Craps Bets Math Question

Here is the Wiz's explanation from WOO of the house edge on the iron cross bet. So my question is this: If as he says the house edge on this bet is 1.136 as he states and a person makes 1,000 bets of $22 covering this bet he would have bet $22,000. Does this mean that his expectation is to lose only 1.136% or $249.92? Or am I missing something? What am I missing?

The American Mensa Guide to Casino Gambling has the following "anything but seven" combination of craps bets that shows a net win on any number except 7. Here's how much MENSA advises to bet in the "Anything but 7" system:

5- place $5
6- place $6
8- place $6
field- $5
total= $22


They claim the house edge is 1.136%. How is that possible if every individual bet made has a higher house edge?
ANONYMOUS

"Good question. To confirm their math I made the following table, based on a field bet paying 3 to 1 on a 12. The lower right cell does shows an expected loss of 25 cents over $22 bet. So the house edge is indeed .25/22 = 1.136%.

The reason the overall house edge appears to be less than the house edge of each individual bet is because the house edge on place bets is generally measured as expected player loss per bet resolved.

However, in this case the player is only keeping the place bets up for one roll. This significantly reduces the house edge on the place bets from 4.00% to 1.11% on the 5 and 9, and from 1.52% to 0.46% on the 6 and 8."

Spending time calculating the edge on bet combinations: that way madess lies, as this all illustrates.

Especially true when alternating between 'bets per roll' and 'bets resolved'

Your post is not clear, but apparently at the end you are quoting the Mensa material. The Wizard gave a similar answer in an ask-the-w:

Quote: wiz, see link

why is this lower than the individual house edge of each bet made? It’s not. The reason it seems that way is the result of comparing apples to oranges. The house edge of place bets is usually expressed as the expected loss per bet resolved. Looking at the individual bets on a per-roll basis ... [blah, blah]

https://wizardofodds.com/ask-the-wizard/219/

TLDR: The place bets' house edges are typically (almost always) written as a "per resolved" wager. The "per resolved" wager HE method gives the 5 a HE of 4% (4 ways to win 7, 6 ways to lose 5: -2 out of $50 in action = 0.04 or 4%). On a per roll basis, there are only 10/36 ways for the bet to be resolved. 0.04 * 10 / 36 = 1.111%. Do the same process for the 6 and 8. Then multiply the wagers by their respective house edges to get the expected loss. Figure out what the expected loss is per total wager ($22). And bingo -- you've found out the combined HE for iron cross on a per roll basis.

Quote: RS

TLDR: The place bets' house edges are typically (almost always) written as a "per resolved" wager. The "per resolved" wager HE method gives the 5 a HE of 4% (4 ways to win 7, 6 ways to lose 5: -2 out of $50 in action = 0.04 or 4%). On a per roll basis, there are only 10/36 ways for the bet to be resolved. 0.04 * 10 / 36 = 1.111%. Do the same process for the 6 and 8. Then multiply the wagers by their respective house edges to get the expected loss. Figure out what the expected loss is per total wager ($22). And bingo -- you've found out the combined HE for iron cross on a per roll basis.

Thanks, got it. Should have thought of that before I posted. It's a high house edge bet. My math says the house edge on this bet is 4.54% if you make the bets that I stated in my OP. The "per roll" basis consideration doesn't mean that the player's expected loss is any less.