Lucky Cards
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
After an intense week of playdowns, the National Draw Poker Championship has come down to the final hand between two fierce competitors: Tex Holden, the defending champion; and Rae Sonheim, the unknown of the tournament, who has amazed the spectators with her cool bluffs. In what is the first of a number of amazing coincidences of the night, both players have an equal stake.
Although Tex kept his poker face intact, his excitement mounted as he examined each card he was dealt. The first card he received was his lucky card, the King of Clubs, which he considered a very good omen. Perhaps it was, as each of the next four cards were of the same value as each other, giving Tex one of the best possible hands: four of a kind.
Meanwhile, across the table, Rae waited until all the cards were dealt before looking at her cards. She, too, was pleasantly surprised as she had received four of a kind as well, along with the Queen of Spades, which is her lucky card.
Each player, believing he or she had an unbeatable hand, prepared to end the tournament then and there with a careful strategy designed to win all the chips.
After many rounds of raises and counterraises, at long last, all the chips were riding on this one hand. The better hand would take the pot, and the trophy.
Tex, having had his last raise called, gleefully displayed his hand first. All eyes turned to Rae, but, as usual her face showed no emotion.
Before the outcome is revealed, can you determine the probability that Tex has retained his title?
Answer
As a single King and Queen have been identified in the hands, neither player can have 4 Kings or 4 Queens. That leaves 11 possibilities (in decreasing value):
4 Aces
4 Jacks
4 Tens
4 Nines
4 Eights
4 Sevens
4 Sixes
4 Fives
4 Fours
4 Threes
4 Twos
The probability that Tex has any of these hands is the same 1/11.
If Tex has 4 Aces, Rae's 4 of a kind cannot outrank his, so the probability that he would win the hand is 1.
If Tex has 4 Jacks, then of the remaining 10 possible hands that Rae holds, only 1 can beat him. Therefore the probability that Tex would win with 4 Jacks is 9/10.
With 4 Tens, the probability that Tex would win is 8/10, and so on.
If Tex has 4 Twos, any four of kind held by Rae would win, so the probability of Tex winning would be 0.
Therefore, the probability that Tex will win can be expressed as:
P = 1/11(10/10 + 9/10 + 8/10 +7/10 + 6/10 + 5/10 + 4/10 + 3/10 + 2/10 + 1/10 + 0/10)
P = 55/110
P = 0.5
Tex's poker face finally fell when he saw Rae smile sweetly and slowly fan out her 4 Eights, which beat his 4 Sixes. He quickly recovered his composure, and congratulated Rae on a hand well played.
Of course, Tex no longer considers the King of Clubs to be his lucky card.
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Comments
blondebookworm
Aug 21, 2006
 nice job. you put a TON of work into that one. 
flub92
Aug 26, 2006
 I enjoyed that one, keep up the good work. 
pistons_32_rip
Aug 26, 2006
 awesome job ppl can tell you thought that one out 
jsdodgers
Sep 13, 2006
 why cant there be a four of a kind with the kings or queens? 
Gizzer
Sep 14, 2006
 Tex has 4 of a kind PLUS a King. Rae has four of a kind PLUS a Queen. Neither can have 4 Kings or 4 Queens. 
jsdodgers
Sep 15, 2006
 OH, but still in texas holdem there are five cards in the middle and each person has two cards. If one card from each person is not included, than that is only 7 cards in all left not counting the king and queen but two different pairs of four would need 8 cards and only 7 are left. 
Gizzer
Sep 16, 2006
 This is DRAW poker, not Texas Holdem. 
jsdodgers
Sep 16, 2006
 his name makes it seem like it is texas holdem 
Gizzer
Sep 16, 2006
 What else can I say? That's just his name. He is playing in the National Draw Poker Championship. 
unstumped
Nov 28, 2006
 i KNOW I'M LATE ON THIS BUT;
THE ODDS ARE 50/50
He either drew a higher 4 or didn't. 
sumit
Mar 09, 2007
 nice question  it was not that difficult as it appeared in the beginnning  long questions always seem s to be tough. 
Jimbo
Mar 24, 2007
 As with many probabilty questions, the answer when calculated suggests a simpler explanation. Since they both have four of a kind, the winner will be the one that has the larger 4oak. Now since they are both equally likely to have been dealt any of the hands it comes down to a 50/50 chance for either of them to have either a larger 4 oak or a smaller 4 oak. Therefore 0.5. Amazinigly simple after the pieces of the puzzle have been fitted together. Thanks for that one! 
Eshootzi_scrs
Apr 24, 2007
 Just read this now but thought the same. It was 50 /50 from the beginning . The second time through though I noticed it said Rae had been dealt four of a kind with the Q of spades it didnt say in addition to, so it could have been one of the quads. 
tpg76
Apr 28, 2008
 Nice teaser  and fun writeup!
But in my opinion (only), whereas it was obvious that Tex did not have four Kings (the next four cards were all the same), it was not obvious that Rae didn't have four Queens ("4 of a kind along with the Qofspades"). This would have made the teaser harder, and in fact, makes it much more likely that Rae has won. If the interpretation (as you've said) is that she does NOT have 4 queens, then of course each player's fifth card is irrelevant and it is a tossup (as it turned out). So maybe it could be clearer that Rae's face card, too, is her odd card.
Thanks. 
javaguru
Dec 14, 2008
 So much irrelevant information. Try this one: Tex has a hand he thinks is good; Rae has a hand she thinks is goodwho wins? The 50/50 answer was so obvious that I reread the problem figuring I must have missed something. The explanation and calculation is laborious when the answer is: we have no relevant information regarding the relative strength of the two hands, so the probabiltiy is 50/50. 
Zag24
Jan 24, 2009
 By the description, it wasn't clear that the challenger couldn't have 4 queens. It just said that she had 4 of a kind, and that she had the queen of spades. I assumed that was the trick, making her a favorite. That problem is actually very tricky to figure out, since her remaining four cards could either by three queens and any one other card, or 4 of another rank.
If you're assuming that neither player can have quad kinds or queens, then all the analysis is silly. The problem is clearly symetrical  both players have quads drawn from the same list, with no advantage to either. You don't have to do the complex calculation. 
opqpop
Sep 24, 2010
 The answer is obviously 1/2. You don't need any calculations to see this if you just notice the symmetry. 
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