For instance, if I am a casino, and I have a game that pays out $100,000 on a $1 bet, but you have a 1 in 120,000 chance of winning, it is obvious that the house is much better odds in the long run.
What I am wondering is how the house odds (or gamblers odds) are calculated in such an example. Is it 60% for the house for example?
sandynlily on Jan 6, 2011 Said:
its a measurement that expresses on average the percentage over 100 they will make for every dollar placed in a bet.
You first figure the total dollar amount of all losses for every possible outcome
and deduct from it the total dollar amount of all wins for every possible outcome.
then divide it by the total dollar amount of all bets made for both wins and losses in all possible outcomes (the amounts placed initally as a bet.)
This is to calculate the “house advantage” if for some reason this works out to be a negetive number it means that in the particular game you are calulating, the player actually has an advantage instead of the house
in your example there is 119,999 in losses – 100,000 in wins (19,999) divided by 120,000 dollars bet over 120,000 trials at 1 dollar
19,999/120,000=0.166658 or a 16.6658% house advantage
LegFuJohnson on Jan 6, 2011 Said:
In your scenario, on average, if 120,000 people played the game, the casino would show a profit of $20,000.
I believe you’d divide 20000 into 120,000 and get a 16.67% house advantage.
I could be wrong, but that sounds about right for that game.