March 28th, 2014 at 5:23:10 AM
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Scenario a : 1 hand x $B, variance = V1, expectation= EV1
Scenario b : n hand x $B/n, combine variance for n hands = V2, combine expectation for n hands = EV2.
EV1 = EV2, how about the relationship between V1 and V2 ?
Scenario b : n hand x $B/n, combine variance for n hands = V2, combine expectation for n hands = EV2.
EV1 = EV2, how about the relationship between V1 and V2 ?
March 28th, 2014 at 6:53:21 AM
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If the results of the hands in scenario b are independent of each other:
* V2_one mini hand = V1/(n^2)
* Total variance of scenario b, V2 = n*V2_one mini hand = n*V1/(n^2) = V1/n
scenario a: V1
scenario b: V1/n
However in most casino game the results of the hands at a single table in a single round are significantly correlated. This means the problem is way more complex and varies by game.
* V2_one mini hand = V1/(n^2)
* Total variance of scenario b, V2 = n*V2_one mini hand = n*V1/(n^2) = V1/n
scenario a: V1
scenario b: V1/n
However in most casino game the results of the hands at a single table in a single round are significantly correlated. This means the problem is way more complex and varies by game.
March 28th, 2014 at 7:15:16 AM
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Are you sure V2 = n*V2_one mini hand ?
March 28th, 2014 at 7:24:22 AM
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Yes, given the hands are independent. http://en.wikipedia.org/wiki/Variance#Basic_propertiesQuote: ssho88Are you sure V2 = n*V2_one mini hand ?
March 28th, 2014 at 12:02:03 PM
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But they aren't independent in blackjack, so unfortunately it's not that easy.
March 28th, 2014 at 12:33:49 PM
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The key formula here is:
Var(X + Y) = Var(X) + Var(Y) + 2Cov(X, Y)
where:
Cov(X, Y) = E(XY) - E(X)E(Y)
and
Var(X) = Cov(X,X) = E(X^2) - (E(X))^2.
If Cov(X,Y) = 0 then the variables are said to be uncorrelated, and Var(X + Y) = Var(X) + Var(Y). Independent variables are uncorrelated.
Var(X + Y) = Var(X) + Var(Y) + 2Cov(X, Y)
where:
Cov(X, Y) = E(XY) - E(X)E(Y)
and
Var(X) = Cov(X,X) = E(X^2) - (E(X))^2.
If Cov(X,Y) = 0 then the variables are said to be uncorrelated, and Var(X + Y) = Var(X) + Var(Y). Independent variables are uncorrelated.
February 1st, 2017 at 9:03:59 PM
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Quote: ssho88Scenario a : 1 hand x $B, variance = V1, expectation= EV1
Scenario b : n hand x $B/n, combine variance for n hands = V2, combine expectation for n hands = EV2.
EV1 = EV2, how about the relationship between V1 and V2 ?
i need this kind of data, so i tried to test it by myself.
the relationship is N is bigger, the V2 is smaller. in other words, the more hands(tables, investments...) you play, the covariance is smaller. or playing more hands or do more different investments+team-money-sharing will reduce the variance/increase the kelly/multiply the profit.
let's say:
1%EV
2hands 3hands 4hands 5hands 6hands 7hands
1.4 1.8 2.1 2.3 2.4 2.8 (here it is strange, i used 500,000 random numbers, each 100 hands/set, 10 times, then choose average).
here note if EV is different, the result is different.
1.2%EV
2hands 3hands 4hands 5hands 6hands 7hands
1.4 1.8 2.0 2.2 2.4 2.5
5%EV
2hands 3hands 4hands 5hands 6hands 7hands
1.4 1.6 1.8 1.9 1.9 1.9
10%EV
2hands 3hands 4hands 5hands 6hands 7hands
1.2 1.2 1.0 1.3 1.1 1.1
when 10% EV, the covariance is even smaller than 2 hands.
by this kind of the result, we can say we need to cooperate with others no matter how rich or capable we are.