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2 comments :
Very good post chhota bheem and the curse of damyaan
Card are drawn randomly from hat, so probability of each card is 1/10
So expected value is :
1*1/10 + 2*1/10 + 3*1/10 + ... + 9*1/10 + 10*1/10 = (1+2+....+9+10)/10 = 5.5
for 2nd case
A rational user draws a second card in exactly 50% of cases, i.e. only if the first card is 1/2/3/4/5. If the cases where he holds the first card, his expected return is 8, i.e the average of 6/7/8/9/10. In the cases where he draws a second card, his expected return is 5.5. His overall expected return is the mean of 8 and 5.5, which is 6.75. If the user were allowed to return up to two cards to the hat (i.e. up to three turns), then on the first turn, he only holds the 7/8/9/10, knowing the result of the two-turn game. His expected outcome then becomes 0.4*8.5 + 0.6*6.5, or 7.3. In the four-turn game, he only settles with 8/9/10 on the first hand, and his expected return is 0.3*9 + 0.7*7.3, or 7.81. You can obviously write a program to determine the strategy for an arbitrary number of turns. For a high enough n, the user's strategy is eventually to return anything other than a 10, but for low values of n, he needs to be careful about throwing away 6s, 7s, 8s, etc., for fear of getting worse cards on the redraws.
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