I Estimating a mean from games of ruin

[lalekl]

Camilla plays a casino game with fixed bet, fixed odds, no skill, and a starting bankroll, $M_0$. She plays until she can no longer afford to bet and records in a journal only her bankroll and how many bets she was able to make, $N_0$, until she could not afford to bet. She has amnesia from drinking too many free drinks at the casino so remembers nothing else about the game. Each day she goes back to the casino with a larger bankroll, $M_i$, plays the same game until she can no longer afford to bet, records her bankroll and how many bets she made, $N_i$ , drinks too much and suffers amnesia again forgetting everything else about the game. Can she determine the true return of the game using only the information in her journal ( her sequences of bankrolls, $\{M_i\}$, and $\{N_i\}$, how many bets she was able to make on day $i$ before she could no longer afford to bet that day)?

Additional info: the distribution of the game has non-negative discrete support, potentially unbounded, potentially including any given real number >= 0, has zero weight on infinity, mean less than 1, and finite variance. Camilla's method of estimating the mean may not use information about the support she could have learned from playing the game unless she intuits it from the only information about the game she is allowed to use (the sequences $\{M_i\}$, and $\{N_i\}$) and the additional assumptions listed here. She does not even know the bet of the game.

 

[StoneTemplePython]

Look, in the predecessor thread to this I suggested using Wald's Equality. With some insight related to rescaling, you can easily apply it here. Your response then was

The problem with using $E[J]$ in practice for my games is that distributions like slot machines with very heavy tails tend to give absurdly high $E[J]$ with very low chances of ever experiencing it, whereas the order statistics have empirically very tantalizing properties just out of reach of my analytic abilities.

which was a non-sequitor and misleading. The game is in the house's favor, each play is iid, has finite mean and in fact a finite variance. There's simply not much more we can ask for. You asked to estimate the expected payoff of the game, and this is the simplest most direct way. Tails are in fact part of the payout mix of an unfair game. The question was about estimating how unfair the game is, and tails must be taken into account. When we have a finite second moment there probability distributions ultimately behave in an intelligible and intuitive way. High variance may necessitate a lot of sampling, but that's life. When you have a only a finite first moment they behave a bit strangely. And when you have no first moment -- e.g. for the symmetric random walk (i.e. null recurrent chains) -- you get bizarre behavior. Thankfully we have a second moment and don't need to worry about any of this.

The order statistics were irrelevant to what was asked then and now.

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I'm increasingly wondering whether there is a real question underlying these postings -- you've repeatedly used nonstandard terminology and changed the problem each time someone tries to answer it. I've seen fairer games in a casino. As a result, I rather like @andrewkirk 's idea of loading you up with integrals and sending you on your way. If you want to try and apply Wald's Equality to solve your question, you're welcome to do so.

 

[lalekl]

Sorry to mislead with the non-sequitor...I didn't think it was, I just don't think estimating $E[J]$ is practical but I think other methods are and empirically they converge faster but I need a strong statement to make progress on them...

the question in the other thread was meant to be more general and definitely I was not 100% and am not 100% sure whether it is a coherent statement. discussing with you and the others pushed me to try to be very specific and I got as far there as I thought I could. The question was always straightforward though, is the limit of that sequence well defined and equal to the unconditional mean...sounds like everyone believed it was incoherent or undefined at best.

Not sure how much of that applies to the question in this thread. My criticisms of the other answers here is that they are not real answers: they make stronger assumptions. There is no demonstration that picking arbitrary supports or parameters will ever get you to the right answer for an unknown distribution. I don't even believe probability statements can be assigned a priori in that way see for instance (https://en.wikipedia.org/wiki/Freiling's_axiom_of_symmetry), and focus on the criticisms of Freiling's intuition. I feel like you should be someone who is saying this.

