Probability of Winning in Poker with Three Jacks and Two High Cards - Week #79

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SUMMARY

The discussion centers on calculating the probability of winning in poker when dealt three Jacks and two high cards (2 of diamonds and 5 of hearts). The primary focus is on determining the likelihood that an opponent holds a higher three of a kind, specifically three Queens, Kings, or Aces. No participants provided correct solutions, indicating a challenging problem for players. The analysis emphasizes the importance of understanding poker hand probabilities and strategic decision-making.

PREREQUISITES
  • Basic understanding of poker hand rankings
  • Familiarity with probability calculations
  • Knowledge of a standard 52-card deck
  • Experience with strategic poker play
NEXT STEPS
  • Research poker probability calculations for various hand combinations
  • Study the concept of expected value in poker
  • Learn about advanced poker strategies and bluffing techniques
  • Explore simulations for different poker scenarios using tools like PokerStove
USEFUL FOR

This discussion is beneficial for poker players, game theorists, and anyone interested in enhancing their understanding of poker probabilities and strategic gameplay.

Jameson
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You are playing some poker with a friend. Say you are dealt 5 cards from a 52 card deck and have three jacks plus the 2 of diamonds and the 5 of hearts. What is the probability that your friend has a higher 3 of a kind, that is three Q's, K's or A's?
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There were no correct solutions this week. :(

Solution:
5 cards from the 52 are already accounted for thus our hand is chosen from a pile of 47 cards. For each situation - 3 Kings, Queens or Aces - they all have the same probability so we can find one of them and multiply by 3.

There are 4 Queens from which to choose 3 and 43 cards left over from which to choose the other 2 cards. However these last two cards also cannot be the same, so for the first card we have 43 choices and for the second card we have 39 choices.

$$P[\text{3 Queens}]=\frac{\binom{4}{3} \left(\frac{43 \times 39}{2!}\right)}{\binom{47}{5}} \approx 0.0021865$$

$$P[\text{3 Queens}] \times 3 = \boxed{0.00656}$$
 

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