3rd grade math....

stupid101 · #1 06-21-2005, 09:26 AM

I aint the brightest in the world, hence the name. I have 2 computers with 2 PT databases. If I have 1000 hands on 1 computer and 3000 hands on the other computer, what is the formula to combine the hands and figure out bb/100 using 30-60 as the big bet.

kiddj · #2 06-21-2005, 09:36 AM

BB won on #1 + BB won on #2, divide by 40.

stupid101 · #3 06-21-2005, 10:16 AM

Sorry I dont understand...
If i have 4257 hands on comp 1 with +9796 on 1 comp 3.84 bb/100, and (-3741) on 1 comp with 6232 hands (-1.00)bb/100. Using 30-60 as the bb per hand what is the formula for that. could u use the formula exactly how u would figure it out.
ty in advance

kiddj · #4 06-21-2005, 10:30 AM

You play 30/60 and you can't figure this out?????

Are the different win/losses from the different computers at different limits?

Give me:
Computer 1 - amt won, limit, # hands
Computer 2 - amt won, limit, # hands

stupid101 · #5 06-21-2005, 10:40 AM

4257 hands 9796 won 3.84 bb/100 30-60
6232 hands 6232 won -1.00 bb/100 30-60

can u explain the formula used

ty

kiddj · #6 06-21-2005, 10:43 AM

[ QUOTE ]
4257 hands 9796 won 3.84 bb/100 30-60
6232 hands 6232 won -1.00 bb/100 30-60

can u explain the formula used

ty

[/ QUOTE ]
How did you win 6232 @ -1BB/100?????????????

stupid101 · #7 06-21-2005, 10:49 AM

4257 hands 9796 won 3.84 bb/100 30-60
6232 hands -3741 won -1.00 bb/100 30-60

can u explain the formula used

ty

sorry

kiddj · #8 06-21-2005, 11:00 AM

[ QUOTE ]
4257 hands 9796 won 3.84 bb/100 30-60
6232 hands -3741 won -1.00 bb/100 30-60

can u explain the formula used

ty

sorry

[/ QUOTE ]
Total won on #1 = a
Total won on #2 = b
Total hands on #1 = c
Total hands on #2 = d
Total BB won = (a+b)/60
# of 100 hands = (c+d)/100

Formula = [(a+b)/60]/[(c+d)/100]
Solution = [(9796 - 3741)/60]/[(4257 + 6232)/100]
= 0.962119045 BB/100

stupid101 · #9 06-21-2005, 11:03 AM

ty for ur time