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Tuesday, March 12, 2013
We're playing a game. Cards numbered from 1-10 are placed in a hat. whatever card is drawn is the paid out amount. how much should i charge you to play this game? If you can redraw by placing the first card back into the hat, what is the new price?
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.
Very good post chhota bheem and the curse of damyaan
ReplyDeleteCard are drawn randomly from hat, so probability of each card is 1/10
ReplyDeleteSo 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.