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Probability

 
 
Reply Sun 7 Oct, 2012 07:24 am
The card game Ace-Jack-Two is played between a player and the bank. The
game goes as follows: a deck of 52 card is shuffled and then the bank reveals
three cards on a table. If an Ace, a Jack or a Two is among the three cards, the
bank gets a point, otherwise the player gets the point. This process is repeated 17 times. At this time, the one (bank or player) who has more points wins the game.What is the probability of the bank winning the game? What is the average
number of points scored by the bank per game?
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Type: Question • Score: 1 • Views: 1,176 • Replies: 5
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markr
 
  2  
Reply Sun 7 Oct, 2012 12:50 pm
@angelrocksvid,
0.67158
9.4
angelrocksvid
 
  1  
Reply Mon 8 Oct, 2012 07:18 am
@markr,
can i get the solution plzz ???
engineer
 
  1  
Reply Mon 8 Oct, 2012 08:11 am
@angelrocksvid,
First calculate the possibility of the player getting a point. There are forty good cards for the players (52 - 4 aces - 4 jacks - 4 deuces) so the probability of the player winning on any deal is:

P = (40x39x38) / (52x51x50)

The probility of the house winning is 1-P. You can find out the average number of points the house will get by multiplying the probability the house will get a point by the number of trials (17).

Once you know P, the chance that a player will get exactly N points is

(P)^N * (1-P)^(17-N) * 17! / N! / (1-N)!

The player loses if the number of points is 8 or lower so just compute the first eight terms and add them together. If you don't get Mark's numbers, you did it wrong.
DrewDad
 
  1  
Reply Mon 8 Oct, 2012 08:28 am
@angelrocksvid,
Is the deck shuffled between deals?
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engineer
 
  1  
Reply Mon 8 Oct, 2012 08:35 am
@engineer,
Quote:
(P)^N * (1-P)^(17-N) * 17! / N! / (1-N)!

Sorry, there is a typo in that last term. It should be (17-N)! instead of (1-N)!
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