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Casino Profit/Loss (Expected Value) [closed]
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I have this question in mind for a long time, it is almost certain that if one gambles long enough he/she will lose over long number of trials(the expected value is in the favor of casino). That's the fundamental idea on which casinos make profit. (And it works!)
Then over long trials it is almost certain that there will be more losses than wins (hence profit for the casino). In such a scenario, then commonsense implies one should keep his/her trials low to increase chances of profit. If the expected value for the casino is positive(profit), how would you formulate the expectation in which average player's expected value is negative, meaning that she/he is losing.
probability probability-theory statistics
asked Aug 1 at 17:32
nmd_07
136
closed as too broad by Math1000, Claude Leibovici, Mostafa Ayaz, Shailesh, Simply Beautiful Art Aug 2 at 16:54
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
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I have this question in mind for a long time, it is almost certain that if one gambles long enough he/she will lose over long number of trials(the expected value is in the favor of casino). That's the fundamental idea on which casinos make profit. (And it works!)
Then over long trials it is almost certain that there will be more losses than wins (hence profit for the casino). In such a scenario, then commonsense implies one should keep his/her trials low to increase chances of profit. If the expected value for the casino is positive(profit), how would you formulate the expectation in which average player's expected value is negative, meaning that she/he is losing.
probability probability-theory statistics
asked Aug 1 at 17:32
nmd_07
136
closed as too broad by Math1000, Claude Leibovici, Mostafa Ayaz, Shailesh, Simply Beautiful Art Aug 2 at 16:54
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
What are you asking? On each play at a casino, the house has positive expectation and the player has a negative expectation. True, you can minimize your expected losses by not playing often, or get them down to $0$ by not playing at all.
– lulu
Aug 1 at 17:44
I can see some questions that you might have in mind: for example, suppose you have $$100$ and want to double it. Is it better to place one double or nothing bet or to bet $$1$ a time? (assuming the house's edge is constant as a percent of the wagered stakes). But as it stands, it's not clear what you have in mind.
– lulu
Aug 1 at 17:47
I want to know whether I can increase chances of profit without having to cheat with a game plan? - besides not playing at all.
– nmd_07
Aug 1 at 17:49
That's too vague. Find the most probable bet the house offers and bet on that one time.
– lulu
Aug 1 at 17:52
"chances of profit" is a dangerous metric. There can be egregiously unfair games in which the chance of profit is high. For example, roll a fair pair of dice... I'll pay you $$1$ if you get anything other than $(1,1)$ but you have to pay me $$1000$ if you get $(1,1)$. In that game you will profit with probability $frac 3536approx 97.22%$, but the expected value of the game is quite negative.
– lulu
Aug 1 at 17:58
add a comment |Â
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up vote
0
down vote
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I have this question in mind for a long time, it is almost certain that if one gambles long enough he/she will lose over long number of trials(the expected value is in the favor of casino). That's the fundamental idea on which casinos make profit. (And it works!)
Then over long trials it is almost certain that there will be more losses than wins (hence profit for the casino). In such a scenario, then commonsense implies one should keep his/her trials low to increase chances of profit. If the expected value for the casino is positive(profit), how would you formulate the expectation in which average player's expected value is negative, meaning that she/he is losing.
probability probability-theory statistics
asked Aug 1 at 17:32
nmd_07
136
I have this question in mind for a long time, it is almost certain that if one gambles long enough he/she will lose over long number of trials(the expected value is in the favor of casino). That's the fundamental idea on which casinos make profit. (And it works!)
Then over long trials it is almost certain that there will be more losses than wins (hence profit for the casino). In such a scenario, then commonsense implies one should keep his/her trials low to increase chances of profit. If the expected value for the casino is positive(profit), how would you formulate the expectation in which average player's expected value is negative, meaning that she/he is losing.
probability probability-theory statistics
asked Aug 1 at 17:32
nmd_07
136
asked Aug 1 at 17:32
nmd_07
136
asked Aug 1 at 17:32
nmd_07
136
asked Aug 1 at 17:32
nmd_07
136
136
closed as too broad by Math1000, Claude Leibovici, Mostafa Ayaz, Shailesh, Simply Beautiful Art Aug 2 at 16:54
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as too broad by Math1000, Claude Leibovici, Mostafa Ayaz, Shailesh, Simply Beautiful Art Aug 2 at 16:54
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
What are you asking? On each play at a casino, the house has positive expectation and the player has a negative expectation. True, you can minimize your expected losses by not playing often, or get them down to $0$ by not playing at all.
– lulu
Aug 1 at 17:44
I can see some questions that you might have in mind: for example, suppose you have $$100$ and want to double it. Is it better to place one double or nothing bet or to bet $$1$ a time? (assuming the house's edge is constant as a percent of the wagered stakes). But as it stands, it's not clear what you have in mind.
– lulu
Aug 1 at 17:47
I want to know whether I can increase chances of profit without having to cheat with a game plan? - besides not playing at all.
– nmd_07
Aug 1 at 17:49
That's too vague. Find the most probable bet the house offers and bet on that one time.
