August 19th, 2020 at 7:35:40 AM
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I am trying to calculate player EV in Blackjack in an arbitrary set of cards (not in a depleted normal deck).

For instance you can have a deck that contains 18 Aces, 10 9s etc.

Let n be the number of cards in the set.

Lets put indexes to BJ values (2->0,3->1,4->2,…,T->8,A->9)

Then let's define: C (0<=i<10) as the number of cards with index i in the set.

Then let's define:

P = C/n

Of course: Sum(P, for 0<=i<10) = 1

If the probability of cards with BJ value of index i appearing.

Also let’s assume E is the EOR for 6 deck shoe of the BJ value of index i

Also let’s assume that b is the EV off the top.

Then the EV of a player facing this set of cards is:

EV = b + Sum{ -E*P, for 0<=i<10}

Is that calculation correct?

For instance you can have a deck that contains 18 Aces, 10 9s etc.

Let n be the number of cards in the set.

Lets put indexes to BJ values (2->0,3->1,4->2,…,T->8,A->9)

Then let's define: C (0<=i<10) as the number of cards with index i in the set.

Then let's define:

P = C/n

Of course: Sum(P, for 0<=i<10) = 1

If the probability of cards with BJ value of index i appearing.

Also let’s assume E is the EOR for 6 deck shoe of the BJ value of index i

Also let’s assume that b is the EV off the top.

Then the EV of a player facing this set of cards is:

EV = b + Sum{ -E*P, for 0<=i<10}

Is that calculation correct?

August 19th, 2020 at 9:04:52 PM
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CharlieGamer,

Your reasoning is not correct.

Here are some simple examples where it breaks down.

If the deck contains only 2's, the player will win 4 times his initial wager simply by splitting to 4 hands and then hitting each one until his total exceeds 18, because the dealer will always end with 18. This same strategy will also work for a deck composed of only 3's.

If the deck contains only 8's, again the player will win 4 times his initial wager by splitting to 4 hands and then standing on each 16, since the dealer is guaranteed to bust with 24.

Your proposed formula would predict that the player's EV is negative for each of these cases, when in fact it is +400%.

The problem is that the EOR values are simply linear approximations.

Hope this helps!

Dog Hand

Your reasoning is not correct.

Here are some simple examples where it breaks down.

If the deck contains only 2's, the player will win 4 times his initial wager simply by splitting to 4 hands and then hitting each one until his total exceeds 18, because the dealer will always end with 18. This same strategy will also work for a deck composed of only 3's.

If the deck contains only 8's, again the player will win 4 times his initial wager by splitting to 4 hands and then standing on each 16, since the dealer is guaranteed to bust with 24.

Your proposed formula would predict that the player's EV is negative for each of these cases, when in fact it is +400%.

The problem is that the EOR values are simply linear approximations.

Hope this helps!

Dog Hand

August 19th, 2020 at 11:22:02 PM
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If you have a weird pack the quickest way to get an approximation is to create a spreadsheet using infinite deck and working out the best strategy. This usually gets close to the composition dependent figure for finite decks which is probably good enough for the first stage.

As has been suggested with just 3s and 2s that method will suggest the EV for being on 21 or 20 is 1, so you split and hit until then. There have been Blackjack games with multiple Aces (Californian Blackjack) but check as sometimes AA is the best hand for these.

Also your EORs method might not work, for instance with fewer 3s in the pack you might split 3s vs 8 (or other similar single deck strategies).

As has been suggested with just 3s and 2s that method will suggest the EV for being on 21 or 20 is 1, so you split and hit until then. There have been Blackjack games with multiple Aces (Californian Blackjack) but check as sometimes AA is the best hand for these.

Also your EORs method might not work, for instance with fewer 3s in the pack you might split 3s vs 8 (or other similar single deck strategies).

August 20th, 2020 at 5:35:04 PM
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Hey,

Your answer is correct. I forgot to mention that the deviations will not be so extreme so as to be concerned by non-linearity.

Thanks anyway.

Your answer is correct. I forgot to mention that the deviations will not be so extreme so as to be concerned by non-linearity.

Thanks anyway.

August 20th, 2020 at 8:19:11 PM
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Over the long term, Blackjack is statistically congruent to a bet with a 0.43 chance of winning that pays 1.31 to 1.

House edge 0.67%. Standard deviation 1.14

House edge 0.67%. Standard deviation 1.14

It’s all about making that GTA

August 20th, 2020 at 8:22:33 PM
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True to the second moment, but not beyond i hazard to guess.Quote:Ace2Over the long term, Blackjack is statistically congruent to a bet with a 0.43 chance of winning that pays 1.31 to 1.

House edge 0.67%. Standard deviation 1.14

The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

August 20th, 2020 at 8:35:03 PM
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It should work very well for large sample size within 4 SDs of the mean.Quote:unJonTrue to the second moment, but not beyond i hazard to guess.

It’s all about making that GTA

August 20th, 2020 at 8:37:21 PM
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Quote:Ace2It should work very well for large sample size within 4 SDs of the mean.

I just meant I would expect the skewness and kurtosis of the distributions would be a bit different.

The race is not always to the swift, nor the battle to the strong; but that is the way to bet.