Stat Trivial: Beat the house...round two for Nicky

pico · 05-18-08, 11:05 PM

link this to the think tank for the morning poeple...

Ganchrow · 05-19-08, 03:22 AM

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%*Σⁿ_i=1(b_i) and a variance of V(n) = 0.99984375*Σⁿ_i=1(b_i²), where b_i refers to the bet size for the i^th 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 b_i's as large as possible. The largest possible b_i 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 B_k is the value of the Mr. L's bankroll after the k^th then we want to calculate two probabilities:

P(B_k - B₀ ≥ $20,000,000 = 2,000 units)
P(B_k - B₀ ≤ -$2,000,000,000 = -200,000 units)

for all values of k ranging from 1 to infinity.

If we define Y_k as the result of the k^th bet (either +1 or - unit) and further define S_k = Y_k + 1.25% then the first inequality is satisfied if and only if:

Σ^k_i=1(S_i) ≥ 2,000 + 0.0125*k

You'll note that E(S_k) = 0 and Var(S_k) = 0.99984375*k.

We can approximate the distribution of S_n using the Brownian motion equation above where P(X(t) ≥ 2,000 + t*0.0125/0.99984375, for 1 ≥ t ≥ n * 0.99984375). S_n in each case is an X(t) process with drift 0 and variance 0.99984375, where the k^th 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 n^th 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 n^th bet.

P(B_n - 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 n^th trial may (for large n) be approximated by:

P = N(0.01250098*sqrt(n)-200015.6268/sqrt(n)) + e^5000.781372*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.

pico · 05-19-08, 10:52 AM

nicely done, ganch.

now i am kind of curious to figure out what is the min limit you need to take down the house with at least 50% prob with 100 to 1 bankroll.

Ganchrow · 05-19-08, 11:13 AM

Originally posted by picoman

now i am kind of curious to figure out what is the min limit you need to take down the house with at least 50% prob with 100 to 1 bankroll.

With an ∞:1 bankroll (and in this case it's about close enough) we'd need:

e^{-2*0.0125 / 0.99984375 * (total bet units in casino bankroll)} = 50%

total bet units in casino bankroll = ln(50%) / -2 * 0.99984375 / 0.0125
≈ 27.7216

So with a total casino bankroll of $20,000,000, a player with an infinite bankroll would need a to be able to wager $20,000,000 27.7216 ≈ $721,460.25 per bet to have a 50% probability of eventually breaking the bank.

pico · 05-19-08, 11:25 AM

i never seen a limit higher than 25k...but i am not sure about the super VIP rooms...those comes with personal dealer, butler and a suite.

you can bust the table, but you can't take down the house. there is a point to my random question after all

Iwinyourmoney · 05-19-08, 11:27 AM

Ganch are you a robot?

Ganchrow · 05-19-08, 11:49 AM

Originally posted by Ganchrow

With an ∞:1 bankroll (and in this case it's about close enough) we'd need:

e^{-2*0.0125 / 0.99984375 * (total bet units in casino bankroll)} = 50%

total bet units in casino bankroll = ln(50%) / -2 * 0.99984375 / 0.0125
≈ 27.7216

So with a total casino bankroll of $20,000,000, a player with an infinite bankroll would need a to be able to wager $20,000,000 27.7216 ≈ $721,460.25 per bet to have a 50% probability of eventually breaking the bank.

FWIW, if we drop the infinite player bankroll assumption in favor of the 100:1 player:casino bankroll ratio the casino bust probability given a total casino stake as above (about 27.7216 unit bets), the win probability would drop from 50% to (very) roughly 50% - 9.14*10^-10.

At 10:1 it would drop down to roughly 49.24%

pico · 05-19-08, 11:54 AM

1:1 bankroll ratio will have win % of 48.75%...that is the lower limit.

bankroll doesn't have much influence once to reaches a certain threshold, but the table limit is the key factor.

Ganchrow · 05-19-08, 05:27 PM

Originally posted by picoman

1:1 bankroll ratio will have win % of 48.75%...that is the lower limit.

I'm actually getting a value of roughly 33% at 1:1, which I believe to be correct.

How did you come up with 48.75%?

pico · 05-19-08, 05:39 PM

Originally posted by Ganchrow

I'm actually getting a value of roughly 33% at 1:1, which I believe to be correct.

How did you come up with 48.75%?

oh, i just thought that suppose the player and the casino each has million dollars, and they just play one hand for all the marbles.

the reason i start this thread is because i think the house always wins because 1) they have more money than the player and 2) the house edge.

there is no argument about house edge, so i start thinking about a -EV player with a much bigger bankroll than the house...how big of an edge does that give him? i guess the correct answer is none...even if the player manage to break one casino, other casinos will still take his action...so in the long run, bankroll doesn't make a difference. the only difference the bankroll size makes is when Tiki Casino goes bust, they can't recoop their losses...a mere transfer of profit from one casino to another.

Ganchrow · 05-19-08, 05:47 PM

Originally posted by picoman

oh, i just thought that suppose the player and the casino each has million dollars, and they just play one hand for all the marbles.

1:1 refers to the ratio of player to casino wealth, but still assuming casino wealth equal to about 28 big bets.

Had both player casino only 1 big bet each then the player would break the casino with probability 49.375% (his probability of winning a single hand). Brownian motion, is obviously of no use in determining that answer.

Good question, btw.

pico · 05-19-08, 05:58 PM

thanks.

this also applies to NL holdem as well. the big stack players at the cash tables always has a edge over the short stacks, but that is assuming both players has equal skills. a good poker player might sit down and start playing with table-limit buy in that is way less than the big stack player. the good player has an edge over the big stack, but can the big stack player use his chips to compensate his short comings in poker skills?

i think that is really similar to the casino question.