I didn't say they won't understand it, they'll just consider the applicant too stupid to understand the problem they presented: grow the bankroll.
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What is your strategy?
- Thread starter kut2k2
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I didn't say they won't understand it, they'll just consider the applicant too stupid to understand the problem they presented: grow the bankroll.
The original question didn't ask "What's the optimal strategy to grow the bankroll?". It's more open-ended. As markettimer pointed out, no utility function was specified, either. It could be quadratic, logarithmic, or it might as well be risk-neutral (i.e. maximization of absolute bankroll, or even maximization of the life roll).
I didn't say they won't understand it, they'll just consider the applicant too stupid to understand the problem they presented: grow the bankroll.
Then we won't exactly be a fit. Because it is THE most relevant aspect of the question as originally asked in terms of what your allocation will be. Markettimer is correct on that. If they can't phrase an intelligent question, they probably can't make an intelligent trade either.
I have seen this type of game offered as a promotion from betting sites like Bodog. For example, "Open a new account and bet up to $1000 on the Super Bowl money line. If you lose, we'll cover your loss."Try to remember: this is not a real casino game. It exists only because the interviewers made it up. I doubt any real-life casino anywhere has offered anything like this game Ever. So good luck explaining to the interviewers how you blew your bankroll on the last spin because you were trying to maximize your at-home net worth instead of maximizing your bankroll.
Also, if you think about this more abstractly, calling "bank roll" to be "what you have saved now" and "life roll" to be "what you expect to save in the future," this question has implications for how much leverage to use for investing when you are young vs. old. A couple Yale Law / School of Management professors asked this question in the following book: http://www.amazon.com/Lifecycle-Investing-Audacious-Performance-Retirement-ebook/dp/B003GYEGK2 Incidentally, my own dismal trading experience was discussed in one of its chapters. My "bank roll" was $210K, "life roll" was expected to be seven or eight figures, and I lost my entire bank roll of $210K in the 2008 crash before ever having a job.
Hey markettimer, I have a question. What you have advocated is maximizing the log of the sum of the liferoll and the bankroll:
Utility = Max[E(Log(L+B))]
where
L is the liferoll,
B is the bankroll
But the notion of geometric growth is meaningful in terms of getting from the starting wealth W(0) to the terminal wealth W(N), such as in this:
Utility = Max[E(Log(W(N) / W(0)))]
So, shouldn't our utility function look like this instead:
Utility = Max[E(Log((L+B) / L))] = Max[E(Log(1 + B/L))]
Utility = Max[E(Log(L+B))]
where
L is the liferoll,
B is the bankroll
But the notion of geometric growth is meaningful in terms of getting from the starting wealth W(0) to the terminal wealth W(N), such as in this:
Utility = Max[E(Log(W(N) / W(0)))]
So, shouldn't our utility function look like this instead:
Utility = Max[E(Log((L+B) / L))] = Max[E(Log(1 + B/L))]
Hey markettimer, I have a question. What you have advocated is maximizing the log of the sum of the liferoll and the bankroll:
Utility = Max[E(Log(L+B))]
where
L is the liferoll,
B is the bankroll
But the notion of geometric growth is meaningful in terms of getting from the starting wealth W(0) to the terminal wealth W(N), such as in this:
Utility = Max[E(Log(W(N) / W(0)))]
So, shouldn't our utility function look like this instead:
Utility = Max[E(Log((L+B) / L))] = Max[E(Log(1 + B/L))]
They are equivalent. W(0) is just a constant.
Max E[log(W(N)/W(0))] is the same problem as max E[log(W(N))]. You can pull W(0) out of the logarithm as follows:
E[log(W(N)/W(0))] = E[log(W(N)] - log(W(0))
Note that argmax f(x) + constant is the same as argmax f(x).
They are equivalent. W(0) is just a constant.
Max E[log(W(N)/W(0))] is the same problem as max E[log(W(N))]. You can pull W(0) out of the logarithm as follows:
E[log(W(N)/W(0))] = E[log(W(N)] - log(W(0))
Note that argmax f(x) + constant is the same as argmax f(x).
Okay, makes sense now, thanks much.
In other words, on spin ten, you have bankroll B_9 (B_9 being the bankroll after the ninth spin) and outside wealth W (total wealth W+B_9). You want to maximize E[log(W+B_10)]. B_10 is a distribution that depends on how B_9 is allocated. This is a straightforward maximization problem where you simply choose allocation fractions xj_10, how to allocate B_9 across the j possible bets on the wheel for the 10th spin.
