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

  • Thread starter Thread starter clarkd25
  • Start date Start date
  • Tags Tags
    Probability
Click For Summary
In a standard 52-card deck, the probability of drawing at least one King, Queen, or Ace in three attempts without replacement is calculated by first determining the odds of not drawing any of these cards. The probability of flopping none of the desired cards is 44.7%, leading to a 55.3% chance of drawing at least one. This calculation is crucial for making informed prop bets in poker, as a greater than 50% chance indicates a positive expected value (EV) for the bet. Understanding these probabilities can enhance strategic decision-making in games like Texas Hold'em. Thus, betting on drawing at least one of these cards is statistically favorable.
clarkd25
Messages
2
Reaction score
0
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.
 
Physics news on Phys.org
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.
 
The standard _A " operator" maps a Null Hypothesis Ho into a decision set { Do not reject:=1 and reject :=0}. In this sense ( HA)_A , makes no sense. Since H0, HA aren't exhaustive, can we find an alternative operator, _A' , so that ( H_A)_A' makes sense? Isn't Pearson Neyman related to this? Hope I'm making sense. Edit: I was motivated by a superficial similarity of the idea with double transposition of matrices M, with ## (M^{T})^{T}=M##, and just wanted to see if it made sense to talk...

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
2
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
Replies
41
Views
17K