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Card game (Posted on 2006-08-12) Difficulty: 3 of 5
Jack has challenged you to play a card game with him. The idea of the game is that a player gets a random card and after seeing it it's placed back in the deck. The player may get as many cards as he/she wants and the sum of the values of these cards represents the points this player got. However if the player gets an ace he/she gets zero points as total and the other player may try. How should one play this game (how many cards should be picked) for maximum chance of winning against Jack?
Face cards can be interpreted so that king is 13 points, queen is 12 points and jack is 11 points.

No Solution Yet Submitted by atheron    
Rating: 4.0000 (2 votes)

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Some Thoughts SLAP JACK!!!! | Comment 1 of 6

I shall assume that the I must make all my draws before Jack. 

Each draw has the odds of 1 in 13 that an ace is drawn.  The probability of drawing a non-ace n-times, with replacement  is then (12/13)n.
        Probability of
Draw    non-ace drawn
1st     0.923076923
2nd     0.852071006
3rd     0.786527082
4th     0.726024999
5th     0.670176922
6th     0.618624851
7th     0.571038324
8th     0.527112299
9th     0.486565199
10th    0.449137107

Playing the odds, as can be seen in the above table, 8 draws may be taken for better than 0.5 (50%) odds.  Yet, one must also take into account the odds by value of points drawn.  The average non-ace draw in points will be 7.5 points (summation of 2 to 13, divided by 2.)  In 8 draws, therefore, the average number of points should be 60.  If one draws 60 or more points before 8 draws and stops drawing, then one should have better than 50% odds of winning.  The question is then, how many more draws must one make if after 8 draws the player has less than 60 points?

Edited on August 12, 2006, 4:11 pm
  Posted by Dej Mar on 2006-08-12 12:46:22

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