y1066380
  • y1066380
A man plays a game in the following manner: he pays $20 to throw a die and he will be paid $(n^2+5) if the number n turns up. What is his expected gain in each trial and what is his expected gain in 30 trials? Is it correct to calculate the expected gain in each trial by: (1/6)*(5*6+1^2+2^2+3^2+4^2+5^2+6^2) =121/6 ? And what is the difference between one trial and 30 trials?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
if i recall correctly ... expected value is: $won * P(win) + $lost * P(lose)
amistre64
  • amistre64
30 trials might indicate a discrete probability function as opposed to a normal distribution
amistre64
  • amistre64
winning/losing any particular instance does not affect the other outcomes, so the trials are independant ... which is mostly a side thought

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amistre64
  • amistre64
1/6 (36+5) + 1/6 (25+5) + 1/6 (16+5) + 1/6 (9+5) + 1/6 (4+5) + 1/6 (1+5) + 5/6 (-20) ... he loses $20 if his number doesnt come up seems to me as tho the expected value of the game is $3.50
amistre64
  • amistre64
should be the same outcome (need to verify) if we subtract 20 from the winnings eh, then the 5/6 is redundant
amistre64
  • amistre64
scratch that last thought ... im just not a gambling man lol
amistre64
  • amistre64
we would expect 30 times the expected outcome of 1 trial, if we played 30 times
amistre64
  • amistre64
In the book, the authors refer to two types of expected values, only one of which is called an expected value The expected winnings or loses on any one draw (average of the box) The expected net gain, the total amount won or lost across draws This is found by multiplying the number of draws times the average of the box Example: if we played the shell 30 times, the net gain is 30 X $0 = $0. http://www.stat.ucla.edu/~cochran/stat10h/lectures/lect9.html
y1066380
  • y1066380
+ 5/6 (-20) ... he loses $20 if his number doesnt come up << There should be a number come up, right? That means: 1/6 (36+5) + 1/6 (25+5) + 1/6 (16+5) + 1/6 (9+5) + 1/6 (4+5) + 1/6 (1+5) -20 (The fee for gambling) =20.166666... Am I right?
amistre64
  • amistre64
there should be a number that comes up ... maybe i read it wrong and assumed he had to pick a number and then hope for it to come up.
amistre64
  • amistre64
0.16 is his expected outcome (since he paid 20 to play the game)
amistre64
  • amistre64
the outcome seems to favor the gambler :) so its not a fair game .. whose outcome would be 0
y1066380
  • y1066380
Haha...
y1066380
  • y1066380
So expected gain in 30 trials is $0.166666*30 = $5
amistre64
  • amistre64
seems legit to me
y1066380
  • y1066380
Thanks for your help!
amistre64
  • amistre64
yep

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