[Phil]

Need some help if someone is willing to do it:

Run a Monte Carlo sim on the following:

6 player Blackjack table. Players 1-5 are dealt a 12 to 16 (but not 77 or 88). Player 6 is dealt a HARD (edit) 16 (not 88).

6 deck shoe.

Dealer shows a 6.

Calculate the table winnings for the table if:

A) Player 6 hits
B) Player 6 stands

Repeat 1,000 times

Obviously this to either prove/debunk the idea that Player 6's strategy can affect the profitability of the entire table.

There's a small bet on this, so Ill play a couple of speedballs with whomever puts this together.

[TylerE]

I might actually have a go at this later.

I suspect 1,000 trials isn't near enough for a low edge game like Blackjack though. 1 million is probably more like it.

Just to make sure I understand the conditions...

Player 6 is last to act?

How deep is shoe penetration?

Rules in play? Doubling down allowed? Players 1-5 assumed to play standard basic strategy? Dealer hits soft 17?

[TylerE]

The problem with this, as I see it, is that's it's impossible. You either are simulating whole shoes, and thus might see an effect, or you're constraining the hands. I don't think you can do both.

[Phil]

 TylerE, on 2017-January-03, 16:35, said:

The problem with this, as I see it, is that's it's impossible. You either are simulating whole shoes, and thus might see an effect, or you're constraining the hands. I don't think you can do both.

Why not? Its to sim for a specific use case.

I do agree 1,000 probably isnt sufficient.

[hrothgar]

Why is this necessary?

I would expect the answer to be obvious...

[Kaitlyn S]

 hrothgar, on 2017-January-03, 17:15, said:

Why is this necessary?

I would expect the answer to be obvious...

I would too. I'm surprised there is two sides to this bet.

[hrothgar]

In all seriousness, if I am understanding the problem properly, the coding and the result are both trivial.

1. By the time that player 6 has chosen whether to hit or to stand, players 1-5 have already committed to their line of player.
Presumably they are all standing (basic strategy says that you stand with 12-16)

2. We now move to player 6

Depending on player 6's decision, dealer will either receive the next card in the shoe or the second card in the shoe.

Player 6's decision whether or not to draw will determine which card dealer receives, but it will not change the expectation of the result...

(You might be able to force a slight change in the odds if player six continues to draw with a soft 17, followed by a soft 18, followed by a soft 19)

[mikeh]

 hrothgar, on 2017-January-03, 18:22, said:

In all seriousness, if I am understanding the problem properly, the coding and the result are both trivial.

1. By the time that player 6 has chosen whether to hit or to stand, players 1-5 have already committed to their line of player.
Presumably they are all standing (basic strategy says that you stand with 12-16)

2. We now move to player 6

Depending on player 6's decision, dealer will either receive the next card in the shoe or the second card in the shoe.

Player 6's decision whether or not to draw will determine which card dealer receives, but it will not change the expectation of the result...

(You might be able to force a slight change in the odds if player six continues to draw with a soft 17, followed by a soft 18, followed by a soft 19)

Your qualification is based, it seems to me, on 'knowing' which card is next in the shoe, and second, etc. The whole point is that player 6's decision cannot affect the probabilities of the outcome unless and until one specifies the value of the card next to be played. Knowing that value changes the odds for the remaining cards in the shoe. For example, if it were a face card, then that slightly reduces the expectation that dealer will go bust with the next two cards, and so on.

Until the card is dealt, it is a variant of Schrodinger's cat. One may be able to calculate the relative likelihood of it being a particular value, based on the known cards played, but it is in precisely the same state of uncertainty as every other card in the shoe. Only when the card is observed do we know what it is, and only then can 'what it is' be used to reassess the remaining cats/cards.

Ok, I do actually know that the card isn't actually collapsing into its value only on exiting the shoe But how often do we get to (mis)use quantum theory in a card game?

