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Thread: Sonny: EMFH team ROR question

  1. #1
    Sonny
    Guest

    Sonny: EMFH team ROR question

    Let's say that I have two team members using different systems in different casinos. They have, respectively, win rates of W1 and W2 with per hand standard deviations of SD1 and SD2. If they share a common bankroll, how do I determine what their combined "team" ROR is? Do I need more information to complete the formula?

    This is probably the first question that everybody asks, right?

    -Sonny-

  2. #2
    Don Schlesinger
    Guest

    Don Schlesinger: Re: EMFH team ROR question

    > Let's say that I have two team members using
    > different systems in different casinos. They
    > have, respectively, win rates of W1 and W2
    > with per hand standard deviations of SD1 and
    > SD2. If they share a common bankroll, how do
    > I determine what their combined
    > "team" ROR is? Do I need more
    > information to complete the formula?

    You might. Is there an assumption that each plays 50% of the hands (hours)? Otherwise, you would have to weight the above values according to what percentage of the total hands each plays.

    In any event, win rates are simply additive, so that shouldn't pose a problem. Standard deviations are not additive, but variances (the squares of the s.d.s) are. So, once you square the s.d.s to get the variances of each player, you may then add those variances to get the cumulative value. I believe that you could then use the "blended" per-hand win rate and (recalculated) s.d. (or variance) to find the combined ROR -- but, I'm not certain that the averaging process is 100% valid. I have a feeling that the riskier of the two players might carry some extra weight in the formulation of the cumulative ROR.

    > This is probably the first question that
    > everybody asks, right?

    No, in fact, it isn't, or I'd already be certain of the answer. :-)

    Don

    P.S. What does "EMFH" mean?


  3. #3
    Sonny
    Guest

    Sonny: EMFH

    I had a feeling about the variance, but I always get nervious when I start squaring things. The potential to amplify small mistakes makes me begin to question my answers. Thanks Don.

    > P.S. What does "EMFH" mean?

    It is a Snyder-ism for an "Every Man for Himself" team. It refers to a group of players who share a common bankroll and play as individuals at different tables as opposed to BP-type methods.

    -Sonny-

  4. #4
    Don Schlesinger
    Guest

    Don Schlesinger: Re: EMFH

    > I had a feeling about the variance, but I
    > always get nervious when I start squaring
    > things. The potential to amplify small
    > mistakes makes me begin to question my
    > answers. Thanks Don.

    My pleasure.

    > It is a Snyder-ism for an "Every Man
    > for Himself" team. It refers to a group
    > of players who share a common bankroll and
    > play as individuals at different tables as
    > opposed to BP-type methods.

    Never heard it referred to that way, but that's exactly how we played for a couple of years in A.C., with our 12-man team.

    Don

  5. #5
    Karel Janecek
    Guest

    Karel Janecek: Re: EMFH team ROR question


    Best regards,

    Karel

    > You might. Is there an assumption that each
    > plays 50% of the hands (hours)? Otherwise,
    > you would have to weight the above values
    > according to what percentage of the total
    > hands each plays.

    > In any event, win rates are simply additive,
    > so that shouldn't pose a problem. Standard
    > deviations are not additive, but variances
    > (the squares of the s.d.s) are. So, once you
    > square the s.d.s to get the variances of
    > each player, you may then add those
    > variances to get the cumulative value. I
    > believe that you could then use the
    > "blended" per-hand win rate and
    > (recalculated) s.d. (or variance) to find
    > the combined ROR -- but, I'm not certain
    > that the averaging process is 100% valid. I
    > have a feeling that the riskier of the two
    > players might carry some extra weight in the
    > formulation of the cumulative ROR.

    > No, in fact, it isn't, or I'd already be
    > certain of the answer. :-)

    > Don

    > P.S. What does "EMFH" mean?

  6. #6
    Sonny
    Guest

    Sonny: For hourly results

    If both players are playing at the same time, am I correct in thinking that you would take the sum of the win rates and the square root of the sum of the variances (sd^2)? It doesn't seem like I should have to average them if they are taking place concurrently. I suppose my ROR would be the same regardless of which numbers I use, but I like to have an idea of what the short-term ranges will look like.

    -Sonny-

  7. #7
    Don Schlesinger
    Guest

    Don Schlesinger: Re: For hourly results

    > If both players are playing at the same
    > time,

    Same time or same table??? Big difference! If at the same table, this method is not applicable, because of covariance.

    > am I correct in thinking that you
    > would take the sum of the win rates and the
    > square root of the sum of the variances
    > (sd^2)? It doesn't seem like I should have
    > to average them if they are taking place
    > concurrently.

    Two players playing one hour each at separate tables is the same thing as one player playing two hours. You add their win rates and multiply one of their s.d.s by the square root of 2, to get their combined e.v. and s.d.

    > I suppose my ROR would be the
    > same regardless of which numbers I use, but
    > I like to have an idea of what the
    > short-term ranges will look like.

    OK. Clear?

    Don

  8. #8
    Sonny
    Guest

    Sonny: Re: For hourly results

    > Same time or same table??? Big difference!
    > If at the same table, this method is not
    > applicable, because of covariance.

    Of course. My question pertains only to players at different tables. I understand that the variance will be significantly greater if they play at the same table. It would be similar to one player playing at twice the stakes, except that the covariance between the two hands would reduce the variance slightly. I'm glad that you brought this up.

    > Two players playing one hour each at
    > separate tables is the same thing as one
    > player playing two hours. You add their win
    > rates and multiply one of their s.d.s by the
    > square root of 2, to get their combined e.v.
    > and s.d.

    Exactly, but the problem is that the two players are using different systems and different bet spreads. Therefore each will have a different hourly win rate and s.d. By combining the two variances and looking at the square root I am attempting to predict what one s.d. will look like for the two players combined after one hour (each) of play. Certainly the variance must be smoothed to some degree as the results of their individual play begin to cancel each other out.

    To be on the safe side I could multiply the higher of the two s.d.s by SQRT(2) in order to see the "worst case scenario", but for ROR and bankroll management I am looking for a more accurate estimate of their combined hourly s.d.

    I believe I got the correct formulas from your first post, but just to summarize:

    Hourly win rate = EV1+EV2
    Hourly S.D. = SQRT(SD1^2 + SD2^2) * SQRT(100)

    Where EV1 and EV2 are the hourly win rates for players 1 and 2, and SD1 and SD2 are their per hand standard deviations. It also assumes 100 hands per hour.

    -Sonny-


  9. #9
    Don Schlesinger
    Guest

    Don Schlesinger: Re: For hourly results

    > I believe I got the correct formulas from
    > your first post, but just to summarize:

    > Hourly win rate = EV1+EV2
    > Hourly S.D. = SQRT(SD1^2 + SD2^2) *
    > SQRT(100)

    > Where EV1 and EV2 are the hourly win rates
    > for players 1 and 2, and SD1 and SD2 are
    > their per hand standard deviations. It also
    > assumes 100 hands per hour.

    Looks right to me.

    Don

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