# Why does same percentage of house edge earn the game less/more money?

Let's say I have a game (1000 slots roulette) with 60 slots that you can win with. So the probability of winning is 0.06;

Now to attract players, I have a discount system. That is if discount is 30%, you pay 1 dollar to play, you only need to pay 0.7 dollars. But if you win the payout is 10 dollars.

GAME A

So the expected house edge on $1 should be (1-D) - 60 / 1000 * P = (1 - 0.3) - 60 / 1000 * 10 = 0.1 GAME B I can modify the game to maintain the same house edge, with different discount and payout D = 0.24, P = 11 (1-D) - 60 / 1000 * P = (1 - 0.24) - 60 / 1000 * 11 = 0.1 So from player point of view both games are the same (probability wise). But if a player can choose he probably will choose to play Game B. Why? Player 1 play Game A for 1320, after discount: 924, potential winning: 13200, Player 2 play Game B for 1200, after discount: 912, potential winning: 13200, As you can see player 2 is better off, same potential winning, but he pays lesser! I cannot figure out what causes this inconsistency when the house edge is obviously the same, it seems one game earn more money than the other. ## 1 Answer The problem is in calculating the edge of the house. In the first case, the house is expected to make \$0.1 for every \$0.7 bet, or approximately 14.2% of the initial bet. In the second case, the house is expected to make \$0.1 for every \$0.76 bet, or approximately 13.2% of the initial bet. Indeed, from the gambler's perspective, the second scenario is the most interesting. Note that you can take this a few steps further, by considering no discount and a payout of \$15. In this case, the edge of the house is \$0.1 for every \$1 bet, or 10% of the initial bet. From the gambler's perspective, this solution is the best approach.

• Thank you. I see my mistake now, so do the house edge formula above mean anything at all? – Zanko Feb 3 at 12:25
• @Zanko They certainly do: they determine the expected profit given an initial bet. As stated above, both the \$0.7 and \$0.76 bet respectively result in a profit of $0.1. – jvdhooft Feb 3 at 12:40 • Hi, sorry but something still doesn't add up for me.$0.1 for every $0.7 bet, or approximately 14.2% vs$0.1 for every $0.76 bet, or approximately 13.2%. Obviously house has better edge for the first case. But looking at expected profit given an initial bet both are$0.1. So both games are actually the same? – Zanko Feb 3 at 12:48
• @Zanko No, the expected profit is not the same. In the first case, the house wins $\frac{1}{7}$ of the amount bet, in the second case $\frac{1}{7.6}$ of the amount bet. Say a gambler bets \$532 in both scenarios, then the house is expected to win \$76 in the first scenario, and \\$70 in the second. – jvdhooft Feb 3 at 13:04
• Thank you I understand that. I guess I just can't figure out what (1-D) - 60 / 1000 * P = (1 - 0.3) - 60 / 1000 * 10 = 0.1 and (1-D) - 60 / 1000 * P = (1 - 0.24) - 60 / 1000 * 11 = 0.1 meant. They both give value of 0.1. It is expected profit GIVEN initial bet. But this information is useless to casino right? – Zanko Feb 3 at 13:08