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2015-08-20T19:57:35+05:30

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1.   to play : 1 $ each game
2.    2 H + 2 T  gets 3 $    on four coins game
3.    3 dice are rolled.  Every 6 , gets  $2.

we have to decide which game is more winning ????

game 1:
   number of outcomes when four coins are tossed once :  2^4 = 16
   Number of outcomes that are favourable = 6 
             ( HHTT, HTHT, HTTH, TTHH, THTH, THHT  )  = 4! /(2 * 2)
   arrangements of 4 letters, with 2 being identical and another 2 being identical.

  now  Probability of winning = 6/16 = 0.375
   Expected winnings =  0.375 * $3 =  $ 1.125

game 2:

   number of out comes:  6^3 = 216
     favourable out comes = 
        1.  outcomes with 1 six only :  1 * 5 * 5 + 5 * 1 * 5 + 5 * 5 * 1 = 75 
         2. outcomes with 2 sixes :   1 * 1 * 5 + 1 * 5 * 1 + 1 * 1 * 5 = 15
         3. outcomes with 3 sixes :   1 * 1 * 1 = 1

   expected winnings =  Sum of P( $2) * $2 + P($4) * $4 + P($6) * $6 + P($0) * $ 0
           = ( 75/216)  * $2  + (15/216) * $4 + (1/216) * $6
           = $ 1

  So  Game 1 is better to play as the expected winnings are more.

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