What Are the Odds of Drawing at Least One King, Queen, or Ace in Three Attempts?

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SUMMARY

The probability of drawing at least one King, Queen, or Ace from a standard 52-card deck in three attempts without replacement is 55.3%. This calculation is derived from first determining the probability of not drawing any of these cards, which is 44.7%, using the formula (40/52) * (39/51) * (38/50). This insight is particularly relevant for poker players, especially in Texas Hold'em, as it informs the expected value (EV) of prop bets involving drawing these high-ranking cards.

PREREQUISITES
  • Understanding of basic probability concepts
  • Familiarity with card games, specifically Texas Hold'em
  • Knowledge of combinatorial mathematics
  • Ability to perform calculations involving fractions and percentages
NEXT STEPS
  • Study advanced probability theory, focusing on card games
  • Learn about expected value (EV) calculations in gambling
  • Explore combinatorial analysis techniques
  • Research strategies for making prop bets in poker
USEFUL FOR

This discussion is beneficial for poker players, mathematicians interested in probability, and anyone involved in gambling strategy, particularly those looking to enhance their understanding of betting odds in card games.

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In a 52 card deck if you choose 3 random cards such as any King any queen or any ace
(12 cards in total) what is the probability of being able to draw at least one of these cards with 3 attempts and not replacing a drawn card. Basically my question is regarding the odds in Texas Holdem of choosing 3 random cards and flopping at least one of them. It seems to me that on the first card the odds of success are 12/52 and then on the second draw if not successful would be 12/51 and then 12/50, but I just don't think I know how to do the math for the problem. Please Help.
 
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Duh!

I went about solving the problem the wrong way. Instead of trying to figure out the probability of flopping at least one of the cards I should have solved for the probability of flopping none of the cards which would be (40/52)*(39/51)*(38/50) which results in 44.7%. This would mean that the probability of flopping at least one of my randomly selected cards woul be 55.3%. My reason for figuring out this problem is that I like to make prop bets when I play poker and wanted to know if this was a good even money bet for me to select 3 cards and betting that one of them would come on the flop. By having chance greater than 50% of hitting one of my cards this bet has a +EV.
 

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