On the other hand the problem stated in this thread is unambiguous, but has maybe a lot of subtle details. You are suggesting for this I try using: $$E[X_1+...+X_J]=E[J]E[X_1]$$
So obviously $E[X_1]$ is what we want to find. How do we get the other two terms from the information in the problem? Each day I see $J_i$ and $M_i$, and I know that she experienced a return bounded by $1-M_i/J_i$ and $1-M_i/J_i+1/J_i$, can I take one of them as an estimator for $E[X_1+...+X_J]/E[J]$ and say that converges to the same mean? It is conditional on a different information set...so that isn't obvious to me...

 

[StoneTemplePython]

Not sure how much of that applies to the question in this thread. My criticisms of the other answers here is that they are not real answers: they make stronger assumptions. There is no demonstration that picking arbitrary supports or parameters will ever get you to the right answer for an unknown distribution. I don't even believe probability statements can be assigned a priori in that way see for instance (https://en.wikipedia.org/wiki/Freiling's_axiom_of_symmetry), and focus on the criticisms of Freiling's intuition. I feel like you should be someone who is saying this.

this axiom of symmetry really is a different thread in my view. It feels awfully close to philosophy though, so watch out. I generally try hard to avoid getting in the weeds of individual distributions and in the process sidestep a lot of issues. This seems to be one of them -- though not one of the usual suspects I had in mind to be honest.

All I need here for the below is that the game has a finite second moment and is in the House's favor. That's it.

On the other hand the problem stated in this thread is unambiguous, but has maybe a lot of subtle details. You are suggesting for this I try using: $$E[X_1+...+X_J]=E[J]E[X_1]$$
So obviously $E[X_1]$ is what we want to find. How do we get the other two terms from the information in the problem? Each day I see $J_i$ and $M_i$, and I know that she experienced a return bounded by $1-M_i/J_i$ and $1-M_i/J_i+1/J_i$, can I take one of them as an estimator for $E[X_1+...+X_J]/E[J]$ and say that converges to the same mean? It is conditional on a different information set...so that isn't obvious to me...

So these rescaling approach / linearity exploit just jumps off the page at me. You basically have a memo or table with results of a full 'day' of play from Bob or whomever:

$\begin{bmatrix} \text{starting bankroll} = S_n& \Big \vert & \text{number of plays} \\ 1 & & j^{(1)}\\ 1 + a & & j^{(2)} \\ 1 + b & & j^{(3)}\\ \vdots & \vdots \\ \end{bmatrix}$

Or something like that. (Note I'm using the change of variable from that prior post #13.)
For starters the table has $r$ entries in it. Maybe it ends up having countably infinite number. Maybe not.

The point is your gambler (Bob?) has different starting bankrolls each day. But $\bar{Y}$ is constant because the same games are played each day. We can rescale the starting bankroll to homogenieze the $J$ that we are estimating an expected value for. That's a backwards take at it. The forward take is, $\bar{Y}$ is fixed, and as you proportionately increase the starting bankroll $S_n$, I know that the associated $E\big[J\big]$ grows linearly with this change -- this is exactly what Wald's Equality tells me.

But we have to work backward at things because this is a curious kind of inference problem. The idea is that each row in there should be rescaled to have a constant (let's go with constant = 1) in the left hand entry -- so for instance your second row, right hand entry would have $\frac{j^{(2)}}{1+a}$ in it.

With this you'll have standardised your table to in effect have the same amount of starting capital. Each right hand side entry thus has the same expected number of turns until failure -- which you don't directly see. But you can still sum all of the results on the RHS and divide by $n$. Weak Law of Large numbers still applies -- even though the trials aren't cleanly iid -- they are still independent. And the average on the RHS tends to $E\big[J\big]$. The Left Hand side is fixed. This allows you to estimate $\bar{Y}$ as the scaling needed to convert your $E\big[J\big]$ to the constant on LHS. It's a bit unpleasant to have these two be a product but that's life. Assymptotically the estimate is right, though because it is in product form, it may be biased for any finite number of estimates. (I can feel Jensen's Inequality possibly be an issue as we have $E\big[J\big] = \frac{\text{constant}}{\bar{Y}}$ in this setup and the mapping of $u \to \frac{1}{u}$ is strictly convex for $u \gt 0$ though maybe it isn't an issue, in any case it's not something I'm inclined to dwell on.)