[hrothgar]

 mikeh, on 2017-January-03, 18:37, said:

Your qualification is based, it seems to me, on 'knowing' which card is next in the shoe, and second, etc. The whole point is that player 6's decision cannot affect the probabilities of the outcome unless and until one specifies the value of the card next to be played. Knowing that value changes the odds for the remaining cards in the shoe. For example, if it were a face card, then that slightly reduces the expectation that dealer will go bust with the next two cards, and so on.

Until the card is dealt, it is a variant of Schrodinger's cat. One may be able to calculate the relative likelihood of it being a particular value, based on the known cards played, but it is in precisely the same state of uncertainty as every other card in the shoe. Only when the card is observed do we know what it is, and only then can 'what it is' be used to reassess the remaining cats/cards.

Ok, I do actually know that the card isn't actually collapsing into its value only on exiting the shoe But how often do we get to (mis)use quantum theory in a card game?

I'm glad that you added that last sentence. I was very worried for a bit.

[Kaitlyn S]

 hrothgar, on 2017-January-03, 18:22, said:

In all seriousness, if I am understanding the problem properly, the coding and the result are both trivial.

1. By the time that player 6 has chosen whether to hit or to stand, players 1-5 have already committed to their line of player.
Presumably they are all standing (basic strategy says that you stand with 12-16)

2. We now move to player 6

Depending on player 6's decision, dealer will either receive the next card in the shoe or the second card in the shoe.

Player 6's decision whether or not to draw will determine which card dealer receives, but it will not change the expectation of the result...

(You might be able to force a slight change in the odds if player six continues to draw with a soft 17, followed by a soft 18, followed by a soft 19)

You might not also.

For this to make any sense, you would have had to draw to A-5. Say the expectation for each other player before the draw was X.

Now player 6 draws an ace giving him A-5-A for a soft 17. Now the new expectation for each other player is Y, but that Y (along with all the other cards player 6 could have drawn) is factored into that X.

Now let's say that player 6 hits his soft 17, planning to stop whatever he draws, and we don't see his card. The dealer takes the next card(s) and the expectation hasn't changed; it is still Y. So whether player 6 hits his 17 or not, the expectation Y doesn't change, and therefore the original expectation X doesn't change.

I could carry out the same thought process no matter how many times you want to hit player 6's hand.

A couple of side notes:

(a) This isn't helping to keep my Water Cooler posts down, and

(b) Phil's "reward" seems like a no-win situation - if the recipient does well, Phil carried him, but if the recipient does poorly, he's deemed to be so bad that not even Phil can carry him.

[hrothgar]

 Kaitlyn S, on 2017-January-03, 19:01, said:

You might not also.

For this to make any sense, you would have had to draw to A-5. Say the expectation for each other player before the draw was X.

Now player 6 draws an ace giving him A-5-A for a soft 17. Now the new expectation for each other player is Y, but that Y (along with all the other cards player 6 could have drawn) is factored into that X.

Now let's say that player 6 hits his soft 17, planning to stop whatever he draws, and we don't see his card. The dealer takes the next card(s) and the expectation hasn't changed; it is still Y. So whether player 6 hits his 17 or not, the expectation Y doesn't change, and therefore the original expectation X doesn't change.

I could carry out the same thought process no matter how many times you want to hit player 6's hand.

A couple of side notes:

(a) This isn't helping to keep my Water Cooler posts down, and

(b) Phil's "reward" seems like a no-win situation - if the recipient does well, Phil carried him, but if the recipient does poorly, he's deemed to be so bad that not even Phil can carry him.

Player 6 is specified to be drawing to both a hard and a soft 16.
Player 6 is already departing from "rational play" so there is no knowing what they "might" do.

Here's where life gets complicated:

The only time that player 6 is able to make multiple draws is if he is repeatedly drawing low cards
If player is able to make more than one draw then we're in the world of conditional probability and the odds are going to start shifting...

