That is why I bring out the question and hope to get the answer.
Just try to learn something from here.
Maybe someone is willing to give the answer.
Agree with you, this is a theoretical discussion, not much practical purposes.
7up
That is why I bring out the question and hope to get the answer.
Just try to learn something from here.
Maybe someone is willing to give the answer.
Agree with you, this is a theoretical discussion, not much practical purposes.
7up
> May be my former post wasn't clear enough.I understand
> you're investigating into specific situations and this
> is fine.But your method of removing the seven only
> after determining the subset doesn't reflect the
> proper probabilities.
> An extreme example will make this clear.Let's assume
> we have a deck consisting only of 2 sevens,2 tens and
> 2 sixes, the representative subset with 3 cards would
> be 1 seven,1 ten and 1 six.Now, if we remove a seven
> according to your method,the probability for one of
> the last 2 cards to be a seven would be 0 and this is
> obviously wrong.In fact there is still a 40% chance
> that one of the 2 remaining cards will be a seven(A
> seven is missing from the original deck,so the
> probability for each of the other 5 cards to be the
> remaining 7 is 1/5).
> Card removals for known dealer or player cards must be
> performed on the original decck with a much more
> limited effect on resulting subsets.If you remove the
> cards on the subsets, you get biased results.
> Francis Salmon
It's not important but that's not quite what I meant. Using your example, I just assumed player already had a hand of 16 and that hand was already included in the intitial shoe count. Then a 7 was removed for up card, which would leave 1 remaining in your example. I acknowledge that this is not the best analysis, but I did it to better reference the shoe comp listed in 7up's initial post. btw, your figures for an infinite deck for 16 vs 7 in your other post are correct :-).
kc
> who is willing to put penetration as a factor in your
> sim.
> Could you give me the figures in this case:
> 6 deck, S17, 3 cards surrender
> Dealer 7, after doubling, I got 13, now I have to
> decide whether to surrender or not.
> I guess I will get different TC values to surrender
> against dealer??s 7, on 208 and 52 cards penetrations.
> Thanks
> 7up
This is a very difficult question to answer without doing a sim. My method is not a sim. It is a mathematical way of computing a shoe comp by getting a weighted average of all possible subsets using HiLo for a given run count/pen/removal combo. It is well suited to analyzing insurance because all that is needed to know whether to insure or not is a computation of the percentage of tens remaining in the shoe. It's not well suited to do an organized analysis of strategy indices.
Sorry I can't be of more help.
kc
> Using your example, I just assumed player already had
> a hand of 16 and that hand was already included in the
> intitial shoe count.
That's fine with me and it makes your calculation fit also for multicard 16s.
> Then a 7 was removed for up card,
> which would leave 1 remaining in your example.
No, with your method of removing the 7 on the resulting subset of only three cards, there will be no 7 left.
> acknowledge that this is not the best analysis, but I
> did it to better reference the shoe comp listed in
> 7up's initial post. btw, your figures for an infinite
> deck for 16 vs 7 in your other post are correct :-).
The best analysis would be to account for the dealer upcard as 1 seven missing out of 24 which would result in a probability for the seven of 23/311 =7.4%.This contrasts with your 3/51=5.9%.You did the same thing also with the 10 and the 6 in your last calculation for 10,6 v 7 which made the error even worse.
I'm sure that with correct removal practice, the results will only differ slightly from the infinite deck method.
Francis Salmon
> That's fine with me and it makes your calculation fit
> also for multicard 16s.
> No, with your method of removing the 7 on the
> resulting subset of only three cards, there will be no
> 7 left.
> The best analysis would be to account for the dealer
> upcard as 1 seven missing out of 24 which would result
> in a probability for the seven of 23/311 =7.4%.This
> contrasts with your 3/51=5.9%.You did the same thing
> also with the 10 and the 6 in your last calculation
> for 10,6 v 7 which made the error even worse.
> I'm sure that with correct removal practice, the
> results will only differ slightly from the infinite
> deck method.
> Francis Salmon
This is the output of my finite deck CA for a deck comp of 3,3,3,3,3,4,4,4,20,5 for 10,6 vs 7. The computation is made after cards are dealt. Shoe comp at that point is 3,3,3,3,2,3,4,4,19,5. This is exact.
Number of Decks: 6
Shoe comp: (1) 5 (2) 3 (3) 3 (4) 3 (5) 3
(6) 2 (7) 3 (8) 4 (9) 4 (T) 19
Computing optimal EVs for an individual hand (input below)....
