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Finacial Expectation (1 Viewer)

Paj20

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In a dice game, two dice are rolled.

*The player wins $1 for rolling a total of 7 or 11

* The player loses $1 for rolling a total of 2,3 or 12

*If any other total is rolled the dice are rolled again

What is the finical expectation???
 

PC

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Paj20 said:
In a dice game, two dice are rolled.
*The player wins $1 for rolling a total of 7 or 11
*The player loses $1 for rolling a total of 2,3 or 12
*If any other total is rolled the dice are rolled again
What is the finical expectation???
When two dice are rolled, there are 36 possible outcomes.

P(7) = 6/36
P(11) = 2/36
So P(7 or 11) = 8/36

P(2) = 1/36
P(3) = 2/36
P(12) = 1/36
So P(2, 3 or 12) = 4/36

So financial expectation = (8/36 x +$1) + (4/36 x –$1) + (24/36 x $0)
= 1/9
= $0.11

That's for the first roll anyway.

An interesting question because the game could go on forever!
 

aakash

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PC said:
When two dice are rolled, there are 36 possible outcomes.

P(7) = 6/36
P(11) = 2/36
So P(7 or 11) = 8/36

P(2) = 1/36
P(3) = 2/36
P(12) = 1/36
So P(2, 3 or 12) = 4/36

So financial expectation = (8/36 x +$1) + (4/36 x –$1) + (24/36 x $0)
= 1/9
= $0.11

That's for the first roll anyway.

An interesting question because the game could go on forever!
hmmm...but when the sum is 4,5,6,8,9,10 the gain is not $0...rather the whole process is repeated again.

my solution: let the expected gain = $x
$x = (8/36 x +$1) + (4/36 x –$1) + (24/36 x $x)

explanation of the last term : (24/36 x $x)
When the sum is 4,5,6,8,9,10 then the dice are rolled again and we are back to the start of the process and hence the expected gain = $x (thts wat we assumed)

solving for x gives x=$(1/3)
 

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