Expected Value at the end of an equal probability game [closed]
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Suppose we play a game where we start with $c$ dollars at each play you either double or halve your money, what is your expected fortune after $n$ trials?
My line of thinking went along the way that this is a Binomial distribution so my expected fortune will be $n/2$ at the end of the game. I couldn't verify this solution but I have a feeling that it might be wrong
ref: All of statistics by L.Wasserman
Update : Assume we start with c amount of dollars then after one game the expected amount of dollars we have is 5/4c , if we sum over a series of game we end up with n*(5/4)*c.
probability expected-value
asked Nov 22 at 12:23
rafi
33
closed as off-topic by amWhy, Chinnapparaj R, KReiser, user10354138, Rebellos Nov 23 at 9:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Chinnapparaj R, KReiser, user10354138, Rebellos
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Suppose we play a game where we start with $c$ dollars at each play you either double or halve your money, what is your expected fortune after $n$ trials?
My line of thinking went along the way that this is a Binomial distribution so my expected fortune will be $n/2$ at the end of the game. I couldn't verify this solution but I have a feeling that it might be wrong
ref: All of statistics by L.Wasserman
Update : Assume we start with c amount of dollars then after one game the expected amount of dollars we have is 5/4c , if we sum over a series of game we end up with n*(5/4)*c.
probability expected-value
asked Nov 22 at 12:23
rafi
33
closed as off-topic by amWhy, Chinnapparaj R, KReiser, user10354138, Rebellos Nov 23 at 9:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Chinnapparaj R, KReiser, user10354138, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose we play a game where we start with $c$ dollars at each play you either double or halve your money, what is your expected fortune after $n$ trials?
My line of thinking went along the way that this is a Binomial distribution so my expected fortune will be $n/2$ at the end of the game. I couldn't verify this solution but I have a feeling that it might be wrong
ref: All of statistics by L.Wasserman
Update : Assume we start with c amount of dollars then after one game the expected amount of dollars we have is 5/4c , if we sum over a series of game we end up with n*(5/4)*c.
probability expected-value
asked Nov 22 at 12:23
rafi
33
Suppose we play a game where we start with $c$ dollars at each play you either double or halve your money, what is your expected fortune after $n$ trials?
My line of thinking went along the way that this is a Binomial distribution so my expected fortune will be $n/2$ at the end of the game. I couldn't verify this solution but I have a feeling that it might be wrong
ref: All of statistics by L.Wasserman
Update : Assume we start with c amount of dollars then after one game the expected amount of dollars we have is 5/4c , if we sum over a series of game we end up with n*(5/4)*c.
probability expected-value
probability expected-value
asked Nov 22 at 12:23
rafi
33
asked Nov 22 at 12:23
rafi
33
edited Nov 23 at 12:20
asked Nov 22 at 12:23
rafi
33
asked Nov 22 at 12:23
rafi
33
asked Nov 22 at 12:23
rafi
33
33
closed as off-topic by amWhy, Chinnapparaj R, KReiser, user10354138, Rebellos Nov 23 at 9:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Chinnapparaj R, KReiser, user10354138, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, Chinnapparaj R, KReiser, user10354138, Rebellos Nov 23 at 9:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Chinnapparaj R, KReiser, user10354138, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
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1 Answer
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Fortune after n-th trials : $X_n=2^{xi_n}2^{xi_{n-1}}ldots2^{xi_1}c$ where $xi_1,xi_2,ldotsxi_nstackrel{iid}{sim}Bernoulli(p)$ for $0<p<1$. So, $E(X_n)=E(2^{xi_n}2^{xi_{n-1}}ldots2^{xi_1}c)=E(2^{xi_n})E(2^{xi_{n-1}})ldots E(2^{xi_1})c=(E(2^{xi_1}))^nc=(2p+(1-p)/2)^nc.$ If $p=1/2$ then $E(X_n)=(5/4)^nc.$
answered Nov 22 at 13:15
John_Wick
1,134111
Oh I redid the same as you got the same result.
– rafi
Nov 22 at 13:38
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Fortune after n-th trials : $X_n=2^{xi_n}2^{xi_{n-1}}ldots2^{xi_1}c$ where $xi_1,xi_2,ldotsxi_nstackrel{iid}{sim}Bernoulli(p)$ for $0<p<1$. So, $E(X_n)=E(2^{xi_n}2^{xi_{n-1}}ldots2^{xi_1}c)=E(2^{xi_n})E(2^{xi_{n-1}})ldots E(2^{xi_1})c=(E(2^{xi_1}))^nc=(2p+(1-p)/2)^nc.$ If $p=1/2$ then $E(X_n)=(5/4)^nc.$
answered Nov 22 at 13:15
John_Wick
1,134111
Oh I redid the same as you got the same result.
– rafi
Nov 22 at 13:38
add a comment |
up vote
1
down vote
accepted
Fortune after n-th trials : $X_n=2^{xi_n}2^{xi_{n-1}}ldots2^{xi_1}c$ where $xi_1,xi_2,ldotsxi_nstackrel{iid}{sim}Bernoulli(p)$ for $0<p<1$. So, $E(X_n)=E(2^{xi_n}2^{xi_{n-1}}ldots2^{xi_1}c)=E(2^{xi_n})E(2^{xi_{n-1}})ldots E(2^{xi_1})c=(E(2^{xi_1}))^nc=(2p+(1-p)/2)^nc.$ If $p=1/2$ then $E(X_n)=(5/4)^nc.$
answered Nov 22 at 13:15
John_Wick
1,134111
Oh I redid the same as you got the same result.
– rafi
Nov 22 at 13:38
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Fortune after n-th trials : $X_n=2^{xi_n}2^{xi_{n-1}}ldots2^{xi_1}c$ where $xi_1,xi_2,ldotsxi_nstackrel{iid}{sim}Bernoulli(p)$ for $0<p<1$. So, $E(X_n)=E(2^{xi_n}2^{xi_{n-1}}ldots2^{xi_1}c)=E(2^{xi_n})E(2^{xi_{n-1}})ldots E(2^{xi_1})c=(E(2^{xi_1}))^nc=(2p+(1-p)/2)^nc.$ If $p=1/2$ then $E(X_n)=(5/4)^nc.$
answered Nov 22 at 13:15
John_Wick
1,134111
Fortune after n-th trials : $X_n=2^{xi_n}2^{xi_{n-1}}ldots2^{xi_1}c$ where $xi_1,xi_2,ldotsxi_nstackrel{iid}{sim}Bernoulli(p)$ for $0<p<1$. So, $E(X_n)=E(2^{xi_n}2^{xi_{n-1}}ldots2^{xi_1}c)=E(2^{xi_n})E(2^{xi_{n-1}})ldots E(2^{xi_1})c=(E(2^{xi_1}))^nc=(2p+(1-p)/2)^nc.$ If $p=1/2$ then $E(X_n)=(5/4)^nc.$
answered Nov 22 at 13:15
John_Wick
1,134111
answered Nov 22 at 13:15
John_Wick
1,134111
answered Nov 22 at 13:15
John_Wick
1,134111
answered Nov 22 at 13:15
John_Wick
1,134111
1,134111
Oh I redid the same as you got the same result.
– rafi
Nov 22 at 13:38
add a comment |
Oh I redid the same as you got the same result.
– rafi
Nov 22 at 13:38
Oh I redid the same as you got the same result.
– rafi
Nov 22 at 13:38
Oh I redid the same as you got the same result.
– rafi
Nov 22 at 13:38
add a comment |