– lulu
Aug 1 at 17:52
"chances of profit" is a dangerous metric. There can be egregiously unfair games in which the chance of profit is high. For example, roll a fair pair of dice... I'll pay you $$1$ if you get anything other than $(1,1)$ but you have to pay me $$1000$ if you get $(1,1)$. In that game you will profit with probability $frac 3536approx 97.22%$, but the expected value of the game is quite negative.
– lulu
Aug 1 at 17:58
add a comment |Â
What are you asking? On each play at a casino, the house has positive expectation and the player has a negative expectation. True, you can minimize your expected losses by not playing often, or get them down to $0$ by not playing at all.
– lulu
Aug 1 at 17:44
I can see some questions that you might have in mind: for example, suppose you have $$100$ and want to double it. Is it better to place one double or nothing bet or to bet $$1$ a time? (assuming the house's edge is constant as a percent of the wagered stakes). But as it stands, it's not clear what you have in mind.
– lulu
Aug 1 at 17:47
I want to know whether I can increase chances of profit without having to cheat with a game plan? - besides not playing at all.
– nmd_07
Aug 1 at 17:49
That's too vague. Find the most probable bet the house offers and bet on that one time.
– lulu
Aug 1 at 17:52
"chances of profit" is a dangerous metric. There can be egregiously unfair games in which the chance of profit is high. For example, roll a fair pair of dice... I'll pay you $$1$ if you get anything other than $(1,1)$ but you have to pay me $$1000$ if you get $(1,1)$. In that game you will profit with probability $frac 3536approx 97.22%$, but the expected value of the game is quite negative.
– lulu
Aug 1 at 17:58
What are you asking? On each play at a casino, the house has positive expectation and the player has a negative expectation. True, you can minimize your expected losses by not playing often, or get them down to $0$ by not playing at all.
– lulu
Aug 1 at 17:44
What are you asking? On each play at a casino, the house has positive expectation and the player has a negative expectation. True, you can minimize your expected losses by not playing often, or get them down to $0$ by not playing at all.
– lulu
Aug 1 at 17:44
I can see some questions that you might have in mind: for example, suppose you have $$100$ and want to double it. Is it better to place one double or nothing bet or to bet $$1$ a time? (assuming the house's edge is constant as a percent of the wagered stakes). But as it stands, it's not clear what you have in mind.
– lulu
Aug 1 at 17:47
I can see some questions that you might have in mind: for example, suppose you have $$100$ and want to double it. Is it better to place one double or nothing bet or to bet $$1$ a time? (assuming the house's edge is constant as a percent of the wagered stakes). But as it stands, it's not clear what you have in mind.
– lulu
Aug 1 at 17:47
I want to know whether I can increase chances of profit without having to cheat with a game plan? - besides not playing at all.
– nmd_07
Aug 1 at 17:49
I want to know whether I can increase chances of profit without having to cheat with a game plan? - besides not playing at all.
– nmd_07
Aug 1 at 17:49
That's too vague. Find the most probable bet the house offers and bet on that one time.
– lulu
Aug 1 at 17:52
That's too vague. Find the most probable bet the house offers and bet on that one time.
– lulu
Aug 1 at 17:52
"chances of profit" is a dangerous metric. There can be egregiously unfair games in which the chance of profit is high. For example, roll a fair pair of dice... I'll pay you $$1$ if you get anything other than $(1,1)$ but you have to pay me $$1000$ if you get $(1,1)$. In that game you will profit with probability $frac 3536approx 97.22%$, but the expected value of the game is quite negative.
– lulu
Aug 1 at 17:58
"chances of profit" is a dangerous metric. There can be egregiously unfair games in which the chance of profit is high. For example, roll a fair pair of dice... I'll pay you $$1$ if you get anything other than $(1,1)$ but you have to pay me $$1000$ if you get $(1,1)$. In that game you will profit with probability $frac 3536approx 97.22%$, but the expected value of the game is quite negative.
– lulu
Aug 1 at 17:58
add a comment |Â
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What are you asking? On each play at a casino, the house has positive expectation and the player has a negative expectation. True, you can minimize your expected losses by not playing often, or get them down to $0$ by not playing at all.
– lulu
Aug 1 at 17:44
I can see some questions that you might have in mind: for example, suppose you have $$100$ and want to double it. Is it better to place one double or nothing bet or to bet $$1$ a time? (assuming the house's edge is constant as a percent of the wagered stakes). But as it stands, it's not clear what you have in mind.
– lulu
Aug 1 at 17:47
I want to know whether I can increase chances of profit without having to cheat with a game plan? - besides not playing at all.
– nmd_07
Aug 1 at 17:49
That's too vague. Find the most probable bet the house offers and bet on that one time.
– lulu
Aug 1 at 17:52
"chances of profit" is a dangerous metric. There can be egregiously unfair games in which the chance of profit is high. For example, roll a fair pair of dice... I'll pay you $$1$ if you get anything other than $(1,1)$ but you have to pay me $$1000$ if you get $(1,1)$. In that game you will profit with probability $frac 3536approx 97.22%$, but the expected value of the game is quite negative.
– lulu
Aug 1 at 17:58