DW44 Bankroll Requirements

[Quad Deuces]

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[New2vp]

First thing is finding the covariance for a particular game using VPFW.  Keeping with APDW, here's what I did.I ran the bankroll calculator on the single-line game.  It gave me a variance of 25.4989 (25.5 as discussed above).Then I ran the calculator on the 10-line game.  It gave me a variance  of 5.2942.  I multiplied this by 10 to get 52.942.We are then presented with a basic algebra problem.  C is covariance and is the unknown we need to solve for:As per New2VP's formula:Variance(4-play) = Variance(1-play) + 3 x Covarianceor, more generically:Variance(n-play) = Variance(1-play) + (n-1) x CovarianceSubstitute the two variance values from VPFW and the proper number of lines (n = 10) and we get:52.942 = 24.4989 25.4989 + 9CDivide both sides by 95.88244 = 2.833211 + CSubtract 2.833211 from both sides3.049233 = CC = 3.0492 which is the same value New2VP gave.I do not know why the value given by FPDW VPFW has to be multiplied by the number of lines.  Still a little fuzzy on per coin, per-line, and per-bet values.Hey 2222,At first I was confused as to why you were switching from Almost Positive Deuces to Full Pay Deuces, but then it dawned on me that you might have unconsciously typed in the wrong acronym.  My answer is based on the corrections I made to your post above.You did find a way to determine one form of the covariance that I calculated a year ago.  I purposely kept my answer there shorter so that fewer eyes would glaze over from a longer post.  To answer your questions, I must issue a warning to others:  EVERYONE EXCEPT QUAD DEUCES MAY WISH TO SKIP TO THE NEXT THREAD AT THIS POINT.  I AM NOT RESPONSIBLE FOR THE DEBRIS FROM EXPLODING HEADS!The formula that I gave [ Variance(n-play) = Variance(1-play) + (n-1) x Covariance ], allows one to compare the effect of n-play against playing n independent hands of another game.  So, as you noted above the variance for one hand of 10-play APDW, whose single-hand variance is about 25.5, would be bigger than say 10 hands of single-line 15-12 Double Deuces (99.78% ER) that pays 2000 coins for quad deuces, whose variance is about 50.  So, this gives us an idea of the level of typical differences from expected return that we get from 1 hand of n-play vs. n hands of single-line.Thus, this formula gives a variance that can be compared to the variance that is given by standard video poker software for single-line play of one hand that is independent of all the other hands.With 10-play, you have 10 hands, thus this number (52.942 for single-line play) needs to be to be multiplied by 10 to get it in units of 10-coin bets, but that is not the whole story here and does not fully explain what you see in VPFW.VPFW gives you two variances.  If you relook at what you did, you will see that the bankroll calculator for 10-play gives two figures:  5.2942 and 13235.52.  The first one is in units of "bets squared" and the second is in the units of "coins squared."  You can see that if you multiply the first value by the bet size squared (50 x 50 = 2500), you will get the second value, within rounding.This can be compared with the single line value that is typically given from standard vp software.  WinPoker gives 25.49886.  This number is in the units of bets squared, so if you want coins squared, you must multiply by 25 (5 x 5 = 25).  You can see that the Bankroll Calculator for VPFW gives values of 25.4989 for bets squared and 637.4715 for coins squared.Finally, you can note that I earlier said the covariance of 3.0492 was in the same units as the single-line variance, which was in (5-coin) bets squared.  If we need the covariance in terms of coins squared, we also must multiply that by 25 (5 x 5 = 25).  Doing that, 3.049249 x 25, gives us 76.2312 coins squared.  (Here, I kept more significant digits to get a more precise answer.)
Part of the problem in understanding all of these calculations is realizing that the bet sizes in
the comparison of single-line vs. multiline play are different.  Single line is 5 coins; 10-play is 50 coins; n-play is 5n coins.
If you keep everything in terms of coins squared, you can see the equivalence, but the variance formula must change slightly (in this case, we need to multiply by 10).  So, I would modify the "coins squared" formula to be as follows:Variance(10-play) = 10 x [Variance(1-play) + 9 x Covariance]or, for n-play:Variance(n-play) = n x Variance(1-play) + n x (n-1) x CovarianceChecking the above example, you can see that, in terms of coins squared:Variance(n-play) = (10 x 637.4715) + (10 x 9 x 76.2312) = 13235.523, which checks to within rounding of the number given by VPFW.Second, why use 2.5 standard deviations for the
bankroll calculations?  Using that value, if I was to want to play
10-line nickel NSUD for an hour, I calculate I'd need $1019.NSUD 10-line nickels:1-line Variance: 25.7803Covariance: 3.043633  (obtained using the procedure outlined above)25.7803 + 9 * 3.043633 = 53.173997500 bets using the above-mentioned formula:sqrt(500) x (2.5) x SqRT(53.17) = 407.63 bets x $2.50/bet or $1019.VPFW
says after 500 10-line games the worst I could possibly expect to be
down is approximately $250.  500 games is only $1250 coin-in.
I know I'm missing something...<>You are correct in questioning the use of 2.5 standard deviations in bankroll calculations.  You will note that I answered a similar question earlier in this thread.  I will repeat that answer, with slight modifications, here.  (I was using a slightly different example, but I think you should be able to follow the logic, especially noting the bold blue portion below.)
And when you say, "So after 700 deals, you can expect 68% of the
population to be within  [1 standard deviation] of true expectation",
you are making a fairly large error.  A video poker distribution is
significantly different than the normal distribution which has the 68%
property that I expect you are using.A normal distribution is
symmetric.  A video poker distribution is not.  With a normal
distribution you can expect 50% of the time to be over the mean and 50%
to be under.  With this game and 700 single-line plays, you will be
under the mean 67% of the time.  You often hear experienced vp players
say that they lose more than they win even when they are playing with an
edge and that's related to this property.With a normal
distribution, the mode (or most typical observation) would be the same
as the mean:  here 3500 coins rounded to the nearest 5 coins.  With this
game, the mode is 3280 or a 220-coin loss.With this game and 700 plays, you will be within +/-
1 standard deviation about 88% of the time, with only 2% worse and 10%
of the time better.  It has to be that way so that you can offset the
typical 220-coin loss on what is essentially a break-even game.With
only 700 plays, you would essentially NEVER be more than 2 standard
deviations lower than the expectation (odds of more than 45 million to
1), whereas with a normal distribution, that would occur about once
every 44 times.  But you would be more than 2 standard deviations higher with this game about 4.6% of the time.Unfortunately,
you cannot carry those inferences exactly to other games or for other
numbers of plays and the calculations are somewhat involved to do this
in general. 

[faygo]

Huh!  Boops do not read this. I was warned but ....    

[New2vp]

Sorry, but I couldn't figure out an easier way to answer this, other than to say, "Carefully study what I have already said and be careful to make necessary supplements where necessary."


Huh!  Boops do not read this. I was warned but ....    

[Quad Deuces]

Sorry for the typos.  I blew out the edit window and was a little P.O'd I had to do it over again.  Did it again offline and copied but I was still too hasty.I also noticed after posting that the OP had tossed out 2.5 standard deviations as a suggested value, so that's probably what led to it being used in the calculations.I am considering stepping outside my bankroll for an upcoming promotion and feel a 50-line nickel game is probably the best way for me to get the coin-in up but since VPFW doesn't do bankroll for 50 lines I am trying to figure out how much I should take in with me, and what I should reasonably expect.This is a lot to chew on.  Thanks.