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?

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.

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.

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.Quote:MichaelBluejayI'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?

link to original post

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.

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

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.Quote:MentalAt 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.

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.

UTH and betting progressions are also examples where more than one unit is bet.

When the (0 - 1) is not the right value to use in the expectation value calculation, your formulation breaks down.

Any game, including blackjack and craps, can be easily expressed in a binary “p(lose 1) or (1 - p) win x” format exactly matching the variance and edge of the actual game. I believe this is what OP is attempting