Mixing
Why does same percentage of house edge earn the game less/more money?
$begingroup$
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.
probability
asked Feb 3 at 11:29
ZankoZanko
1556
$endgroup$
add a comment |
$begingroup$
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.
probability
asked Feb 3 at 11:29
ZankoZanko
1556
$endgroup$
add a comment |
$begingroup$
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.
probability
asked Feb 3 at 11:29
ZankoZanko
1556
$endgroup$
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.
probability
probability
asked Feb 3 at 11:29
ZankoZanko
1556
asked Feb 3 at 11:29
ZankoZanko
1556
asked Feb 3 at 11:29
ZankoZanko
1556
asked Feb 3 at 11:29
ZankoZanko
1556
asked Feb 3 at 11:29
ZankoZanko
1556
1556
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Feb 3 at 12:03
jvdhooftjvdhooft
5,65961641
$endgroup$
$begingroup$
Thank you. I see my mistake now, so do the house edge formula above mean anything at all?
$endgroup$
– Zanko
Feb 3 at 12:25
$begingroup$
@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.
$endgroup$
– jvdhooft
Feb 3 at 12:40
$begingroup$
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?
$endgroup$
– Zanko
Feb 3 at 12:48
$begingroup$
@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.
$endgroup$
– jvdhooft
Feb 3 at 13:04
$begingroup$
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 = 0.1and(1-D) - 60 / 1000 * P = (1 - 0.24) - 60 / 1000 * 11 = 0.1meant. They both give value of 0.1. It is expected profit GIVEN initial bet. But this information is useless to casino right?
$endgroup$
– Zanko
Feb 3 at 13:08
|
show 3 more comments
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1 Answer
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1 Answer
1
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$begingroup$
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.
answered Feb 3 at 12:03
jvdhooftjvdhooft
5,65961641
$endgroup$
$begingroup$
Thank you. I see my mistake now, so do the house edge formula above mean anything at all?
$endgroup$
– Zanko
Feb 3 at 12:25
$begingroup$
@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.
$endgroup$
– jvdhooft
Feb 3 at 12:40
$begingroup$
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?
$endgroup$
– Zanko
Feb 3 at 12:48
$begingroup$
@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.
$endgroup$
– jvdhooft
Feb 3 at 13:04
$begingroup$
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 = 0.1and(1-D) - 60 / 1000 * P = (1 - 0.24) - 60 / 1000 * 11 = 0.1meant. They both give value of 0.1. It is expected profit GIVEN initial bet. But this information is useless to casino right?
$endgroup$
– Zanko
Feb 3 at 13:08
|
show 3 more comments
$begingroup$
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.
answered Feb 3 at 12:03
jvdhooftjvdhooft
5,65961641
$endgroup$
$begingroup$
Thank you. I see my mistake now, so do the house edge formula above mean anything at all?
$endgroup$
– Zanko
Feb 3 at 12:25
$begingroup$
@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.
$endgroup$
– jvdhooft
Feb 3 at 12:40
$begingroup$
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?
$endgroup$
– Zanko
Feb 3 at 12:48
$begingroup$
@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.
$endgroup$
– jvdhooft
Feb 3 at 13:04
$begingroup$
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 = 0.1and(1-D) - 60 / 1000 * P = (1 - 0.24) - 60 / 1000 * 11 = 0.1meant. They both give value of 0.1. It is expected profit GIVEN initial bet. But this information is useless to casino right?
$endgroup$
– Zanko
Feb 3 at 13:08
|
show 3 more comments
$begingroup$
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.
answered Feb 3 at 12:03
jvdhooftjvdhooft
5,65961641
$endgroup$
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.
answered Feb 3 at 12:03
jvdhooftjvdhooft
5,65961641
answered Feb 3 at 12:03
jvdhooftjvdhooft
5,65961641
answered Feb 3 at 12:03
jvdhooftjvdhooft
5,65961641
answered Feb 3 at 12:03
jvdhooftjvdhooft
5,65961641
5,65961641
$begingroup$
Thank you. I see my mistake now, so do the house edge formula above mean anything at all?
$endgroup$
– Zanko
Feb 3 at 12:25
$begingroup$
@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.
$endgroup$
– jvdhooft
Feb 3 at 12:40
$begingroup$
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?
$endgroup$
– Zanko
Feb 3 at 12:48
$begingroup$
@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.
$endgroup$
– jvdhooft
Feb 3 at 13:04
$begingroup$
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 = 0.1and(1-D) - 60 / 1000 * P = (1 - 0.24) - 60 / 1000 * 11 = 0.1meant. They both give value of 0.1. It is expected profit GIVEN initial bet. But this information is useless to casino right?
$endgroup$
– Zanko
Feb 3 at 13:08
|
show 3 more comments
$begingroup$
Thank you. I see my mistake now, so do the house edge formula above mean anything at all?
$endgroup$
– Zanko
Feb 3 at 12:25
$begingroup$
@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.
$endgroup$
– jvdhooft
Feb 3 at 12:40
$begingroup$
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?
$endgroup$
– Zanko
Feb 3 at 12:48
$begingroup$
@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.
$endgroup$
– jvdhooft
Feb 3 at 13:04
$begingroup$
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 = 0.1and(1-D) - 60 / 1000 * P = (1 - 0.24) - 60 / 1000 * 11 = 0.1meant. They both give value of 0.1. It is expected profit GIVEN initial bet. But this information is useless to casino right?
$endgroup$
– Zanko
Feb 3 at 13:08
$begingroup$
Thank you. I see my mistake now, so do the house edge formula above mean anything at all?
$endgroup$
– Zanko
Feb 3 at 12:25
$begingroup$
Thank you. I see my mistake now, so do the house edge formula above mean anything at all?
$endgroup$
– Zanko
Feb 3 at 12:25
$begingroup$
@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.
$endgroup$
– jvdhooft
Feb 3 at 12:40
$begingroup$
@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.
$endgroup$
– jvdhooft
Feb 3 at 12:40
$begingroup$
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?
$endgroup$
– Zanko
Feb 3 at 12:48
$begingroup$
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?
$endgroup$
– Zanko
Feb 3 at 12:48
$begingroup$
@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.
$endgroup$
– jvdhooft
Feb 3 at 13:04
$begingroup$
@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.
$endgroup$
– jvdhooft
Feb 3 at 13:04
$begingroup$
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?$endgroup$
– Zanko
Feb 3 at 13:08
$begingroup$
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?$endgroup$
– Zanko
Feb 3 at 13:08
|
show 3 more comments
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