Is there a better concept than expectation for one time play?
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Given a simple lottery game like
- Guess the right (random generated) number $in [0,1000]$.
- Stake = 1€
- Win= 2001€
the expected outcome is $frac{1}{1001}cdot2001 + frac{1000}{1001}cdot(-1) = 1$.
Hence in the limit, you will win 1€ per play. So far everything is clear.
But what happens if I am only allowed to play the game once? The most probable outcome is not winning.
My main question is:
Is there a mathematical concept that says the expected outcome is -1? (since it is the most probable outcome?)
Going further, adding a second choice of not playing the game but instead always getting 0.80€.
Is there a concept that favors the safe 0.80€ choice over playing the game?
probability game-theory
asked Nov 16 at 12:19
Jobi
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up vote
1
down vote
favorite
Given a simple lottery game like
- Guess the right (random generated) number $in [0,1000]$.
- Stake = 1€
- Win= 2001€
the expected outcome is $frac{1}{1001}cdot2001 + frac{1000}{1001}cdot(-1) = 1$.
Hence in the limit, you will win 1€ per play. So far everything is clear.
But what happens if I am only allowed to play the game once? The most probable outcome is not winning.
My main question is:
Is there a mathematical concept that says the expected outcome is -1? (since it is the most probable outcome?)
Going further, adding a second choice of not playing the game but instead always getting 0.80€.
Is there a concept that favors the safe 0.80€ choice over playing the game?
probability game-theory
asked Nov 16 at 12:19
Jobi
83
New contributor
Jobi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Another criteria to assess a game is the variance which could be regarded as a risk measure. The higher the variance the worse the game for a risk neutral or risk averse player.
– callculus
Nov 16 at 12:32
You might be interested in "expected utility": en.wikipedia.org/wiki/Expected_utility_hypothesis
– awkward
Nov 16 at 13:43
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given a simple lottery game like
- Guess the right (random generated) number $in [0,1000]$.
- Stake = 1€
- Win= 2001€
the expected outcome is $frac{1}{1001}cdot2001 + frac{1000}{1001}cdot(-1) = 1$.
Hence in the limit, you will win 1€ per play. So far everything is clear.
But what happens if I am only allowed to play the game once? The most probable outcome is not winning.
My main question is:
Is there a mathematical concept that says the expected outcome is -1? (since it is the most probable outcome?)
Going further, adding a second choice of not playing the game but instead always getting 0.80€.
Is there a concept that favors the safe 0.80€ choice over playing the game?
probability game-theory
asked Nov 16 at 12:19
Jobi
83
New contributor
Jobi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Given a simple lottery game like
- Guess the right (random generated) number $in [0,1000]$.
- Stake = 1€
- Win= 2001€
the expected outcome is $frac{1}{1001}cdot2001 + frac{1000}{1001}cdot(-1) = 1$.
Hence in the limit, you will win 1€ per play. So far everything is clear.
But what happens if I am only allowed to play the game once? The most probable outcome is not winning.
My main question is:
Is there a mathematical concept that says the expected outcome is -1? (since it is the most probable outcome?)
Going further, adding a second choice of not playing the game but instead always getting 0.80€.
Is there a concept that favors the safe 0.80€ choice over playing the game?
probability game-theory
probability game-theory
asked Nov 16 at 12:19
Jobi
83
New contributor
Jobi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 16 at 12:19
Jobi
83
New contributor
Jobi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 16 at 12:19
Jobi
83
New contributor
Jobi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 16 at 12:19
Jobi
83
asked Nov 16 at 12:19
Jobi
83
83
New contributor
Jobi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Jobi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Jobi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Another criteria to assess a game is the variance which could be regarded as a risk measure. The higher the variance the worse the game for a risk neutral or risk averse player.
– callculus
Nov 16 at 12:32
You might be interested in "expected utility": en.wikipedia.org/wiki/Expected_utility_hypothesis
– awkward
Nov 16 at 13:43
add a comment |
Another criteria to assess a game is the variance which could be regarded as a risk measure. The higher the variance the worse the game for a risk neutral or risk averse player.
– callculus
Nov 16 at 12:32
You might be interested in "expected utility": en.wikipedia.org/wiki/Expected_utility_hypothesis
– awkward
Nov 16 at 13:43
Another criteria to assess a game is the variance which could be regarded as a risk measure. The higher the variance the worse the game for a risk neutral or risk averse player.
– callculus
Nov 16 at 12:32
Another criteria to assess a game is the variance which could be regarded as a risk measure. The higher the variance the worse the game for a risk neutral or risk averse player.
