"Two G's" funny house edges

The Wizard presented another interesting example of how progressive betting does (or does not?) alter the house edge: "Two G's" as introduced at the Global Game Protection and Table Games Conference 2022.

The "Two G's" bet offers a win of 350 to 1.
The house edge for this bet is 2.77%, nearly as good as single zero roulette.


As the Wizard explains, a player could wager one dollar on the 0-00-split and after a hit could parlay that bet.
The win would be 323 to 1.


If the casino offered a win of 323 to 1, the house edge would be 10.25%.


QUESTION
If the player wagered the parlay as proposed by the Wizard, how would the house edge now be 5.26%, while the house, offering the same bet, would nearly have twice the edge worth 10.25%?

Quote: ThomasK

If the player wagered the parlay as proposed by the Wizard, how would the house edge now be 5.26%, while the house, offering the same bet, would nearly have twice the edge worth 10.25%?
link to original post

If the player bet a pair of numbers in roulette with the intent to parlay everything if he won the first bet, then if he won the second bet he would have 18*18=324 times as much money as he started. In other words, his original bet would have been paid 323 to 1.

The expected return of such a strategy is (2/38)*(2/38)*324 = 89.75%. In other words, a house edge of 10.25%.

The question asks how is this possible if the individual bets have a house edge of 5.26%.

The answer is that the player is betting more than 1 unit under this strategy. He is actually betting, on average, 1+(2/38)*18 = 1.9474 units.

The ratio of the expected loss to the expected bet is 0.1025/1.9474 = 5.26%.

*****





***** Does not apply to EvenBob

Quote: Wizard

Quote: ThomasK

If the player wagered the parlay as proposed by the Wizard, how would the house edge now be 5.26%, while the house, offering the same bet, would nearly have twice the edge worth 10.25%?
link to original post

If the player bet a pair of numbers in roulette with the intent to parlay everything if he won the first bet, then if he won the second bet he would have 18*18=324 times as much money as he started. In other words, his original bet would have been paid 323 to 1.

The expected return of such a strategy is (2/38)*(2/38)*324 = 89.75%. In other words, a house edge of 10.25%.

The question asks how is this possible if the individual bets have a house edge of 5.26%.

The answer is that the player is betting more than 1 unit under this strategy. He is actually betting, on average, 1+(2/38)*18 = 1.9474 units.

The ratio of the expected loss to the expected bet is 0.1025/1.9474 = 5.26%.
link to original post

Thanks for your reply.

Does the player actually wager 1.9474 units?

If the house offered the bet, there would be no intermediate wager of 18 units. Only the initial 1 unit would be at stake.

Just because the the offer is not made by the house, the player needs to use the intermediate chips in order to be eligible to participate in the second roll. These chips are of no value to the player as long as she decides to parlay (also see here). So the player's effective wager should remain the initial 1 unit bet and her house edge therefore should be 10.25%.

If, on the other hand, the player decided to spend the 17 units won for some amenities, she would not play the parlay and therefore her house edge on this first roll would be 5.26%.

The experiment I usually suggest in these cases is the following: The player hands the 1 unit to a friend who actually does the wagering at the table. The player herself stays away and only waits for the friend to return with either nothing (= 1 unit lost) or 324 units, i.e. the initial unit plus the 323 units won. The probability of this happening remains the (2/38)*(2/38) and leads to the 10.25% house edge.