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As another aside: I've thought about re-framing as a simple linear regression -- there's a closely related point about minimizing squared errors though for the linear relationship between expected results that Wald's Equality tells us, though doing this requires potentially a lot more work to fully justify and I'm not going to go that route.
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There's various ways to interpret/ justify the rescaling done above. Hopefully this is enough to play around with for a while.

There can of course be other nits about 'chunkiness' of betting stakes -- e.g. minimum bet size of 1 and after a long day at the casino you have $30 \text{ cents}$ in your pocket. Boundary issues are a nuisance but not something I'd worry too much about. The solution of course is to gather more data farther from the boundary to smooth the result out. Alternatively, and what I'd suggest doing to start, is for convenience assume there are not meaningful boundary issues, and come back and add them later on to complicate the model.

 

[lalekl]

Very interesting! So much to learn and explore here.

The forward take is, ¯YY¯\bar{Y} is fixed, and as you proportionately increase the starting bankroll SnSnS_n, I know that the associated E[J]E[J]E\big[J\big] grows linearly with this change -- this is exactly what Wald's Equality tells me.

I think I am missing something to get $S_n$ and $E[J]$ increase linearly together. I see $E[J]$ has to increase linearly with $E[Y_1...+Y_J]$. So does that rely on $E[Y_1...+Y_J]$ and $S_n$ increasing linearly together, and is that obvious?

 

[StoneTemplePython]

Very interesting! So much to learn and explore here.I think I am missing something to get $S_n$ and $E[J]$ increase linearly together. I see $E[J]$ has to increase linearly with $E[Y_1...+Y_J]$. So does that rely on $E[Y_1...+Y_J]$ and $S_n$ increase linearly together, and is that obvious?

depending on how you interpret it, yes, because $S_J = \sum_{i=1}^J Y_i$

... it belatedly occurs to me that I should have been using $S_J$ or given the lack of homogeneity, $S_J^{(n)}$ or something like that to denote the $S_J$ on the nth "day". I think I merged notations in a not so great way.
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so, remember what lurks in the background for anything Wald Equality related is (a) a valid stopping rule and (b) $E\big[J\big] \lt \infty$. These are both satisfied here though the latter takes some work to show. (If you're interested in this sort of stuff, may I suggest https://ocw.mit.edu/courses/electri...62-discrete-stochastic-processes-spring-2011/ ? It's a lot of work, but not a lot of wasted work...)

Ignoring nuisances around boundary conditions with minimum bet size, for the time being, we stop when our cumulative bankroll = 0 or equivalently, when accumulated losses are equal to the bankroll we walked into the casino with on that day. Otherwise the game never stops. Hence we can say $E\big[S_J\big] =\text{bankroll at start of the day}$

The equation of Wald tells us $E\big[S_J\big] = E\big[J\big]\bar{Y}$ and the same game is being played over and over (and independently) so $\bar{Y}$ which is expected payoff per game, is constant.

Now let's have some fun with scaling. Choose some $\alpha \gt 0$ and scale

$\alpha E\big[S_J\big] = E\big[\alpha S_J\big] = E\big[(\alpha J)\big]\bar{Y} = \big(\alpha E\big[J\big]\big)\bar{Y} = \alpha E\big[J\big]\bar{Y}$

And then repeatedly choosing the right scalars along the way to homogenize your starting bankroll and hence homogenize expectation of J as well. Then back into expected value of Y after a lot of trials. That's really it.

 

[jim mcnamara]

@lalekl Please choose either A or B
A: accept suggestions
B: stop bringing up non-sequitur requirements and ad-hoc "spray", without complete counterexamples or proof.