Now, working recursively, you would solve for the allocation fractions on the ninth spin, starting with bankroll B_8. And so onâ¦
I thought about this again, and I am still not clear about this algo.
To maximize E[log(W+B_10)], I need to try different allocations of B_9. But as we have established, the optimal allocation would depend on B_9. That is, the optimal 10th allocation is a function of B_9. I was hoping to solve this numerically, but that doesn't seem possible to me with this algo.
I can still solve it numerically by going forward (from spin 1 to spin 10), instead of going backwards, but this leads to combinatorial explosion. If I take all possible allocations across R16, R14, RedColor, and BlackNumbers, from 0% to 100%, with an increment of 1%, there would be 4598126 such combinations. Then for each of those, there would be 4598126 combinations on the next spin. So, for 10 spins, I am looking to evaluate 4598126^10 total paths. That would take the age of the Universe to compute.
So, I don't see an efficient way of doing it numerically, unless I replace the "brute-force" search with something like gradient ascent optimization.
Nonlinear5, the recursive solution avoids the combinatorial explosion. The way it is implemented is via a Bellman equation (more here: http://en.wikipedia.org/wiki/Bellman_equation).
Essentially, the goal is to solve for the optimal allocation on the final spin as a function of W and B_9. Then plug this solution into the objective function, E[log(x)], to calculate expected terminal log wealth as a function of B_9 and W. Let's call this V_10(W, B_9).
In other words, V_10(W,B_9) = max E[log(W+B_10)], given W and B_9.
On the final spin, the solution is to bet everything on R-16 when B_9 is small relative to W. However, the optimal allocation changes (becomes lower risk, lower return) as B_9 grows larger relative to W. Therefore, for any given W, V_10(W,B_9) is concave in B_9.
Once V_10(W,B_9) is known, you can solve the 9th spin problem. I'll use V_9(W,B_8) to denote the maximization problem on the 9th spin, and here you can see the recursion begin:
V_9(W,B_8) = max E[V_10(W,B_9)]
In summary, on any spin, the goal is to find the optimal allocation as a function of bankroll, then solve for the expected log wealth given this bankroll. Using this Bellman solution avoids the combinatorial explosion, since the computation time grows linearly, not exponentially, with number of spins.
Essentially, the goal is to solve for the optimal allocation on the final spin as a function of W and B_9. Then plug this solution into the objective function, E[log(x)], to calculate expected terminal log wealth as a function of B_9 and W. Let's call this V_10(W, B_9).
In other words, V_10(W,B_9) = max E[log(W+B_10)], given W and B_9.
On the final spin, the solution is to bet everything on R-16 when B_9 is small relative to W. However, the optimal allocation changes (becomes lower risk, lower return) as B_9 grows larger relative to W. Therefore, for any given W, V_10(W,B_9) is concave in B_9.
Once V_10(W,B_9) is known, you can solve the 9th spin problem. I'll use V_9(W,B_8) to denote the maximization problem on the 9th spin, and here you can see the recursion begin:
V_9(W,B_8) = max E[V_10(W,B_9)]
In summary, on any spin, the goal is to find the optimal allocation as a function of bankroll, then solve for the expected log wealth given this bankroll. Using this Bellman solution avoids the combinatorial explosion, since the computation time grows linearly, not exponentially, with number of spins.
Nonlinear5, the recursive solution avoids the combinatorial explosion. The way it is implemented is via a Bellman equation (more here: http://en.wikipedia.org/wiki/Bellman_equation).
Essentially, the goal is to solve for the optimal allocation on the final spin as a function of W and B_9. Then plug this solution into the objective function, E[log(x)], to calculate expected terminal log wealth as a function of B_9 and W. Let's call this V_10(W, B_9).
In other words, V_10(W,B_9) = max E[log(W+B_10)], given W and B_9.
On the final spin, the solution is to bet everything on R-16 when B_9 is small relative to W. However, the optimal allocation changes (becomes lower risk, lower return) as B_9 grows larger relative to W. Therefore, for any given W, V_10(W,B_9) is concave in B_9.
Once V_10(W,B_9) is known, you can solve the 9th spin problem. I'll use V_9(W,B_8) to denote the maximization problem on the 9th spin, and here you can see the recursion begin:
V_9(W,B_8) = max E[V_10(W,B_9)]
In summary, on any spin, the goal is to find the optimal allocation as a function of bankroll, then solve for the expected log wealth given this bankroll. Using this Bellman solution avoids the combinatorial explosion, since the computation time grows linearly, not exponentially, with number of spins.
Thanks again, markettimer. I'll see if I can figure it how to code that in.
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