Help with 2-7 triple draw probability
I am struggling with calculating the probability for a 2 card draw in 2-7 triple draw. I understand the mechanics of the game and how to play (i think) however i am trying to learn the math behind this so i can get comfortable with pot odds etc etc.
here is the scenario:
you are dealt 2,3,7,10,k
you discard 10,k
1. what is the probability that you catch both a 4 and a 5 over the next 3 draws?
2. What is the probability of catching both by the second draw?
Thank you very much for your help!!
2 Replies
Assuming no blocker effects, this is a straight combination problem from high school probability class.
There are other issues, however. First is that say you catch 8/4 on the first draw...You aren't going to break the 8 and go for the wheel.
Here is how you calculate the odds on the first draw: There are 16 combos of 5/4. There are a total of (47*46)/2 combos of cards you can get. You are 16/1081 to get the wheel; or about 1.5%.
To calculate the odds you get it on the second draw, you have to take the chance of catching 1, and then you are 4/45 to catch the other one. If you bricked the first draw, then you (45*44)/2, *16 to catch perfect on the second draw...
Don't forget to eliminate the 1.5% from the first draw.
I played triple draw for the first time in a Big Bet Mix tournament recently. I am not sure if I prefer it to NL 2-7 but I think of the probabilities similarly to holdem. Multiply the "outs" by 2 for each street. Like calculating runner-runner probabilities. So what you are asking is what is the probability of runner-runner to the nuts. A three card draw would be runner-runner-runner. Of course you can also draw to the virtual nuts 54, 64, 65, and even 84, 85, 86. Depending how the action goes, even 9 might be good: 96,95,94.
So on your first "street" of your first draw you have 8 cards; so 16% and on the second street of your first draw you have 4 cards so 8% and these two things must happen so they are compounded events. So the chances are 8% of 16% on the first draw. This approximation yields 1.3% (close to what 3for3Poker calculated). Then you can multiply this by the options for the virtual nuts. So you have 6x1.3% or about 8% on each draw of making a strong hand (12% with the strong 9s). And across all three draws you've got 24%/36% chance of making a winning hand - if all you did was draw 2 each time
Of course, that isn't what you would do but you can see that with a strong draw of 2,3,7 you can easily afford 1/3 pot with 2 card draw and multiple streets.
Maybe. I don't really know.
