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Thread: MJ: CE Question for Norm

  1. #1
    MJ
    Guest

    MJ: CE Question for Norm

    Can you please explain why the ratio of CE/WR should always be 0.5? My understanding of CE is that it represents how much money you are willing to either accept/pay in order to forgo/enter an opportunity that involves risk.

    In the context of BJ, suppose a game has a WR of $50/Hr with an SD of $700/Hr. Moreover, say a counter earns $20/Hr in his job, if he is content to take the sure thing of 20 bucks, then his CE = $20/Hr.

    Now say he is not satisfied with his hourly wage, and requires at least $30/Hr at his regular job to forgo the BJ opportunity which has risk. Now his CE = $30/Hr.

    The point I am trying to make is that CE seems like a very subjective concept. Going back to the counter with 50 bucks per hour in EV, why must his CE = $25/Hr? When you suggest ratio of CE/WR = 0.5, that is exactly what you are saying. What if he is risk averse and would rather take the sure thing of 10 bucks at his job? Now the ratio becomes 0.2. Perhaps he is prone to risk and would not settle for anything less than a wage of 40 bucks/Hr to forgo playing BJ. Here the ratio increases to 0.8.

    Note: I realize and know the formula for CE, but it seems rather peculiar a formula can tell us how much money we are willing to accept in order to forgo playing BJ.

    Btw, why is it if the ratio is less than 0.5, you are overbetting? Is the converse true as well?

    Thanks,
    MJ

  2. #2
    Norm Wattenberger
    Guest

    Norm Wattenberger: Re: CE Question for Norm

    You can have a ratio of .5 with whatever risk you choose by changing the unit size. For example, in CVCX, set the risk to 2%, it will calculate the betting ramp that gets as close to 2% as possible, and CE/WR will be close to .5. If you bet less than that, your risk will drop below your target and CE/WR will increase.

  3. #3
    MJ
    Guest

    MJ: Re: CE Question for Norm


    > You can have a ratio of .5 with whatever risk you
    > choose by changing the unit size. For example, in
    > CVCX, set the risk to 2%, it will calculate the
    > betting ramp that gets as close to 2% as possible, and
    > CE/WR will be close to .5. If you bet less than that,
    > your risk will drop below your target and CE/WR will
    > increase.

    Okay, I found this on a website that explains CE in simplistic terms. This is a good site and I suggest all players read it. See link below.

    "A certainty equivalent is the minimum amount of money I would rather have for certain instead of taking some risk. The more risk-averse a person is, the lower is her certainty equivalent."

    Given that definition, what if the amount of money that I would accept to forgo playing BJ is LESS than the CE indicated by CVCX? Does that mean I did something wrong in setting up the parameters on CVCX??? Should I lower the K-f for the game or something?

    In your example, if you bet less than your target 2% RoR, would this mean that you are risk averse or are you simply betting sub-optimally? Surely WR will decrease, but why should it decrease at a greater rate than CE?

    Also, is it possible for CE = WR on CVCX?

    MJ



  4. #4
    Norm Wattenberger
    Guest

    Norm Wattenberger: Re: CE Question for Norm

    > Given that definition, what if the amount of money
    > that I would accept to forgo playing BJ is LESS than
    > the CE indicated by CVCX? Does that mean I did
    > something wrong in setting up the parameters on
    > CVCX??? Should I lower the K-f for the game or
    > something?

    Yes, lower risk. CE and WR will both decrease and CE/WR will remain about the same.

    > In your example, if you bet less than your target 2%
    > RoR, would this mean that you are risk averse or are
    > you simply betting sub-optimally? Surely WR will
    > decrease, but why should it decrease at a greater rate
    > than CE?

    Everyone that is not missing an amygdala is risk-averse to some degree. Yes, underbetting is sub-optimal, and being sub-optimal, WR will decrease more quickly.

    > Also, is it possible for CE = WR on CVCX?

    Yes, but not with any rational settings


  5. #5
    MJ
    Guest

    MJ: Re: CE Question for Norm

    I wrote: Is it possible for CE = WR on CVCX?

    > Yes, but not with any rational settings

    Under what circumstance could this scenario manifest itself? When a player bets to less than target RoR in a suboptimal fashion, you said the ratio of CE to WR is greater than 0.5. If he grossly under-bets (sub-optimally) his BR, would the ratio converge to 1:1?

    What confuses me in this discussion is that under-betting implies risk-averse. But the closer CE is to WR, the greater the level of risk. Going back to the coin example on the website I linked to, there was a coin flip where if heads came up you get $100, tails you get nothing. If you pay $10 for the opportunity to play, you are risk averse. $50 you are neutral. $75 you are a risk taker.