Double Rule: Double on any total, 2 cards only
Dealer stands on soft 17
Hole card rule: None (American rules)
No surrender allowed
Number of allowed splits vs. Ace: 1
Number of allowed splits vs. 2 - 10: 1
NDAS
Player is allowed one card to split ace
Optimal strategy (split strat is fixed optimally on first split hand)
Dealer up card: 7
Player Hand: 10,6....computing....please wait
Player Stand EV: -0.5055174698
Player Hit EV: -0.4748407516
Double EV on 16: -0.9496815033
kc
If your method of removing the dealer's upcard from the subset were correct, we would have to take insurance at a count of +2 with one deck to go because it would be the same situation as in single deck from the top.
But we know that the minimum requirement for insurance in a 6-deck-game is TC+3 at any point in the shoe.
Francis Salmon
> If your method of removing the dealer's upcard from
> the subset were correct, we would have to take
> insurance at a count of +2 with one deck to go because
> it would be the same situation as in single deck from
> the top.
> But we know that the minimum requirement for insurance
> in a 6-deck-game is TC+3 at any point in the shoe.
If pen & hand comp are taken into account, it may be best to insure at a TC < 3 if hand is dealt from 6 decks. For example a hand of A,A vs A is generally insurable at a TC < 3 (but no less than 2.39 depending on pen.) I'm not advocating that a person do this, only that it would be correct. If the game is dealt from fewer decks, the greater the effect of hand comp. For example a hand of A,A vs A dealt from a single deck is insurable at a TC of -2!
A characteristic of an infinite pack is that hand comp has no effect on outcome. This is because if you remove any number of a rank, it is still a trivial number compared to infinity. Below is the output from my finite deck CA for 2,000,000 decks. I used the exact same removal method as before. The starting pack was obtained by removing 5,000,000 each (2-6), 4,000,000 each (7-9), 12,000,000 tens, and 3,000,000 aces. The resulting shoe comp is 3M,3M,3M,3M,3M,4M,4M,4M,20M,5M. Total cards = 52M. Running count = 25M-15M = +10M. True count = +10. From this shoe 1 ten, 1 six, and 1 seven were removed to allow for a player hand of hard 16 vs 7 just as before. It's the identical situation as before except there are 1M times more cards. As you can see, the EVs approach the EVs for an infinite deck. Hand comp does have some effect on EVs. As more and more decks are added, the measurable effect approaches 0. The figures just are what they are.
Number of Decks: 2000000
Shoe comp: (1) 5000000 (2) 3000000 (3) 3000000 (4) 3000000 (5) 3000000
(6) 2999999 (7) 3999999 (8) 4000000 (9) 4000000 (T) 19999999
Computing optimal EVs for an individual hand (shown below)....
Double Rule: Double on any total, 2 cards only
Dealer stands on soft 17
Hole card rule: None (American rules)
No surrender allowed
Number of allowed splits vs. Ace: 1
Number of allowed splits vs. 2 - 10: 1
NDAS
Player is allowed one card to split ace
Optimal strategy (split strat is fixed optimally on first split hand)
Dealer up card: 7
Player Hand: 10,6....computing....please wait
Player Stand EV: -0.4908872645
Player Hit EV: -0.5086332809
Double EV on 16: -1.017266562
I hope this helps.
kc
[i]If pen & hand comp are taken into account, it may be best to insure at a TC < 3 if hand is dealt from 6 decks. For example a hand of A,A vs A is generally insurable at a TC < 3 (but no less than 2.39 depending on pen.) I'm not advocating that a person do this, only that it would be correct.[i]
Agreed! But we arrive at this result only if we use the proper removal practice(from the original pack and not from the subset!).Otherwise with one deck to go, we would get an index of -2 as in your single deck example below.
Francis Salmon
[i] If the game is dealt from fewer decks, the greater the effect of hand comp. For example a hand of A,A vs A dealt from a single deck is insurable at a TC of -2![i]
A
> If pen & hand comp are taken into account, it may
> be best to insure at a TC Agreed! But we arrive at
> this result only if we use the proper removal
> practice(from the original pack and not from the
> subset!).Otherwise with one deck to go, we would get
> an index of -2 as in your single deck example below.
