Select Page
Poker Forum
Over 1,292,000 Posts!
Poker ForumBeginners Circle

Robb's 3000th Post

Results 1 to 34 of 34

Hybrid View

Previous Post Previous Post   Next Post Next Post
  1. #1
    MadMojoMonkey's Avatar
    Join Date
    Apr 2012
    Posts
    10,456
    Location
    St Louis, MO
    Start with 2 identical decks and take all the diamonds out of one of them. Turn the diamonds into suns by adding 4 points with a red pen or marker. Shuffle them into a standard deck. They can always be separated into 2 standard decks later (if you can handle that the diamonds are a little different on one deck). Oh yeah, ask a question:

    If you had a 5-suited deck (65 cards), what would the hand rankings be (including 5-of-a-kind)?

    Same question with the 3-suited deck (39 cards).
  2. #2
    Quote Originally Posted by MadMojoMonkey View Post
    If you had a 5-suited deck (65 cards), what would the hand rankings be (including 5-of-a-kind)?
    For your new home game, just swap the flushes and full houses, and 5-of-a-kind beats everything obv. Here's the breakdown:

    We have 65c5 = 8.3 million hands (standard deck is 2.5 million hands).
    • Exactly 1 Pair = 13c1 * 5c2 * 12c3 * (5c1)^3 / 65c5 = 3575000 / 8259888 ~ 43.38% (standard deck is 42.26%)
    • Exactly 2 Pair = 13c2 * (5c2)^2 * 11c1 * 5c1 / 65c5 = 429000 / 8259888 ~ 5.19% (standard deck is 4.75%)
    • Exactly Trips = 13c1 * 5c3 * 12c2 * (5c1)^2 / 65c5 = 214500 / 8259888 ~ 2.60% (standard deck is 2.11%)
    • Straight (no straight flushes) = (10 * (5c1)^5 -50 ) / 65c5 = 31200 / 8259888 ~ .38% (standard deck is .39%)
    • Flush (no straight flushes) = (5c1 * 13c5 - 50) / 65c5 = 6435 / 8259888 ~ .08% (standard deck is .20%)
    • Full House = 13c1 * 5c2 * 12c1 *5c3 / 65c5 = 15600 / 8259888 ~ .19% (standard deck is .14%)
    • Quads = 13c1 * 5c4 * 12c1 * 5c1 / 65c5 = 3900 / 8259888 ~ .047% (standard deck is .024%)
    • Straight Flush = 50 / 8259888 ~ .00061% (standard deck is .0015%)
    • Quints = 13 / 8259888 ~ .00016%


    Have fun teaching your new game to the feesh
    Last edited by Robb; 05-13-2012 at 11:42 PM.
  3. #3
    Quote Originally Posted by MadMojoMonkey View Post
    Same question with the 3-suited deck (39 cards).
    After your 3-year-old f**ks up your only deck of cards two minutes before your home game starts, you can play 39-card poker by switching the flushes < straights, and no quads obv. Here's the breakdown.

    We have 39c5 = 575757 hands - interesting number - who knew?


    • Exactly 1 Pair = 13c1 * 3c2 * 12c3 * (3c1)^3 / 39c5 = 231660 / 575757 ~ 40.24% (standard deck is 42.26%)
    • Exactly 2 Pair = 13c2 * (3c2)^2 * 11c1 * 3c1 / 39c5 = 23166 / 429000 ~ 4.02% (standard deck is 4.75%)
    • Exactly Trips = 13c1 * 3c3 * 12c2 * (3c1)^2 / 39c5 = 7722 / 575757 ~ 1.34% (standard deck is 2.11%)
    • Straight (no straight flushes) = (10 * (3c1)^5 - 30 ) / 39c5 = 2400 / 575757 ~ .41% (standard deck is .39%)
    • Flush (no straight flushes) = (3c1 * 13c5 - 30) / 39c5 = 3831 / 575757 ~ .67% (standard deck is .20%)
    • Full House = 13c1 * 3c2 * 12c1 * 3c3 / 39c5 = 15600 / 575757 ~ .08% (standard deck is .14%)
    • Straight Flush = 30 / 575757 ~ .01% (standard deck is .0015%)


    Don't ground the 3-year-old or send him to bed without dinner. Easy game.

    Edit: fixed flush calculation MadMoJo pointed out below.
    Last edited by Robb; 05-14-2012 at 01:49 PM.
  4. #4
    MadMojoMonkey's Avatar
    Join Date
    Apr 2012
    Posts
    10,456
    Location
    St Louis, MO
    Quote Originally Posted by Robb View Post
    Found a typo in the 3-suit frequencies: the formula is correct, but the result should be 0.67%

    Also: Is the numerator for the straight flush eq's based on:
    10c1 * [n]c1
    where [n] is the number of suits, or are there missing terms (all of which are 1, so moot)?
  5. #5
    Quote Originally Posted by MadMojoMonkey View Post
    Found a typo in the 3-suit frequencies: the formula is correct, but the result should be 0.67%

    Also: Is the numerator for the straight flush eq's based on:
    10c1 * [n]c1
    where [n] is the number of suits, or are there missing terms (all of which are 1, so moot)?
    You are exactly right. Another way to do it is simply to count them since they're so few. There are 10 possible straights, 5-high through A-high, and there is one of each variety in each suit, so 10 * [n] is the correct number.

    Either way, you're right.

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •