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Scardeycat
Scardeycat
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November 24th, 2019 at 5:16:39 AM permalink
I wonder if anybody has ever calculated this. Assume a head to head game and all cards are dealt face up, the player's 2 cards and the dealer's initial card(s). The player is then told exactly what the next sequence of cards in the shoe is and is allowed to come up with any strategy where he doesn't lose money, the dealer's hand obviously taking cards from where the player leaves off from the same sequence. Anybody ever worked out what percentage of the time it's impossible?

Respliltting, doubling, dealer hole card, dealer behaviour on S17 etc. can all be whatever is simplest.
Mission146
Mission146
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November 24th, 2019 at 5:37:23 AM permalink
I don’t understand what you mean, “Sequence,” do you mean the player gets to know the very next card or the next few cards?
Vultures can't be choosers.
Scardeycat
Scardeycat
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November 24th, 2019 at 6:09:00 AM permalink
I mean with omniscience which hands can you avoid losing, the sequence is just the known fixed order of the next cards in the shoe.

For example Dealer 53 Player Q9. With no knowledge of the shoe you'd stand. If you knew the shoe went 2, K, 7... though you'd realise that you should hit. If the shoe goes 3, 3, 7... the hand is lost. I'm curious about how many hands are impossible to salvage.
Last edited by: Scardeycat on Nov 24, 2019
gordonm888
gordonm888
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November 24th, 2019 at 7:58:10 AM permalink
It's an excellent question. I have never seen this calculated.

Sometimes, given the actual sequence of cards in the shoe, nothing you choose to do can make a difference in the outcome.

Example: (Assuming no surrender.)
T6 vs TT with 6,7,8,9 or T as the next card. Player always loses.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
Mission146
Mission146
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November 24th, 2019 at 8:51:21 AM permalink
Quote: Scardeycat

I mean with omniscience which hands can you avoid losing, the sequence is just the known fixed order of the next cards in the shoe.

For example Dealer 53 Player Q9. With no knowledge of the shoe you'd stand. If you knew the shoe went 2, K, 7... though you'd realise that you should hit. If the shoe goes 3, 3, 7... the hand is lost. I'm curious about how many hands are impossible to salvage.



I would have been interested if the question had been with the player only knowing the very next card. This particular, "Sequence," thing just seems like way too much work for a hypothetical.

With the player to know only the next card, the only situations he would be screwed are dealer BJ v. Anything Else as well as made dealer hand (I'd assume S17 just to make it easier) against a situation where the player could only bust.

Anyway, it is interesting, but the amount of work it would take me exceeds the level of interest this question has for me. Hopefully someone with programming skill comes along and can do this in the time it would take me to do the scenario I would have been willing to do.
Vultures can't be choosers.
IndyJeffrey
IndyJeffrey
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gordonm888
November 24th, 2019 at 8:56:00 AM permalink
Quote: Scardeycat

I wonder if anybody has ever calculated this. Assume a head to head game and all cards are dealt face up, the player's 2 cards and the dealer's initial card(s). The player is then told exactly what the next sequence of cards in the shoe is and is allowed to come up with any strategy where he doesn't lose money, the dealer's hand obviously taking cards from where the player leaves off from the same sequence. Anybody ever worked out what percentage of the time it's impossible?

Respliltting, doubling, dealer hole card, dealer behaviour on S17 etc. can all be whatever is simplest.



Would it be accurate to rephrase this question, "What percentage of time (i.e., hands) does a player have no chance of winning before the cards are dealt?"
michael99000
michael99000
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November 24th, 2019 at 10:21:39 AM permalink
Probably a tough question to calculate.


There’d be times for example where you’re dealt 17, you see next card is a 3, but you also see that by taking that card you cause the dealer to make 21 with his hand.
gordonm888
gordonm888
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November 24th, 2019 at 1:59:25 PM permalink
The mathematics on this is fascinating. It may sometimes involve partition numbers, of all things.

Assumptions: Infinite deck, no surrender, no splitting.

16 v 16 (player and dealer both have a hard 16)

The only sequences of cards for which the player must lose are:

A-5
2-5
3-5
4-5
2-4
3-4
A-2-5
A-3-5
A-2-4

That's it. Let's count Ace as a "1." The rule for 16 v 16 is that it must be a sequence where each number must be larger than the sum of the numbers that precede it and cannot be larger than 5.

Given an infinite deck, the frequency with which the player will encounter one of the above sequences and thus "must lose" is (6*13+30/13^3 = 0.036868.

What about 16 vs 20? With a dealer 20, we must only worry about what the player total would be. Player can win or push (not lose) if he gets a sequence of numbers that add up to either 4 or 5. These are:

A-A-A-A
A-A-2, A-2-A, 2-A-A
A-A-3,
A-3, 3-A
2-2, 2-3, 3-2
4
5

So, for 16 v 0, the player can NOT LOSE only with these sequences which have a frequency (given an infinite deck) of
1/13^4 + 4/13^3 +5/13^2 + 2/13 = 0.185288.

So the probability that player MUST LOSE with a 16 v 20 is 1- 0.185288 =0.814712.
**********
Trivially, for a case like 20 vs 20, 19 vs 19, 18 vs 18 and 17 vs 17 the player can always stand and thus has a 0% probability of a "MUST LOSE" sequence.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

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