There is no C:
We are volunteers here, in effect you abusing our time. Reddit is a nice place to have undirected arguments. PF is educational.

Thank you.

 

[Dale]

Now we’re cooking with whale oil! But how?

I would make a statistical model that could encompass a reasonable set of payout schedules, and have one of the parameters of the model be the EV. Then I would marginalize over all of the other parameters.

 

[lalekl]

depending on how you interpret it, yes, because $S_J = \sum_{i=1}^J Y_i$

... it belatedly occurs to me that I should have been using $S_J$ or given the lack of homogeneity, $S_J^{(n)}$ or something like that to denote the $S_J$ on the nth "day". I think I merged notations in a not so great way.
- - - -

so, remember what lurks in the background for anything Wald Equality related is (a) a valid stopping rule and (b) $E\big[J\big] \lt \infty$. These are both satisfied here though the latter takes some work to show. (If you're interested in this sort of stuff, may I suggest https://ocw.mit.edu/courses/electri...62-discrete-stochastic-processes-spring-2011/ ? It's a lot of work, but not a lot of wasted work...)

Ignoring nuisances around boundary conditions with minimum bet size, for the time being, we stop when our cumulative bankroll = 0 or equivalently, when accumulated losses are equal to the bankroll we walked into the casino with on that day. Otherwise the game never stops. Hence we can say $E\big[S_J\big] =\text{bankroll at start of the day}$

The equation of Wald tells us $E\big[S_J\big] = E\big[J\big]\bar{Y}$ and the same game is being played over and over (and independently) so $\bar{Y}$ which is expected payoff per game, is constant.

Now let's have some fun with scaling. Choose some $\alpha \gt 0$ and scale

$\alpha E\big[S_J\big] = E\big[\alpha S_J\big] = E\big[(\alpha J)\big]\bar{Y} = \big(\alpha E\big[J\big]\big)\bar{Y} = \alpha E\big[J\big]\bar{Y}$

And then repeatedly choosing the right scalars along the way to homogenize your starting bankroll and hence homogenize expectation of J as well. Then back into expected value of Y after a lot of trials. That's really it.

Thank you, between this and some of the inequalities I've mentioned before, I have a complete solution to the above problem. I will program it up and see how fast it converges.

 

[StoneTemplePython]

Thank you, between this and some of the inequalities I've mentioned before, I have a complete solution to the above problem. I will program it up and see how fast it converges.

Looking through some of my notes, it may turn out that some additional structure needs to be imposed.

1.) In particular I don't see the guaranty on finiteness of variance of your recurrence time that I was looking for (in essence $E\Big[\big(J^{(n)}\big)^2\Big]$ as we know $E\Big[\big(J^{(n)}\big)\Big]$ is finite)

My sense is no real problems, baring a pathological setup... and if the payoff mix of a gambling problem is finite (like say a real slot machine) or at least bounded, a lot of this should simplify. In this case you could directly show exponential decay in probabilities of being away from the mean with increasing numbers of bets (via chernoff bound) to assure $E\Big[\big(J^{(n)}\big)^2\Big] \lt \infty$, and not having mess around with more complex machinery like truncation arguments.

2.) it is possible that we may need to bound the range of starting bankrolls for the gambler, so $M\in [a, b]$ -- which after (1) then bounds $\sigma_{J^{(n)}}$. In general there are many different sufficient conditions for WLLN or SLLN for independent non identically distributed random variables like these $J^{(n)}$, but not clean cut necessary and sufficient recipes for the non iid case. My bias is to just pick a sufficient condition like the Kolmogorov Criterion for SLLN which is easy to apply when there is a $\text{max}\big(\sigma_{J^{(n)}}\big)$.

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Again, I wouldn't anticipate problems, but technically there's a few too many ambiguities / degrees of freedom still on the table.

 

[jim mcnamara]

Thanks for participating. The thread seems to have some good answers. Thread closed.