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  #1  
Old 02-02-2005, 12:19 PM
Derek in NYC Derek in NYC is offline
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Default Single 10/20 table vs. two 5/10 multitables

Please discuss the consequence of playing a single 10/20 table vs. two 5/10 tables simultaneously. My gut tells me that the overall hourly rate should be unchanged, but the swings (up and down) should be smaller.

This consequence needs to be reduced, however by the fact that your play quality will suffer somewhat as you add tables; on the other hand, your opponent toughness will increase as you move up in levels. On balance, if you assume these factors cancel each other out, the original conclusion is generally correct.

Am I right here?
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  #2  
Old 02-02-2005, 05:33 PM
hurlyburly hurlyburly is offline
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Default Re: Single 10/20 table vs. two 5/10 multitables

b/r considerations are a big determining factor. Lets say you have 100BB for the 10/20. Playing your "A" game at the 5/10 with 200BB allows you to see double the hands with a smaller risk of ruin for the same rate of return. Add in game softness and multitabling 5/10 actually comes out ahead. If you can reach 400BB inside 50-75 hours, then you've got the b/r to 2 table 10/20 or single 20/40. Personally I'd rather 2-table the 10/20 and see a more consistent rate of return at that stage, and would carry that down to the 5/10.
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  #3  
Old 02-03-2005, 07:02 AM
gaming_mouse gaming_mouse is offline
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Default Re: Single 10/20 table vs. two 5/10 multitables

This is a question which keeps cropping up, and I am still a little confused about it too.

Assuming your decision-making ability is unaffected by multi-tabling, you SD/100 should not change, though your hourly variance will. Your risk of ruin also will not change, which means that, aside from the extra money you need to buy in to each table, you should need the same bankroll.

Is this correct, or does your bankroll need to increase by sqrt(# of tables)? If so, please explain the flaw in the above reasoning.

The reason I have doubts is that, anecdotally, it seems my swings are much bigger when I multi-table. Not just my hourly swings, but the swings themselves.

Thanks for any thoughts,
gm
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  #4  
Old 02-03-2005, 07:16 AM
pzhon pzhon is offline
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Default Re: Single 10/20 table vs. two 5/10 multitables

You need a smaller bankroll to play 2 tables of $5/$10 than one table of $10/$20.

Suppose that at the $10/$20 table, you win on average $20 per hour with a standard deviation of $200.

Suppose at each $5/$10 table, you win on average $10 per hour with a standard deviation of $100. When you play two $5/$10 tables, in each hour you average $20 with a standard deviation of $100*Sqrt(2) ~ $140.

The standard deviation of A+B is squareroot(stddev(A)^2 + stddev(B)^2) when A and B are independent.
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  #5  
Old 02-03-2005, 07:19 AM
gaming_mouse gaming_mouse is offline
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Default Re: Single 10/20 table vs. two 5/10 multitables

[ QUOTE ]
The standard deviation of A+B is squareroot(stddev(A)^2 + stddev(B)^2) when A and B are independent.

[/ QUOTE ]
Yes, but for bankroll (and ROR calcs) aren't we only interested in BB/100 as the units for SD and win rate?
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  #6  
Old 02-03-2005, 07:43 AM
pzhon pzhon is offline
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Default Re: Single 10/20 table vs. two 5/10 multitables

[ QUOTE ]

Yes, but for bankroll (and ROR calcs) aren't we only interested in BB/100 as the units for SD and win rate?

[/ QUOTE ]
There is no contradiction. For a particular comfort level, you might set your bankroll requirement at 3 SD^2/W. You should get essentially the same answer whether you compute the bankroll based on the win rate and standard deviation in an hour, a table-hour, or 100 hands.

For the example of a multitabling player winning $10/table-hour with a standard deviation of $100, the bankroll by the above formula would be 3 (100^2)/10 = $3000. By hour, the recommended bankroll would be 3 (140^2)/20 = $3000. If there are 80 hands at a table in each hour, then the recommended bankroll could be computed by 100 hand periods as 3 ( 112^2) / 12.5 = $3000.

When you increase the length of the period (or the number of tables) by a factor of n, the win rate increases by a factor of n, and the standard deviation increases by a factor of sqrt(n). The quotient SD^2/W doesn't change.
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  #7  
Old 02-03-2005, 07:51 AM
gaming_mouse gaming_mouse is offline
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Default Re: Single 10/20 table vs. two 5/10 multitables

Thanks pzhon.

That was my understanding. Which must mean that the swings I've seen multi-tabling just happen to be larger by chance.

gm
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  #8  
Old 02-03-2005, 10:32 AM
Lost Wages Lost Wages is offline
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Default Re: Single 10/20 table vs. two 5/10 multitables

Or perhaps the assumption that your win rate is independent of the number of tables is incorrect.

Lost Wages
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  #9  
Old 02-03-2005, 10:44 AM
lighterjobs lighterjobs is offline
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Default Re: Single 10/20 table vs. two 5/10 multitables

i would think the swings would be bigger playing two tables.
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  #10  
Old 02-03-2005, 12:07 PM
Duke Duke is offline
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Default Re: Single 10/20 table vs. two 5/10 multitables

[ QUOTE ]
i would think the swings would be bigger playing two tables.

[/ QUOTE ]

No reason the variance would increase.

Though yes, you can lose with aces on 8 tables at once and lose more in "1 time-hand" than is possible in an hour of normal 1 table play.

Multi-tabling is nothing more than time compression.

~D
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