| О рисках... ID:48250 |
Вс, 19 августа 2001 00:00 [#] [») |
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| DeaLeR |
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 (иконки IM)
Форумы Покер.ру
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Что подразумевается под терминам ROR, если он равен 10%, то в девяти случаях достигаешь какой-то суммы (какой именно???) а в одном проигрываешься??? Если это так, то если снизить плюсовой барьер, то и риски снизяться, или я чего-то недопонимаю...
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| Re: О рисках... ID:48251 ответ на 48250 |
Вс, 19 августа 2001 00:00 («] [#] [») |
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| Garry Baldy |
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(иконки IM)
Форумы Покер.ру
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Ты примерно прав. Классический ROR - это риск проиграть все деньги до того как ты удвоишь
свой банк.
Существует также понятие Goal Oriented ROR, то есть, грубо, вероятность потерять X% от банка
раньше чем ты выиграешь Y%.
Они, очевидно, считаются по разным формулам и используются для разных целей. Я, например,
больше внимания уделяю дисперсии, нежели параметру ROR, несмотря на их тесную
взаимосвязь. Мне проще оценить, насколько безумными будут скачки моего банка, глядя на
цифру SD, нежели на цифру ROR.
Если сильно интересно изучение этих двух типов риска, то вот тебе ниже статейка.
Удачи.
Garry Baldy.
Goal Oriented ROR derivation (long)
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Posted by Richard Reid on June 03, 1997:
The General Risk of Ruin formula in Allan Wilson's "The Casino Gambler's Guide" determines the
probability of losing ones entire bankroll before doubling ones bankroll. I've been thinking that perhaps
a Goal Oriented Risk of Ruin (ROR) formula that allows one to determine the ROR for a user specified
bankroll percentage might come in handy. In other words, there may be times when one wishes to
determine the probability of losing z% of their bankroll before winning y%. Equation [19] will allow a
person to calculate a Goal Oriented ROR for these kinds of questions.
We start the proof presuming as Wilson does in "The Casino Gambler's Guide", that the solution to the
difference equation can be shown in the following simplified form:
[1] r (x) = A + B (1/S)^x
where:
p = probability of winning on a single play
q = probability of losing on a single play
S = p/q
A and B are arbitrary constants that depend upon:
1) the initial capital of the player,
2) the amount the player is willing to lose and
3) the amount the player wishes to win
x = the amount of capital the player has at any given point in time
Let's set up a game where a player starts with 'a' units of capital and his opponent starts with 'b' units of
capital.
What we are looking for is the equation that tells us the probability of the player losing 'c' units of his
own capital before winning the opponents 'b' units.
For any a >= b and a>=c, if the player starts with 'a' units and loses 'c' of those units, then he can be
said to have immediately lost the game. In other words, if his capital gets reduced by 'c' units, then the
probability of loss is 1.
If the player starts with 'a' units and wins his opponents 'b' units, then the game is immediately won. In
other words, if the player's capital increases by 'b' units, then the probability of loss is 0.
We can state things mathematically as follows:
[2] r(a-c) = A + B (1/S)^(a-c) = 1
[3] r(a+b) = A + B(1/S)^(a+b) = 0
If we let b = y*a, where y is the percentage of 'a' that we wish the player to win and if we let c = z*a,
where z is the percentage of 'a' that gives us the portion of bankroll we can lose. Then, we can rewrite
the equations as follows:
[4] r(a-za) = A + B (1/S)^(a-za) = 1
[5] r(a+ya) = A + B(1/S)^(a+ya) = 0
Now we have to solve for A and for B.
From equation [4] we get:
[6] B = (1 - A) / (1/S)^(a-za)
Substituting [6] into [5] we get:
[8] A + [(1 - A) / (1/S)^(a-za)] * (1/S)^(a+ya) = 0
Multiplying both sides of [8] by (1/S)^(a-za) gives us
[9] A * (1/S)^(a-za) + (1 - A) * (1/S)^(a+ya) = 0
[10] A * (1/S)^(a-za) + (1/S)^(a+ya) - A * (1/S)^(a+ya) = 0
[11] A * [(1/S)^(a-za) - (1/S)^(a+ya)] = -(1/S)^(a+ya)
[12] A = -(1/S)^(a+ya) / [(1/S)^(a-za) - (1/S)^(a+ya)]
Now, substituting A from [12] into [4] gives us:
[13] -(1/S)^(a+ya) / [(1/S)^(a-za) - (1/S)^(a+ya)] + B (1/S)^(a-za) = 1
Multiplying both sides of [13] by [(1/S)^(a-za) - (1/S)^(a+ya)] gives us
[14] -(1/S)^(a+ya) + B (1/S)^(a-za) * [(1/S)^(a-za) - (1/S)^(a+ya)] = [(1/S)^(a-za) - (1/S)^(a+ya)]
[15] B = [(1/S)^(a-za) - (1/S)^(a+ya) + (1/S)^(a+ya)] / [ (1/S)^(a-za) * [(1/S)^(a-za) - (1/S)^(a+ya)] ]
[16] B = (1/S)^(a-za) / [(1/S)^(a-za + a-za) - (1/S)^(a-za + a + ya)]
[17] B = (1/S)^(a-za) / [(1/S)^(2a-2za) - (1/S)^(2a + ya - za)]
Now if we substitute A from [12] and B from [17] into [1] and with x set to a, we get:
[18] r (a) = -(1/S)^(a+ya) / [(1/S)^(a-za) - (1/S)^(a+ya)] +
[ (1/S)^(a-za) / [(1/S)^(2a-2za) - (1/S)^(2a + ya - za)] ] * (1/S)^a
Rearranging [18] a bit gives us:
[19] ] r (a) = [ (1/S)^(2a-za) / [(1/S)^(2a-2za) - (1/S)^(2a + ya - za)] ] - (1/S)^(a+ya) / [(1/S)^(a-za) - (1/S)^
(a+ya)]
[QED]
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