  1  0 
  1  0 
k
1 q p
M
1 q p
 m  0  1

m 1
  k 1  k 
k 0
m 1
q
  1  0    
k 0  p 
  1  0 
k
1 q p
m
1 q p
By combining these two expressions and using ψ0 = 1 we obtain:
m
M
q p  q p

m 
M
1 q p
We can then take limits as M → ∞. If p > q then ψm → (q/p)m and if p < q then ψ0 → 1. For
p = q the entire process simplifies dramatically and our initial formula becomes
 m1  m  1  0
Since we know that ψM – ψ0 = -1 by definition, we quickly obtain ψ1 – ψ0 = -1/M and deduce
that ψm = 1-m/M. So as M → ∞ then ψm → 1.
Note: It is not immediately obvious whether this random walk can continue forever. However
similar telescoping performed on the probability of hitting M before zero shows that for any
initial value 0 ≤ m ≤ M, the sum of hitting either barrier equal 1.
So we have the final result that if we start with an equal number of red and black spins, then
conditioning on the very first spin the probability of ever returning to this position at any time
in the future is Pr(Xt = 0, t > 0) = p.ψ1 + q.ψ-1 and so:

If p = q then both ψ1 and ψ-1 (by mirror arguments) equal 1 and so Pr(Xt = 0, t > 0) = 1

If p > q then ψ1 = q/p and ψ-1 = 1 and so Pr(Xt = 0, t > 0) = 2q

If p < q then ψ1 = 1 and ψ-1 = p/q and so Pr(Xt = 0, t > 0) = 2p
A more concise way of phrasing this final answer is that the probability of ever returning to a
equal number of red and black spins is 1 – α, where α = |1 – 2p|.
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3.
Method 2: Martingales
q
As suggested in [2], define a new process, Yt   
 p
q
E Yt | Y0   p  
 p
q
 
 p
Xt
q
 
 p
 Yt
Xt
X t 1
Xt
. Then it follows that Yt is a martingale
q
 q 
 p
X t 1
 p  q
Now, call τ the time that Xt first hits zero or M (or equivalently, the time when Yt first hits
(q/p)M or (q/p)0 = 1). τ is therefore a stopping time and we have:

Pr[τ < ∞] = 1 and E[τ] < ∞ (i.e. M is finite and must eventually hit one of the barriers)

There exists a constant c such that |Xt+1 – Xt| ≤ c for all t (i.e. c = 1 will suffice)
Therefore, we can apply the optimal stopping theorem to conclude that if we start from
X0 = m, then E[Yτ] = E[Y0] = (q/p)m. Consequentially, we have
m
q
q
   
 p
 p
M
1  m   m
m
M
q p  q p

m 
M
1 q p
Which is identical to Method 1 and the same arguments then apply except is p = q then Yt = 1
for all t and the stopping time is triggered immediately, hence ψm = 1 as before.
4.
References
[1]
Stephen Woods, “Two Ducks Swimming”, Actuary Australia 111 (2006), 22-23.
[2]
Charles Grinstead & J. Snell, “Introduction to Probability”, Chapter 12, Exercise 9,
http://math.dartmouth.edu/~prob/prob/prob.pdf
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