[Kaitlyn S]

 hrothgar, on 2017-January-03, 19:13, said:

The only time that player 6 is able to make multiple draws is if he is repeatedly drawing low cards
If player is able to make more than one draw then we're in the world of conditional probability and the odds are going to start shifting...

To say that the odds can shift is essentially saying that Player 6 can choose to increase the odds of his accomplices in seats 1-5, or choose not to decrease their odds, which would seem to be quite an advantage to the coalition of players, especially if players 1-5 have huge bets and player 6 has bet a dollar.

This seems hard to believe.

 Phil, on 2017-January-03, 15:12, said:

Obviously this to either prove/debunk the idea that Player 6's strategy can affect the profitability of the entire table.

But wait! Your friend might win this one. (Or you might if you pulled off the word play scam.) Player 6 is part of the table, so if Player 6 plays sub-optimally, his own expectation is decreased, and everyone else's is unchanged, so yes indeed, the profiability of the entire table is decreased.

Unless you go even one step further in the word-play scam and say the dealer is part of the table so it's a zero sum game and the profit for the entire table (all the players and the dealer) is zero by default.

[hrothgar]

 Kaitlyn S, on 2017-January-03, 19:22, said:

To say that the odds can shift is essentially saying that Player 6 can choose to increase the odds of his accomplices in seats 1-5, or choose not to decrease their odds, which would seem to be quite an advantage to the coalition of players, especially if players 1-5 have huge bets and player 6 has bet a dollar.

This seems hard to believe.

But wait! Your friend might win this one. (Or you might if you pulled off the word play scam.) Player 6 is part of the table, so if Player 6 plays sub-optimally, his own expectation is decreased, and everyone else's is unchanged, so yes indeed, the profiability of the entire table is decreased.

Unless you go even one step further in the word-play scam and say the dealer is part of the table so it's a zero sum game and the profit for the entire table (all the players and the dealer) is zero by default.

Please note: Everything that I discussed is really at the margin...

I suspect that radically suboptimal behavior could shift the odds, but the amount would probably be negligible.

[WellSpyder]

I'm not sure I have fully understood the question, so let me answer a simpler question first. What is the probability that the second card in a standard pack is an ace? Well, obviously it depends on what the first card is, since if this is an ace then it lowers the probability that the second card is. But if you multiply together the probabilities for the first card by the conditional probabilities for the second card and add the results together then I think it is obvious that you will just get the same 1 in 13 probability that you would get if you just looked at the second card on its own.

Is this any different in principle from the question we are being asked here?

[Phil]

 hrothgar, on 2017-January-03, 19:13, said:

Player 6 is specified to be drawing to both a hard and a soft 16.

Sorry, the intent was just one card, so only a hard 16. OP edited.

[Phil]

 Kaitlyn S, on 2017-January-03, 17:40, said:

I would too. I'm surprised there is two sides to this bet.

Yes, me too. The cards coming out of the shoe are not deterministic and its a fallacy to think 6's play matters.

I compared the situation to 3rd seat making a sub-optimal play - this would also affect dealer's card - and would affect the result of 1 and 2.

As far as the table's 'profitability' I'm certainly not interested in the dealer because then its a zero sum game. Im not that concerned about Player 6 either because its his play that allegedly is altering the outcome of the other 5.

[Kaitlyn S]

 Phil, on 2017-January-04, 08:03, said:

As far as the table's 'profitability' I'm certainly not interested in the dealer because then its a zero sum game. Im not that concerned about Player 6 either because its his play that allegedly is altering the outcome of the other 5.

You may be looking at the situation sensibly but the reason that your friend might have made such an absurd bet is that he intends to play word games to win - he may agree with you but thinks he tricked you to fall for a sucker's scam.

[Phil]

 Kaitlyn S, on 2017-January-04, 08:13, said:

You may be looking at the situation sensibly but the reason that your friend might have made such an absurd bet is that he intends to play word games to win - he may agree with you but thinks he tricked you to fall for a sucker's scam.

No, he genuinely thinks the strategy of the last players affects the other 5.