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Math guys help me understand this concept: Outs when on a Draw

I understand when you have a FD on the flop you have a 9/47 (9 diamond cards left on the deck / by 47 unseen cards) = 19.1 chance of hitting a diamond OTT and a 9/46 - 19.5 chance of hitting a diamond OTR. Therefore a total of 35% chance of hitting by the river.

The example above would make perfect sense to me if:
1. It was heads-up match (2 players)
2. Zero "burn cards"
3. Opponent has zero diamonds in his/ her hand

Therefore, all the diamonds would be available and we would be able to use the 9/47 and 9/46 method.

However, in the games we play in, its usually a full ring game 10 handed so that means that there are approximately 18 cards ( 9 players) + 1 "burn card" = 19 cards that will not to be seen OTT.


Assuming Full ring game 10 handed and we have a FD on the flop. 19 cards that will not be seen OTT out of 47 cards.

1. What is the probability that there are zero diamonds in the 19 cards that are not in play? I bet it is pretty minuscule.
2. What is the probability that there is 1, 2, 3, etc diamonds?
3. What is the average number of diamonds in 19 random cards?

I did a small sample size experiment based on the above. I took out 4 diamond cards and a non diamond card. I shuffled 47 cards and took out 19 random cards to see how many diamonds in 19 cards. My results were are as follow:

Test 1: 2 diamonds
Test 2: 5 diamonds
Test 3: 5 diamonds
Test 4: 3 diamonds
Test 5: 3 diamonds
Test 6: 5 diamonds
Test 7: 2 diamonds
Test 8: 4 diamonds
Test 9: 5 diamonds
Test 10: 3 diamonds

Average: 37 diamonds/ # of tests = 3.7 diamonds

Therefore, based on data above should we really figure 9 diamonds still available when we are on a FD or a lesser number (9 diamonds - 3.7 diamonds (Average) = 5.3 diamonds) since there will be 19 random cards that will not be seen OTT?

****Correction it should be 20 cards that will not be seen i didnt take into account the burn card before the turn. So even less diamonds available with 20 cards not to be seen OTT.


  • dpbuckdpbuck Posts: 1,989SubscriberProfessional
    Whether in other hands, the deck, or the burn card(s), they are "unseen", and each one is as likely to be a diamond or non-diamond as the other. In essence, each of the three cards on the flop has a 1/47 chance of being a specific card.

    Now, if you want to use your (small) sample to establish that 3.7 of the diamonds are gone, then that's fine, as you now have 5.3 diamonds remaining out of 27 "unseen" cards, or 19.6%, basically what you ended up with when not doing the distribution of the remaining cards.

    Does that make sense?
  • Bandgeek Posts: 140Subscriber
    If you remove the 20 from one side of the equation you have to remove it from the other side, so the answer is still the same.
  • beauregard Posts: 1,592Subscriber
    removing possible outs from your calculations based on the probability that your opponents had/folded them is a bit too goofy for my book. As others have implied, assuming your opponents have your outs is as likely as it is unlikely. So you'd drive yourself crazy making these assumptions and then basing your strategy on them.

    I, personally, use the 2X rule for figuring out my odds. Count your outs and multiply it by 2 for one street or 4 for two streets. So, 9 outs (FD) typically produces 36% for both streets, 18% for just the turn.

    Plus, if you've got the nut flush (e.g., AX) - isn't it possible that your Ace may be good too? +3 more outs?

    BTW, I think most folks use 44 or 45 unseen cards instead of your 47. Because you've got 2 in your hand, there's 3 on flop and and your opponent has 2. 52-7=45.

    Hope this helps.
  • GhostDogGhostDog Posts: 328Subscriber
    I too use the rules of 4 and 2. I think I learned them a long time ago from one of Phil Gordon's books. That's right, I read a Phil Gordon book! lol
  • PocketAceTrader782 Posts: 439Subscriber
    Listen to podcast 30. Simple poker Math

  • FuzzypupFuzzypup Posts: 2,298Subscriber
    2% per out one card to come
    1% for having a 2 gap backdoor straight draw (assuming it's open on both ends like 87x)
    2% for having a 1 gap backdoor straight draw (assuming it's open on both ends like 87x)
    4% for having a 0 gap backdoor straight draw (assuming it's open on both ends like 87x)
    2% for having a backdoor flush draw

    If you have 11 or more outs add 2%

    That is how I do it. Mind you that there are other things to consider like the chance if you have a non-nut BD FD that your opponent has the Ace of that suit. Or their position might have your str8 draw cards.

    So lets say the UTG PR raiser bets a 8h7h2c flop and gets 2 callers. Your 7c6h probably has less outs than you think since it is likely one of them is drawing to a better flush and the other has one of your straight cards since players are more likely yo play 0 and 1 gap connecting cards.
  • FreeLunch Posts: 1,308Pro
    edited July 2015
    Sounds like you should play stud- you might have the right kind of mind to track the dead cards.

    While you are right about the fact that most of the time your outs are not really in the deck, unless you are playing a game like stud you dont get this information. In games of imperfect information a lot of the variance comes from the unknown - you simply have to accept this concept as there is nothing you can do to change it.

    So, as a game of imperfect information we still need to use the best information we have to make decisions and the best information is what goes into the equity calculations the others talked about.

    There is one exception - there are cases multi-way where you think someone else may have have a similar draw or you have someone on a narrow enough range that you can take some cards or some fraction of a card out of the math. In other games this can be even more true - like in 08 when there are players that almost never play without have A2 in their hand. In draw games like 2-7 and badugi there are also times where you have to adjust based on others play.
  • FuzzypupFuzzypup Posts: 2,298Subscriber
    oops forgot to mention one thing. The backdoor percentages are for 2 cards coming not one. so 1/2. Yes I do play stud okish
  • FuzzypupFuzzypup Posts: 2,298Subscriber
    So let me edit


    2% per full out (FH, 2P, OED, FD)
    .5% for having a 2 gap backdoor straight draw (assuming it's open on both ends like 87x)
    1% for having a 1 gap backdoor straight draw (assuming it's open on both ends like 87x)
    2% for having a 0 gap backdoor straight draw (assuming it's open on both ends like 87x)
    1% for having a backdoor flush draw
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