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An Ace for a Pair (Posted on 2005-03-01) Difficulty: 3 of 5
Playing five-card stud poker with two friends one night, one of them, Kevin, accidentaly drops one of his cards on the table, the ace of hearts. My other friend, Nick, laughs and says, "I also have at least one ace in my hand." I have no reason not to believe him. Now, I do not have any aces in my hand, but I do have a pair of kings. Which of my friends is more likely to have at least a pair of aces (that is to say, at least one more ace) in his hand?

See The Solution Submitted by DJ    
Rating: 3.7500 (4 votes)

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Simulation shows different numbers (spoilers present) | Comment 3 of 12 |
(In reply to I think this is it. by Charlie)

The following program simulates the process, and counts only those instances that meet the observed conditions as trials.  Each set of 4 ace positions in the deck account for 5 simulations: one for each possible card that Kevin could have dropped.  The results of all these simulations are totaled for the final result.  In these simulations Kevin has at least one additional ace about 19% of the time and Nick 11 and a fraction %.  This disagrees with my calculated 17% and 9% respectively, but retains Kevin having the larger probability.

DEFDBL A-Z
RANDOMIZE TIMER
k0More = 0: k1More = 0: k2More = 0: k3More = 0: k4More = 0
k0NMore = 0: k1NMore = 0: k2NMore = 0: k3NMore = 0: k4NMore = 0
k0Tries = 0: k1Tries = 0: k2Tries = 0: k3Tries = 0: k4Tries = 0
' 5 sets of experiments at once--Kevin might drop 1st, 2nd,...,5th card
DO
 a1 = INT(52 * RND(1))
 a2 = INT(52 * RND(1))
 a3 = INT(52 * RND(1))
 a4 = INT(52 * RND(1))
 ' Kevin's cards are 0 thru 4: count his aces
 kevins = ABS((a1 < 5) + (a2 < 5) + (a3 < 5) + (a4 < 5))
 ' Nick's cards are 5 thru 9: count his aces
 nicks = ABS((a1 > 4 AND a1 < 10) + (a2 > 4 AND a2 < 10) + (a3 > 4 AND a3 < 10) + (a4 > 4 AND a4 < 10))
 ' My cards are 10 thru 14: count my aces
 mine = ABS((a1 > 9 AND a1 < 15) + (a2 > 9 AND a2 < 15) + (a3 > 9 AND a3 < 15) + (a4 > 9 AND a4 < 15))
 IF mine = 0 THEN
   IF kevins > 0 AND nicks > 0 THEN
    IF a1 = 0 OR a2 = 0 OR a3 = 0 OR a4 = 0 THEN
     k0Tries = k0Tries + 1
     IF kevins > 1 THEN k0More = k0More + 1
     IF nicks > 1 THEN k0NMore = k0NMore + 1
    END IF
    IF a1 = 1 OR a2 = 1 OR a3 = 1 OR a4 = 1 THEN
     k1Tries = k1Tries + 1
     IF kevins > 1 THEN k1More = k1More + 1
     IF nicks > 1 THEN k1NMore = k1NMore + 1
    END IF
    IF a1 = 2 OR a2 = 2 OR a3 = 2 OR a4 = 2 THEN
     k2Tries = k2Tries + 1
     IF kevins > 1 THEN k2More = k2More + 1
     IF nicks > 1 THEN k2NMore = k2NMore + 1
    END IF
    IF a1 = 3 OR a2 = 3 OR a3 = 3 OR a4 = 3 THEN
     k3Tries = k3Tries + 1
     IF kevins > 1 THEN k3More = k3More + 1
     IF nicks > 1 THEN k3NMore = k3NMore + 1
    END IF
    IF a1 = 4 OR a2 = 4 OR a3 = 4 OR a4 = 4 THEN
     k4Tries = k4Tries + 1
     IF kevins > 1 THEN k4More = k4More + 1
     IF nicks > 1 THEN k4NMore = k4NMore + 1
    END IF
   END IF
 END IF
 num = k0More + k1More + k2More + k3More + k4More
 den = k0Tries + k1Tries + k2Tries + k3Tries + k4Tries
 IF den > 0 THEN
   PRINT num / den;
   num = k0NMore + k1NMore + k2NMore + k3NMore + k4NMore
   PRINT num / den
 END IF
LOOP

 


  Posted by Charlie on 2005-03-01 21:16:34
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