– callculus
Nov 16 at 12:32
You might be interested in "expected utility": en.wikipedia.org/wiki/Expected_utility_hypothesis
– awkward
Nov 16 at 13:43
You might be interested in "expected utility": en.wikipedia.org/wiki/Expected_utility_hypothesis
– awkward
Nov 16 at 13:43
add a comment |
1 Answer
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votes
up vote
1
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accepted
For your first part:
The mathematical concept you are looking for is "most probable outcome". The most probable outcome is $-1$. Why would we need another term to describe the most probable outcome? The term we have is perfectly fine. It's descriptive, not too long, and very understandable.
For the second part:
The concept that favors the safe choice would be if you also look at the variance of the random variable. The random variable of playing the game has an expected value of $1$, and a variance of $$E(X^2)-E(X)^2 = frac{1}{1001}2001^2 + frac{1000}{1001}1^2 - 1^2 = 4000$$
which is... well, a lot. The "get $0.8$ for sure" game has a variance of $0$, so it's a much safer bet. Note, however, that if you are allowed to play the first game many many times, the variance decreases as the number of times you play increases.
answered Nov 16 at 12:27
5xum
88.3k392160
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
For your first part:
The mathematical concept you are looking for is "most probable outcome". The most probable outcome is $-1$. Why would we need another term to describe the most probable outcome? The term we have is perfectly fine. It's descriptive, not too long, and very understandable.
For the second part:
The concept that favors the safe choice would be if you also look at the variance of the random variable. The random variable of playing the game has an expected value of $1$, and a variance of $$E(X^2)-E(X)^2 = frac{1}{1001}2001^2 + frac{1000}{1001}1^2 - 1^2 = 4000$$
which is... well, a lot. The "get $0.8$ for sure" game has a variance of $0$, so it's a much safer bet. Note, however, that if you are allowed to play the first game many many times, the variance decreases as the number of times you play increases.
answered Nov 16 at 12:27
5xum
88.3k392160
add a comment |
up vote
1
down vote
accepted
For your first part:
The mathematical concept you are looking for is "most probable outcome". The most probable outcome is $-1$. Why would we need another term to describe the most probable outcome? The term we have is perfectly fine. It's descriptive, not too long, and very understandable.
For the second part:
The concept that favors the safe choice would be if you also look at the variance of the random variable. The random variable of playing the game has an expected value of $1$, and a variance of $$E(X^2)-E(X)^2 = frac{1}{1001}2001^2 + frac{1000}{1001}1^2 - 1^2 = 4000$$
which is... well, a lot. The "get $0.8$ for sure" game has a variance of $0$, so it's a much safer bet. Note, however, that if you are allowed to play the first game many many times, the variance decreases as the number of times you play increases.
answered Nov 16 at 12:27
5xum
88.3k392160
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
For your first part:
The mathematical concept you are looking for is "most probable outcome". The most probable outcome is $-1$. Why would we need another term to describe the most probable outcome? The term we have is perfectly fine. It's descriptive, not too long, and very understandable.
For the second part:
The concept that favors the safe choice would be if you also look at the variance of the random variable. The random variable of playing the game has an expected value of $1$, and a variance of $$E(X^2)-E(X)^2 = frac{1}{1001}2001^2 + frac{1000}{1001}1^2 - 1^2 = 4000$$
which is... well, a lot. The "get $0.8$ for sure" game has a variance of $0$, so it's a much safer bet. Note, however, that if you are allowed to play the first game many many times, the variance decreases as the number of times you play increases.
answered Nov 16 at 12:27
5xum
88.3k392160
For your first part:
The mathematical concept you are looking for is "most probable outcome". The most probable outcome is $-1$. Why would we need another term to describe the most probable outcome? The term we have is perfectly fine. It's descriptive, not too long, and very understandable.
For the second part:
The concept that favors the safe choice would be if you also look at the variance of the random variable. The random variable of playing the game has an expected value of $1$, and a variance of $$E(X^2)-E(X)^2 = frac{1}{1001}2001^2 + frac{1000}{1001}1^2 - 1^2 = 4000$$
which is... well, a lot. The "get $0.8$ for sure" game has a variance of $0$, so it's a much safer bet. Note, however, that if you are allowed to play the first game many many times, the variance decreases as the number of times you play increases.
answered Nov 16 at 12:27
5xum
88.3k392160
answered Nov 16 at 12:27
5xum
88.3k392160
answered Nov 16 at 12:27
5xum
88.3k392160
answered Nov 16 at 12:27
5xum
88.3k392160
88.3k392160
add a comment |
add a comment |
Jobi is a new contributor. Be nice, and check out our Code of Conduct.
Jobi is a new contributor. Be nice, and check out our Code of Conduct.
Jobi is a new contributor. Be nice, and check out our Code of Conduct.
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Another criteria to assess a game is the variance which could be regarded as a risk measure. The higher the variance the worse the game for a risk neutral or risk averse player.
– callculus
Nov 16 at 12:32
You might be interested in "expected utility": en.wikipedia.org/wiki/Expected_utility_hypothesis
– awkward
Nov 16 at 13:43