OK. For simplicity we'll assume a player bet, winning with probability 50.625% and losing with probability 49.375% (we ignore ties). This implies a house edge of 1.25%*b and a variance of (1+1.25%)*(1-1.25%)*b2 = 0.99984375*b2, where b refers to the bet size.
Hence, after n plays we have a drift of E(n) = -1.25%*Σni=1(bi) and a variance of V(n) = 0.99984375*Σni=1(bi2), where bi refers to the bet size for the ith bet.
Because the ostensible Mr. Leeson is attempting to win as much as possible at a -EV game before going broke, he's best served keeping his variance as high as possible. This is accomplished by selecting his bi's as large as possible. The largest possible bi would correspond to a $10,000 maximum bet. For simplicity sake we'll henceforth refer to this as a single unit. This gives us:
E(n) = 0.0125*n
V(n) = 0.99984375*n
For standard Brownian motion (technically a "Wiener process") X(t) we have:
P(LUB[X(t) - (at+b)] ≥ 0, 0 ≥ t ≥ T) = N(-α-β) + e-2ab * N(α-β)
where:
α = a*sqrt(T),
β = b sqrt(T) ,
LUB[] is the Least Upper Bound operator, and
N() is the cdf of the standard normal distribution (mean=0, var=1)
If Bk is the value of the Mr. L's bankroll after the kth then we want to calculate two probabilities:
- P(Bk - B0 ≥ $20,000,000 = 2,000 units)
- P(Bk - B0 ≤ -$2,000,000,000 = -200,000 units)
for all values of k ranging from 1 to infinity.
If we define Yk as the result of the kth bet (either +1 or - unit) and further define Sk = Yk + 1.25% then the first inequality is satisfied if and only if:
Σki=1(Si) ≥ 2,000 + 0.0125*k
You'll note that E(Sk) = 0 and Var(Sk) = 0.99984375*k.
We can approximate the distribution of Sn using the Brownian motion equation above where P(X(t) ≥ 2,000 + t*0.0125/0.99984375, for 1 ≥ t ≥ n * 0.99984375). Sn in each case is an X(t) process with drift 0 and variance 0.99984375, where the kth term of X(t) has (n-1) * 0.99984375 ≥ t ≥ n * 0.99984375.
Hence, in our original Brownian equation above we have:
T = 0.99984375 * n
a = 0.0125/0.99984375 = 0.01250195
b = 2,000
α = 0.0125/0.99984375 * sqrt(0.99984375 * n) = 0.01250098 * sqrt(n)
β = 2,000 / sqrt(0.99984375 * n) = 2001.56268 / sqrt(n)
Hence, Mr. L's probability of netting +2,000 units on by the nth trial may (for large n) be approximated by:
P = N(-0.01250098*sqrt(n)-2001.56268/sqrt(n)) + e-50.00781372*N(0.01250098*sqrt(n)-2001.56268/sqrt(n))
We see here that as n → ∞ the probability approaches N(-∞) + e-50.00781372*N(∞)
= 0 + 1.91374*10-22 * 1
= 1.91374*10-22
This means that even if Mr. L had an infinite bankroll the probability of him ever (in other words, as number of bets approached infinity) bankrupting the casino (i.e., by netting +2,000 units) only approaches 1.91374*10-22. That's pretty slim.
Of course, in the problem given Mr. L's bankroll wasn't infinite but in fact only contained 200,000 betting units. So let's figure the second inequality above in order to determine what's the probability of Mr. L. losing 200,000 units on or before the nth bet.
P(Bn - B0 ≤ -200,000 units)
In this case we have:
T = 0.99984375 * n
a = 0.0125/0.99984375 = -0.01250195
b = 200,000
α = -0.0125/0.99984375 * sqrt(0.99984375 * n) = -0.01250098 * sqrt(n)
β = -200,000 / sqrt(0.99984375 * n) = -200015.6268 / sqrt(n)
Hence, Mr. L's probability of netting -200,000 units on before the nth trial may (for large n) be approximated by:
P = N(0.01250098*sqrt(n)-200015.6268/sqrt(n)) + e5000.781372*N(-0.01250098*sqrt(n)-200015.6268/sqrt(n))
which implies that P → 1 as n → ∞.
Now in theory the way to proceed at this point would be to determine the probability of the player bankrupting the casino at every value of n jointly conditoned on his not already having bankrupted the casino and of his not having already gone broke.
In practice, however this is largely unnecessary as the probability of bankrupting the casino converges much more quickly than the probability of the player going bankrupt himself. In other words, the player's best chance of bankrupting the casino is to do so fairly quickly.
After 1,000,000 bets, for example, the probability of the player having bankrupted the casino has already converged to the limit (to a precision of 14 decimal places) while the probability of the player having bankrupted himself still stands at an incredibly meager 2.122589974*10-7636. (Thanks go out to RickySteve for the incredibly high precision calculation).
As such the probability upper probability bound of 1.91374*10-22, while only strictly accurate as a Brownian approximation for an infinite player bankroll, nevertheless serves as an excellent approximation of the player's probability of bankrupting the casino given a player bankroll of 200,000 units.
Furthermore, the only way for the player to have a probability greater than 1.91374*10-22 of taking out the house would be to increase the maximum bet size. Increasing his bankroll would have virtually no impact.
Stat Trivial: Beat the house...round two for Nicky
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Originally Posted by picoman
