MichaelBluejay
Joined: Sep 17, 2010
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March 1st, 2023 at 9:46:21 AM permalink
I'm having a hard time calculating the house edge in blackjack using the Wizard's data. I'm starting with the Wizard's excellent Blackjack variance article, which includes this table for a particular ruleset:

 Net Win Probability Return -8 0.00000019 -0.00000154 -7 0.00000235 -0.00001643 -6 0.00001785 -0.00010709 -5 0.00008947 -0.00044736 -4 0.00048248 -0.00192993 -3 0.00207909 -0.00623728 -2 0.04180923 -0.08361847 -1 0.40171191 -0.40171191 -0.5 0.04470705 -0.02235353 0 0.08483290 0.00000000 1 0.31697909 0.31697909 1.5 0.04529632 0.06794448 2 0.05844299 0.11688598 3 0.00259645 0.00778935 4 0.00076323 0.00305292 5 0.00014491 0.00072453 6 0.00003774 0.00022646 7 0.00000609 0.00004263 8 0.00000066 0.00000526 Total 1.00000000 -0.00277282

i.e., house edge of -0.277282%

I tried to use that data to calculate the RTP another way: \$ returned to player ÷ total amount bet. For each line where we didn't lose, I tally the \$ the player was paid, plus getting their non-losing bet back. First I'll show how that would look with a -simple, made-up game:

 Net Win] Probability Return -1 0.51 -0.51 +1 0.49 +0.49 Total 1 -0.02

i.e., 2% house edge.

We could get the same result by looking at only the non-losing bets, plus getting our money back on those bets:

 Net Win] Bet returned Raw return Probability Return +1 +1 2 0.49 0.98 Total 0.98

i.e., 98% of money returned to player.

But that method doesn't work with the blackjack data:

 Net win Bet returned Total Probability Return 0 1 1 0.08483290 0.08483290 1 1 2 0.31697909 0.63395818 1.5 1 2.5 0.04529632 0.11324080 2 1 3 0.05844299 0.17532897 3 1 4 0.00259645 0.01038580 4 1 5 0.00076323 0.00381615 5 1 6 0.00014491 0.00086946 6 1 7 0.00003774 0.00026418 7 1 8 0.00000609 0.00004872 8 1 9 0.00000066 0.00000594 Total 0.50910038 1.0228

For starters, the probability of having a non-losing hand is positive, so it seems that it's not possible that we're gonna see a house edge in this game. And indeed, the return shown is 1.0228, a 2.28% player edge.

I'm thinking that the reason for the discrepancy is that blackjack is special in that you can lose more than the original bet (e.g., a loss of -8 if you split 3x to 4 hands and double each one). Classic calculation for BJ house edge is relative to the original wager only, not any extra \$ you put down for doubles and splits. So, my conclusion is that my method would work for most other games, but not in a game where you can lose more than your original bet. Does that seem right?
Ace2
Joined: Oct 2, 2017
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March 1st, 2023 at 12:57:25 PM permalink
Seem right? Actually, that seems all wrong, excluding the unaltered data copied from Wizard
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MichaelBluejay
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March 1st, 2023 at 1:03:45 PM permalink
Okay then, care to explain?
gordonm888
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March 1st, 2023 at 1:30:51 PM permalink
Michael, I believe you have misunderstood Wizard's published table.

When there is net payout of +8, he is providing the probability or frequency with which the following has occured:
- player is dealt a pair that is split and resplit until he has four separate hands
- each of the four hands is then doubled (after split)
- each of the four doubled hands wins versus the dealer's hand
- your net outcome after this action is that you are +8 units wealthier than you were before the action

When there is a net payout of -8 units, the identical action has occured- you have split and resplit to 4 hands and doubled each of the four hands - but each of the four doubled hands has lost to the dealer's hand and you lost a total of 8 chips.

So, when you win or tie the value of the "net payout" already includes the return of your original wager.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
MichaelBluejay
Joined: Sep 17, 2010
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March 1st, 2023 at 1:52:49 PM permalink
I already understood all your bullet points, I'm not sure why you'd think I didn't.

But does the +8 really include the return of the original wager? If you make a bet and split to 4 hands, that's 4 units wagered. If you double each of those, you've now wagered 8 units. If you win all of those, then you're +8 from where you started. In addition to that +8, you also get back everything you put on the table (8 units).

As I said in my OP, with a similar table to the Wizard's for a simple game, I can look at the winning line(s) only to figure the return, but that doesn't work with the blackjack table, and I'm thinking the reason is that blackjack is special in that you can lose *more* than the original wager.
Mental
Joined: Dec 10, 2018
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March 1st, 2023 at 4:33:00 PM permalink
Quote: MichaelBluejay

I'm thinking that the reason for the discrepancy is that blackjack is special in that you can lose more than the original bet (e.g., a loss of -8 if you split 3x to 4 hands and double each one). Classic calculation for BJ house edge is relative to the original wager only, not any extra \$ you put down for doubles and splits. So, my conclusion is that my method would work for most other games, but not in a game where you can lose more than your original bet. Does that seem right?

You cannot determine the number of bets at risk from this table. Therefore, you cannot determine the net W/L divided by the money put at risk.

Just look at the line for 'Net Win' = zero. This represents a large number of hands where you simply push plus a large number of hands where you double and push or split and win one and lose one hand. Also, it includes a smaller number of hands where you spilt/double to four bet units and get back four units, etc.

At no place in this table does the Wizard provide information regarding these partial probabilities. I assume he calculated them separately in his work product, but just chose not to display them here. If you knew what the subfractions were, then you could calculate the denominator for ROI.

Two lines where you know the amount of money at risk are the +8 and -8 lines. These are purely the result of winning or losing all bets with 8 units bet. I think +7 and -7 are also pure. The other lines are a mixture of different numbers of bets at risk.
Last edited by: Mental on Mar 1, 2023
MichaelBluejay
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March 1st, 2023 at 4:54:47 PM permalink
Quote: Mental

You cannot determine the number of bets at risk from this table.

The amount put at risk is arbitrary. He's showing eight decimals. Okay, then \$100,000,000 @ \$1/round would cover the whole table.

Quote: Mental

At no place in this table does the Wizard provide information regarding these partial probabilities. I assume he calculated them separately in his work product, but just chose not to display them here.

How do you figure? My conclusion was the opposite: He *did* include those hands. e.g., Split once, one hand win, the other loses, that's a net of zero, showed on the Net Win = 0 line.

Though no one is addressing my conjecture directly, it seems more and more to me like the reason I can't add the positive lines (like I can with the other game) is because blackjack is screwy in that you can lose *more* than your original wager.
Mental
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March 1st, 2023 at 5:23:21 PM permalink
Yes, your example assumes that 1 unit is wagered and lost on all other outcomes not included in the 0.49 probability. This is why I would never think to use this shortcut in the calculation. This is clearly not a general rule for PDFs even if it is common in gambling situations.

UTH and betting progressions are also examples where more than one unit is bet.
Mental
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March 1st, 2023 at 6:02:53 PM permalink
Your formula is implicitly HE = 0.49 * (2-1) + (1 - 0.49) * (0 - 1) = 0.49 +.49 - 1 = -.02

When the (0 - 1) is not the right value to use in the expectation value calculation, your formulation breaks down.
Ace2
Joined: Oct 2, 2017