Question 1200014
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A simple gambling game involves the throw of a die. 
Players are charged 40 cents if the die shows 1,2 or 3 they are charged nothing 
for 4 and they receive 30 cents or 60 cents respectively for 5 or 6. 
Let the random variable X be the payout to the player.
Calculate the expected value and how much profit would the casino expect to make on 100 games?
If the backer weighted the die so that it is 50% more likely to turn up 
each of the results 1,2,3 as to turn up 4,5 or 6.
What is the probability now of each outcome?
Find expected value now.
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            I agree with the solution of the first problem given by  @math_tutor2020.


            But regarding the second problem in the post,  I read and interpret it differently; 

            therefore,  my solution,  my numbers and my answer are different.



<pre>
In the second problem, we have six events 1, 2, 3, 4, 5 and 6, and we are given that

    P(1) = P(2) = P(3);  P(4) = P(5) = P(6) 

and 
  
    P(1) = 1.5*P(4);  P(2) = 1.5*P(5);  P(3) = 1.5*P(6).


Let x be any of the three equal values  P(4) = P(5) = P(6).

Then any of the three other equal values  P(1) = P(2) = P(3)  is  1.5x  (it is where and how the multiplier 1.5 works in my solution).


The sum of probabilities  P(1) + . . . + P(6) is 1.  It gives this equation

    3*(1.5x) + 3x = 1,  or  4.5x + 3x = 1,  or  7.5x = 1,


which implies 

    x = {{{1/7.5}}} = {{{2/15}}};  1.5x = {{{1.5*(2/15)}}} = {{{3/15}}}.


Thus  P(1) = P(2) = P(3) = {{{3/15}}};  P(4) = P(5) = P(6) = {{{2/15}}}.


Now math expectation of the player/(the gamer) is


      {{{(-40)*(3/15)}}} + {{{(-40)*(3/15)}}} + {{{(-40)*(3/15)}}} + {{{0*(2/15)}}} + {{{30*(2/15)}}} + {{{60*(2/15)}}} = 


   = {{{((-40)*3 + (-40)*3 + (-40)*3 + 0 + 30*2 + 60*2)/15}}} = {{{-180/15}}} cents = -12 cents.


Which means that the gamer loses 12 cents per game, in average. 


<U>ANSWER</U>.  Math expectation is -12 cents: the gamer loses 12 cents per game, in average. 
</pre>

Solved.