Losing much more than HE
January 8th, 2024 at 5:12:28 AM
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The house edge found as a factor within the expected value of lilredrooster's scenario.
The ingredients are rather simple.
1) There is only a single outcome
Player loses eventually: -$10
2) There is only a single probability
Player loses eventually: 1.000000
Constraints.
The probability is 1 and therefore fullfills the constaint.
There is only one probability, this probability already is 1 and so a single hand of the game is sufficient to exactly meet this probability. Nevertheless the player may play many hundred thousands of hands of the game (wagering $10 each hand) in order to adhere to the constraint for the general case of an expected value.
Expected value.
The definition of the random variable for the scenarion would be
So the game represented by the random variable
X2 = Player starts betting $10 on PLAYER in Baccarat and keeps parlaying the PLAYER bet until he eventually loses
has a house edge of 100%
In other words:
The house edge defines the percentage the player loses on average from every dollar bet in the long run.
The scenario dictates that the player loses all eventually, which corresponds to 100%.
In turn the house receives all from the player, which also corresponds to 100%.
The ingredients are rather simple.
1) There is only a single outcome
Player loses eventually: -$10
2) There is only a single probability
Player loses eventually: 1.000000
Constraints.
The probability is 1 and therefore fullfills the constaint.
There is only one probability, this probability already is 1 and so a single hand of the game is sufficient to exactly meet this probability. Nevertheless the player may play many hundred thousands of hands of the game (wagering $10 each hand) in order to adhere to the constraint for the general case of an expected value.
Expected value.
The definition of the random variable for the scenarion would be
X2 = Player starts betting $10 on PLAYER in Baccarat and
keeps parlaying the PLAYER bet until he eventually loses.
E(X2) = 1.000000 * (-$10) expressing as losses (and gains)
= 1.000000 * ($0 - $10) expressing as payouts and wager
= 1.000000 * (-$10) + 1.000000 * $0 separating wager parts and payouts
/ $0 \
= (-$10) * | 1.000000 * 1 - 1.000000 * --- | factoring out the wager completely
\ $10 /
= (-$10) * (1 - 0.000000) revealing the RTP
= 10 * (-$1) * 1.000000 "amount wagered" times "per dollar bet" times "house edge"
So the game represented by the random variable
X2 = Player starts betting $10 on PLAYER in Baccarat and keeps parlaying the PLAYER bet until he eventually loses
has a house edge of 100%
In other words:
The house edge defines the percentage the player loses on average from every dollar bet in the long run.
The scenario dictates that the player loses all eventually, which corresponds to 100%.
In turn the house receives all from the player, which also corresponds to 100%.
"When it comes to probability and statistics, intuition is a bad advisor. Don't speculate. Calculate." - a math textbook author (name not recalled)
