View Full Version : Rare stud/8 situation...

08-26-2002, 10:53 PM
I was playing stud/8 tonight and found myself ina rare situation. I was dealt rolled up 2's and I wasnt the bring-in. What is the probability that on the next hand I play i will be dealt rolled up 2's and wont be the bring in? (8-handed)

08-27-2002, 12:39 AM
The odds or probability are the same for every set of first three cards in seven card stud(in any game) -- asumming things are essentially random. Probability might be 7.5414781297134238310708898944193e-6

08-27-2002, 01:42 AM
More Rare or Really Rare:

About two years ago at the Hawaiian Gardens Casino in Hawaiian Gardens, CA: the following event occurred in a Stud8 game. Two players split the pot with each having a seven high straight flush (for high) and a straight six (for low). Obiviously the suit of the dueces involved was a moot point.

What are the odds (or the probability) of this event?

And what are the odds of this event occurring for three players? 8~)

08-27-2002, 09:09 AM
The probability for the next hand is the same for any given hand.

For this to happen someone needs to be have the 2c as their door card and you need to be dealt the other three deuces.

Let's start with you:

3/52 = chance of your first card being a non-club duece.

2/51 = chance of your second card being a non-club deuce.

1/50 = chance of your door card being the remaining non-club deuce.

Multiplying yields your chance of getting these three deuces is 1/22100.

There are 49 unseen cards left and we'll assume you have seven opponents. So the chance is 7/49 that someone will get the deuce of clubs. Multiply this by our previous result.

Thus you will get rolled up deuces and not be the bring in once every 154,700 hands on average.

08-27-2002, 05:20 PM
Up cards possible:

2s - 2h, 2s - 2d, 2h - 2d, all in addition to the 3 possibilities where the bring in has 2c, so that's 6 ways total to be rolled up without being the bring in.

Now recompute. :-)


08-27-2002, 05:21 PM
The probability for the next hand is the same for any given hand.

yes i know that. i just phrased my question that way so i wouldnt get smartass answers of p=1 because it already happened.

you are missing some possibilities, I could get (2c2d)2s and the person with the 2h up would be the bring in etc...