[LordMushroom2]

[ QUOTE ]
when i tried shortstacking (20bb) as a learning process for nl (over 50k hands), my SD was 23bb per 100. when i played fullstacked, 100bb, over 300,000 hands, my SD was 38bb per 100. these might be PTBB but the comparison stands nonetheless.

[/ QUOTE ]

I think that´s PTBB because I saw someone posting that he had a standard deviation of almost 100 big blinds per 100 hands.

Can some of the difference in shortstack and big stack standard deviation be explained by you having a smaller skill-advantage over your opponents when playing full-stacked (because you were playing at higher stakes), or would any such effect be insignificant?

Why big stacks have higher variance than shortstacks.

I have always thought that the size of the average pot dictated variance, but now I can see how it doesn´t.

Say there are two (non-poker) games. In one every pot is $10 and in the other the pot is $0 50% of the time and $20 50% of the time. Which one has the greatest variance?

I don´t have a way to calculate it mathematically, but by putting it in a poker context, I can illustrate it. Say player A is playing $5 HU SnGs and player B is alternating between playing play-money HU SnGs and $10 HU SnGs. Both have $150 bankrolls and both make $2,5 per hour.

The variance in player B´s case must be higher because his risk of ruin is greater because he is in reality playing only $10 SnGs and we know the risk of ruin is greater when playing $10 SnGs instead of $5 SnGs all else being equal.

As risk of ruin is calculated using bankroll, winrate and standard deviation, and we know bankroll and winrate are equal, the reason for player B´s higher RoR must be a higher standard deviation.

What was the point of all this? When you play with a big stack in a cash-game, the size of the pots will vary to a greater degree from the average pot than it would if you were playing a shortstack, which means a big stack experiences higher variance than a shortstack. And my statement that stacks didn´t matter on variance is false.

The standard deviation of a normal SnG per 100 hands

Thanks to the guy I quoted, we have the standard deviaton per hundred hands in a full buy-in cash-game. Let´s try to find the standard deviation per 100 hands of SnGs.

This will vary depending on how many SnGs you can play in 100 hands and how big your winrate is.

To have something tangible to work with, let´s say you are playing a type of SnG, which lasts 50 hands on average, the fee is 5% and your winrate is 60%.

Since it lasts for 50 hands, we get to play 2 SnGs per 100 hands. Let´s see what the standard deviation for those 2 SnGs is (warning, I almost always mess up the math):

Var=(2^2)*0,6*0,6+[(1-1,05)^2]*0,6*0,4*2+(-2,1^2)*0,4*0,4-(2*0,6*0,6-0,05*0,6*0,4*2-2,1*0,4*0,4)^2
Var=(4*0,36)+(0,0025*0,48)+(4,41*0,16)-(0,72-0,024-0,336)^2
Var=1,44+0,0012+0,7056-0,1296
Var=2,0172
Standard deviation=1,42

If the 38PTBB/100 was at NL100, its standard deviation would be $76/100.
If the buy-in of the SnG was $100+5, its standard deviation would be $142/100.

This would suggest the variance is much higher in SnGs. Now let the slaughter of my math begin. [img]/images/graemlins/laugh.gif[/img]