    When a player under-bets his BR as mentioned above, would this be construed as risk-averse or foolishness (because he is not betting optimally)? I guess if he wanted risk-averse, he could lower his K-f. So maybe this kind of betting is not risk-averse after all.

    Also, if CE = WR, wouldn't that mean that in theory the counter would not forgo BJ unless he was paid no less than his hourly WR? For instance, if his WR/Hr was $25 and his job paid $24/Hr, he would opt BJ.

    Thanks,
    MJ

  6. #6
    Norm Wattenberger
    Guest

    Norm Wattenberger: Re: CE Question for Norm

    > I wrote: Is it possible for CE = WR on CVCX?

    > Under what circumstance could this scenario manifest
    > itself? When a player bets to less than target RoR in
    > a suboptimal fashion, you said the ratio of CE to WR
    > is greater than 0.5. If he grossly under-bets
    > (sub-optimally) his BR, would the ratio converge to
    > 1:1?

    Just play with the custom bets and see what happens.

    > What confuses me in this discussion is that
    > under-betting implies risk-averse. But the closer CE
    > is to WR, the greater the level of risk. Going back to
    > the coin example on the website I linked to, there was
    > a coin flip where if heads came up you get $100, tails
    > you get nothing. If you pay $10 for the opportunity to
    > play, you are risk averse. $50 you are neutral. $75
    > you are a risk taker.

    If you bet $75 to gain $100 on a fair coin toss, you aren't a risk-taker, you're stupid. If you bet $49 to win $100, you are more risk-averse than a similarly bank-rolled person that would bet $98 to gain $200.

    > When a player under-bets his BR as mentioned above,
    > would this be construed as risk-averse or foolishness
    > (because he is not betting optimally)? I guess if he
    > wanted risk-averse, he could lower his K-f. So maybe
    > this kind of betting is not risk-averse after all.

    One should bet optimally if one can. Lower the betting unit if you wish to reduce risk.

    > Also, if CE = WR, wouldn't that mean that in theory
    > the counter would not forgo BJ unless he was paid no
    > less than his hourly WR? For instance, if his WR/Hr
    > was $25 and his job paid $24/Hr, he would opt BJ.

    I really don't like that comparison in general.

  7. #7
    MJ
    Guest

    MJ: Re: CE Question for Norm

    I wrote:

    > Also, if CE = WR, wouldn't that mean that in theory
    > the counter would not forgo BJ unless he was paid no
    > less than his hourly WR? For instance, if his WR/Hr
    > was $25 and his job paid $24/Hr, he would opt BJ.

    You wrote:

    > I don't like that comparison in general.

    Fine, but what metric do you suggest one use to measure the opportunity cost of his play if not CE?

    In BJ, CE = WR - [Variance / (2 * BR * K-f)]

    Suppose a system was devised that had positive EV with 0 variance. This is surely a counter's dream come true. Now CE = WR. Doesn't this imply that the ideal ratio of CE/WR is 1? Yet it is maintained in the community that the ideal ratio is 0.5 when betting optimally.

    MJ

  8. #8
    Norm Wattenberger
    Guest

    Norm Wattenberger: Re: CE Question for Norm

    > I wrote:

    > You wrote:

    > Fine, but what metric do you suggest one use to
    > measure the opportunity cost of his play if not CE?

    > In BJ, CE = WR - [Variance / (2 * BR * K-f)]

    > Suppose a system was devised that had positive EV with
    > 0 variance. This is surely a counter's dream come
    > true. Now CE = WR. Doesn't this imply that the ideal
    > ratio of CE/WR is 1? Yet it is maintained in the
    > community that the ideal ratio is 0.5 when betting
    > optimally.

    If the variance is zero, the optimal bet is your entire bankroll, increasing every round.


  9. #9
    Don Schlesinger
    Guest

    Don Schlesinger: Re: CE Question for Norm

    > Suppose a system was devised that had positive EV with
    > 0 variance. This is surely a counter's dream come
    > true.

    Now, it's fantasy; it isn't blackjack, and it has no relevance in this discussion.

    > Now CE = WR. Doesn't this imply that the ideal
    > ratio of CE/WR is 1? Yet it is maintained in the
    > community that the ideal ratio is 0.5 when betting
    > optimally.

    That's because WHEN you bet optimally, CE turns out to be half of WR, for blackjack. It isn't an "ideal" ratio; it's what comes out of the math, when betting optimally. You have it backwards. FIRST, you bet optimally, and then CE BECOMES one-half of WR. You don't "set" the CE to half. It "sets" itself!

    Don

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