Where/Who the hell is Bryce Carlson?
- Thread starter jack.jackson
- Start date
If an ace is removed from a full shoe then of course the prob of all other ranks increases given no additional information.
If you are given that X number of cards remain and RC = Y and Z number of a specific rank have been removed then it's possible to compute the prob of drawing each rank. If an additional specific removal is made then the probabilities will change again. You can see the effect when 2 cards remain. Assume HiLo count = 0 with no specific removals. If an ace is specifically removed then 1 card remains, RC=-1, and then prob of drawing a 10 (single deck) goes from .343348 to 0.
If a 7,8,9 is removed then 1 card remains, RC=0, and prob of drawing a 7,8,9 goes from .0472103 (single deck) to 3/11 or 4/11 depending upon exactly which card is drawn.
If you make a response and I don't answer it's because it's time for me to leave for work.
Cheers
If you are given that X number of cards remain and RC = Y and Z number of a specific rank have been removed then it's possible to compute the prob of drawing each rank. If an additional specific removal is made then the probabilities will change again. You can see the effect when 2 cards remain. Assume HiLo count = 0 with no specific removals. If an ace is specifically removed then 1 card remains, RC=-1, and then prob of drawing a 10 (single deck) goes from .343348 to 0.
If a 7,8,9 is removed then 1 card remains, RC=0, and prob of drawing a 7,8,9 goes from .0472103 (single deck) to 3/11 or 4/11 depending upon exactly which card is drawn.
If you make a response and I don't answer it's because it's time for me to leave for work.
Cheers
cool
That's what we were discussing
Probability of 10 goes down from 0 to 0-which is same.
Note: 2cards remaing HLRC=0, one is known to be an ACE. Because in our example, we are only considering subsets-which has atleast one ace(we need to remove it). So, we know that there are no tens. So, prob(T) was same as before =0 in this case. Chap was arguing over the above point which was obvious to you::
He was supporing his assumption with the following point::
"After removing an Ace, HI-lo RC=-1, hence probability of Ten decreased compared to before removal of ace as RC=0."
which I think is wrong and I think you have just agreed...
That's what we were discussing
Probability of 10 goes down from 0 to 0-which is same.
Note: 2cards remaing HLRC=0, one is known to be an ACE. Because in our example, we are only considering subsets-which has atleast one ace(we need to remove it). So, we know that there are no tens. So, prob(T) was same as before =0 in this case. Chap was arguing over the above point which was obvious to you::
He was supporing his assumption with the following point::
"After removing an Ace, HI-lo RC=-1, hence probability of Ten decreased compared to before removal of ace as RC=0."
which I think is wrong and I think you have just agreed...
NightStalker said:
That's what we were discussing
Probability of 10 goes down from 0 to 0-which is same.
Note: 2cards remaing HLRC=0, one is known to be an ACE. Because in our example, we are only considering subsets-which has atleast one ace(we need to remove it). So, we know that there are no tens. So, prob(T) was same as before =0 in this case. Chap was arguing over the above point which was obvious to you::
Probability of 10 goes down from 0 to 0-which is same.
Note: 2cards remaing HLRC=0, one is known to be an ACE. Because in our example, we are only considering subsets-which has atleast one ace(we need to remove it). So, we know that there are no tens. So, prob(T) was same as before =0 in this case. Chap was arguing over the above point which was obvious to you::
1) 1 low card, 1 high card - prob from single deck with no specific removals = 20/52*20/51*2 = 800/52/51
2) 2 medium cards - prob from single deck with no specific removals = 12/52*11/51 = 132/52/51
Relative weight subset 1 = 800/932
Relative weight subset 2 = 132/932
Prob low(subset 1) = 1/2, Prob med(subset 1) = 0, Prob high(subset 1) = 1/2
Prob low(subset 2) = 0, Prob med(subset 2) = 1, Prob high(subset 2) = 0
Overall prob low = 800/932*1/2 + 132/932*0 = 400/932
Overall prob med = 800/932*0 + 132/932*1 = 132/932
Overall prob high = 800/932*1/2 + 132/932*0 = 400/932
Prob(2) = 80/932 = .085837
Prob(3) = 80/932 = .085837
Prob(4) = 80/932 = .085837
Prob(5) = 80/932 = .085837
Prob(6) = 80/932 = .085837
Prob(7) = 44/932 = .047210
Prob(8) = 44/932 = .047210
Prob(9) = 44/932 = .047210
Prob(T) = 320/932 = .343348
Prob(A) = 80/932 = .085837
If one card is dealt and removed from the above 2 cards it is possible that it could be 2,3,4,5,6,7,8,9,T,A with the above probabilities.
If the dealt card is a 2,3,4,5,6 then the remaining card must be a high card and last card probs are:
Prob(2) through Prob(9) = 0
Prob(T) = .8, Prob(A) = .2
If the dealt card is a 7,8,9 then the remaining card must be a med card and last card probs are:
If dealt card is specifically a 7
Prob(7) = 3/11, prob(8) = 4/11, prob(9) = 4/11
If dealt card is specifically an 8
Prob(7) = 4/11, prob(8) = 3/11, prob(9) = 4/11
If dealt card is specifically a 9
Prob(7) = 4/11, prob(8) = 4/11, prob(9) = 3/11
If the dealt card is a T or A then the remaining card must be a low card and last card probs are
Prob(2) through Prob(6) = .2
Prob(7) through Prob(9) = 0
Prob(T) = 0
Prob(A) = 0
So as I said the probability of drawing a ten went from .343348 to 0 when an ace is drawn.
You can insist that there is one ace present but in doing that you are creating a compostion first and then figuring out what the count is later. Like I tried to show in my post with the 2 examples if you input composition first then removing 1 of a rank always increases probability of the other ranks on the next draw. This is how a combinatorial analyzer computes EVs and has nothing to do with counting and a count's running count is just an after thought. On the other hand if you want to determine how likely it is to draw a rank given only count parameters then you need to consider all of the possible subsets relative to a given counting system. Counting system math is not quite the same thing as normal combinatorial analysis math.
I don't know. :grin:
What you are doing is imposing an extra condition: shoe must contain an ace. OK fine. How about saying shoe consists of 26 cards and must contain 2 (A,2,3,4,5,6,7,8,9) and 8 tens. You are imposing an exact condition. In that case removing an ace always increases the prob of the other ranks. You are just simply inputting a composition which happens to have a HiLo RC=0. There is nothing wrong with this, by the way. It just has nothing to do with counting. A counter has less information to work with. A counter doing side counts may have additional information but unless he was able to track each and every card his information is less than knowledge of the exact current shoe composition.
Now say all you know is that 26 cards remain, HiLo RC=0, and nothing else (nothing specifically removed.) Is it possible to draw an ace knowing only this information?
Now say all you know is that 26 cards remain, HiLo RC=0, and nothing else (nothing specifically removed.) Is it possible to draw an ace knowing only this information?