> Francis Salmon
> If the game is dealt from fewer decks, the greater
> the effect of hand comp. For example a hand of A,A vs
> A dealt from a single deck is insurable at a TC of -2!
> A
I don't think we have a disagreement. I am making the assumption that a given count/pen can be translated into a shoe comp. My method does that. Luckily, the example 7up listed resulted in a comp where all ranks were very nearly integers. However, this doesn't necessarily have to be the case. That is one reason why I say it would be difficult to apply my method to analyzing strategy indices. Anyway, any shoe comp can be handled by my CA and is handled exactly the same as a full shoe. All I did was apply what a finite CA does to a computed shoe comp. I was not redefining my starting point, which is a full 6 deck shoe.
Now one of the subsets that goes into the computation has no sevens, eights, and nines so it would be impossible to have a hand of 10,6 vs 7 in that subset so I may be fudging in that respect. This may be another problem with using what I do, in conjunction with a CA, to analyze strategy indices.
Thanks for your responses.
kc
> If pen & hand comp are taken into account, it may
> be best to insure at a TC Agreed! But we arrive at
> this result only if we use the proper removal
> practice(from the original pack and not from the
> subset!).Otherwise with one deck to go, we would get
> an index of -2 as in your single deck example below.
> Francis Salmon
> If the game is dealt from fewer decks, the greater
> the effect of hand comp. For example a hand of A,A vs
> A dealt from a single deck is insurable at a TC of -2!
> A
...given that the hand is 10,6 vs 7, what are the EVs? With 52 cards left and 10,6,7 removed, my method gives the folowing comp:
Cards in deck = 52
p(2) = 5.85731783266168E-2
p(3) = 5.85731783266168E-2
p(4) = 5.85731783266168E-2
p(5) = 5.85731783266168E-2
p(6) = 5.61326292296745E-2
p(7) = 7.34839059191627E-2
p(8) = 7.66788583504306E-2
p(9) = 7.66788583504306E-2
p(10) = 3.8537511185012E-1
p(1) = 9.73579229937145E-2
.
This translates to a shoe comp of 3.04,3.04,3.04,3.04,2.91,3.82,3.99,3.99,20.03,5.06 or rounded to integers 3,3,3,3,3,4,4,4,20,5. Given player's hand is hard 16 vs 7:
Hit EV = -.5037
Stand EV = -.4875
.
The way I was looking at it was at the time the player made his bet; 52 cards left, count = +10, no specific removals, hand has yet to be dealt, computed comp is 3,3,3,3,3,4,4,4,20,5, then cards are dealt and removed, player hand could be anything that is possible given the computed comp, player makes his hit/stand decision.
I think this should clear it up.
kc
>
> ...given that the hand is 10,6 vs 7, what are the EVs?
> With 52 cards left and 10,6,7 removed, my method gives
> the folowing comp:
> Cards in deck = 52
> p(2) = 5.85731783266168E-2
> p(3) = 5.85731783266168E-2
> p(4) = 5.85731783266168E-2
> p(5) = 5.85731783266168E-2
> p(6) = 5.61326292296745E-2
> p(7) = 7.34839059191627E-2
> p(8) = 7.66788583504306E-2
> p(9) = 7.66788583504306E-2
> p(10) = 3.8537511185012E-1
> p(1) = 9.73579229937145E-2
> .
> This translates to a shoe comp of
> 3.04,3.04,3.04,3.04,2.91,3.82,3.99,3.99,20.03,5.06 or
> rounded to integers 3,3,3,3,3,4,4,4,20,5. Given
> player's hand is hard 16 vs 7:
> Hit EV = -.5037
> Stand EV = -.4875
Since we already know that a 10,6 and 7 has been dealt (or will be dealt)the composition of the most probable subset is altered in the way you described and that is exactly the situation we are facing when we have to make a playing decision.
Note that the EV-situation is now in line with known hit/stand indices.Keep up the good work!
Francis Salmon
> Since we already know that a 10,6 and 7 has been dealt
> (or will be dealt)the composition of the most probable
> subset is altered in the way you described and that is
> exactly the situation we are facing when we have to
> make a playing decision.
> Note that the EV-situation is now in line with known
> hit/stand indices.Keep up the good work!
> Francis Salmon
Thanks for the compliment. However, I still feel that the method would need more work (possibly more than could realistically be done) for better count/pen/hit-stand EV calculations for reasons stated in my other posts. Thanks again for your responses.
kc
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