2-7 Triple Draw, odds and chances

I have been playing quite a bit of game lately as well as HORSE. I decided to have a look at the maths in some of these and started with 2-7 Triple draw, mostly because i didnt actually know the rules of the game.

Anyway here are my ideas.

1. The single card needed situation. You have something like 2,3,6,7,X where X is no good a pair or A,K.

On the first draw you need a 4 or a 5 which will arrive 17% of the time, which is not good. However if you continue for all three draws you will make your hand 43% of the time, this is very good in a multiway pot. You will always have the odds if 3 are playing and you go all the way. Importantly if you allow the 8 to be an out also your % of making an 8 high low hand goes to 58%. Making you a favourite even in Heads up as long as the other person isnt standing pat.

2. The double card draw. This time you have something like 2,3,7,XX where both the X cards need to be changed.

This time your odds of getting both cards on the first draw is less than 6%. It is also only 16% if you go through all 3 draws.

This has interesting implications. If you are heads up and the villain draws 2 cards or more he will miss 84% of the time. Bluffing opportunities here are large, also if you only need to draw one card you are a an 81% favoutite to win the hand, probably as good a set of odds as you will get anywhere in poker.

any comments much appreciated.

1. Can you explain your calculations a little bit
2. When you are heads up since the first draw 8 is often good and sometimes 9 or T as well

im not great with the numbers but in multiway pots you arent allowing for the amount of lo cards already out there and that will make ur hand.

but obviously as u can see you shud be betting when you are a card or 2 ahead, geting a little more complicated when out of position tho.

Thanks for reading, I have posted maths stuff like this before (in the MTT forum and never had a response). i will try and explain a bit more

1. Drawing a single card, By definition this means you already have 4 cards 7 or less. e.g 2367X. You have 8 outs the 4 and the 5 each available 4 times giving a probability of 8/47= 17%.
You may need to reduce these odds in certain circumstances. If your 4 good hole cards are the same suit you only have 6 outs giving 6/47=13%.

If you commit to all three draws then you get three chances to make your hand so if unsuited 17%+17%+17%=51% and if suited 39% of the time. To get the final odds of 43% we must take account of the times when your hole cards are of the type 2345X and the 6 will make a straight so it isnt an out.

Going throught exactly the same calculations but this time allowing a low hand with an 8 gives the 58% I mention in the original post.

2. Drawing with two cards. For you to hit both cards on one draw and not drawing a pair =0.058 or 6%. If you committ to all three draws then you could hit in the following ways 2,0,0 or 0,2,0 or 0,0,2 or 1,0,1 or 0,1,1 again adjusting for straights and flushes gives us 16%.

Finally if you need one card and villain needs two. By definition you must have at least one of his outs and he may or may not have one of yours. For villain to win he must hit and you musty miss chance is 0.16*0.57=9% he is 9% to have the best hand at show down.

You have discounted the fact that your chances increase in each draw
Lets take the 1st draw example
8/47 = 0.170
2nd draw = 8/46 = 0.174
3rd draw = 8/45 = 0.178

Calculating the chance of hitting in any of the 3 draws in not adding the numbers, it is incorrect statistically - without going through the math process think of it this way - if I gave you 6 draws what is your chance of hitting - if you add it up like you did you will have more then 100% to hit, however since you only picked 12 cards out of the 47 there is a chance you will not hit it.
The correct math procedure is 1-((1-170)(1-0.174)(1-0.178)) which comes down to about 44%

The drawing 2 card example is a bit more complex to do, and I dont really have the time to do it now, since it also depends on your starting hand, however just to start you off assuming you start with 2,3,7 , your chance of drawing perfect 2 (ignoring the other people cards of course) is 12/47*8/46 = 4.44%, your chance of hitting one but not the other is 12/47*38/46+35/47*12/46 = 40.5%, your chance of missing both is 35/47*34/46 = 55.0%

Your maths is correct TLR but you will notice that the answers are the same I worked out 43% you said about 44%. The chances do increase with each draw but by such a small amount (less than 2%) I decided to ignore it. I like to simplify the calculations so that they can be done at the table.

On the two card you are correct i miss counted the outs. Overstating the probability by 1.4%. I will recalculate when i have time

Originally Posted by hopeful

Your maths is correct TLR but you will notice that the answers are the same I worked out 43% you said about 44%. The chances do increase with each draw but by such a small amount (less than 2%) I decided to ignore it. I like to simplify the calculations so that they can be done at the table.

On the two card you are correct i miss counted the outs. Overstating the probability by 1.4%. I will recalculate when i have time

We ended up with the same number by coincidence I assume

No, not coincidence different ways to get to the same answer. One looking at the chance of not loosing the other looking at the chance of winning.

Another situation. Should You bet on the end when the villain draws a card?
We will assume the villain has 4 really good cards (any 4 of 2,3,4,5,6,7)
and decides he needs to draw one to improve.

I will also make an assumption that the hero has a reasonable hand has a few good low cards. The two are likely to be sharing some cards I will assume that they have two in common.

When the villain draws his last card he has the following probabilities

He makes a pair 10/48 = 22% ( remember we discounted two shared cards)

He gets a big card (A,K,Q) 12/48 = 26%

He gets a medium high card (J,T,9) 12/48 = 26%

He gets a good to nut card (the rest of the time) =30%

This is the best case scenario for the villain as we have not looked at his stright or flush possibilities which will exist and we have not discounted his good to nut hands based on our own cards. Consequently it is a conservative estimate from the Heros point of view.

Now I will assume he will allways call with a good to nut hand.
He will call 75% of the time with a medium hand
25% of the time when he has a high card
only 10% of the time with a pair (for many people this would be zero)

this means in total he will call
30%+0.75*26%+.25*26%+.10*.22%= 58% of the time. (we know this is too high based on this maths.

So should we bet regardless of our hand. Well to decide that we look at expected value. I will assume we loose every time we are called which is definately not true but it is our worse case.

E(Bet)=Pot-0.58*bet

So the expected value will be positive as long as the pot is bigger than 0.58*bet. Which it almost always is.

So this suggests if we are first to go, or we are checked to we should always bet if the villain drew a card.

I think your numbers are off again

There are 52 cards in the deck, there are 5 cards in our hand that are accounted for, 4 cards in villian's hand + the 5th one which we know is not a card he needs, and the discards by both us and villian in the first two draws, assuming we both discarded 2 in the first draw and 1 in the second draw there are 16 accounted cards, if he has 10 good cards his chance of hitting are 10/36.

One more thing you have to factor into your calculation is that if he has a good hand he will raise if we bet, usually the pot is big enough for us to call with almost anything, so sometimes we lose 2 bets by betting and not just one

yea you are right, i was working it out as if a reshuffle happens after every draw. I will put that right this morning and add the possibility of a raise to the EV calculation.

Firstly thanks TLR for all the feedback, I hope its helping me get somewhere with this thread. Hopefully other readers will benefit also.

This is a revised version of the previous post trying to take into account the issues raised by TLR. I have come up with an identical answer and the same conclusion. It is interesting because it suggests a pretty clear strategy. The strategy suggested is

If on the third draw you are heads up and the villain draws one (or more cards) then
1. You should always bet if first to act
2. If you get raised you should probably fold without the goods.

Maths and assumptions to follow.

Deciding the odds of success at the final draw is awkward. The remaining cards are a function of the draws and discards that have already taken place plus the known cards in the hero’s hand. I will have to make a lot of assumptions to get somewhere near the actual probabilities.

I will accept TLR’s assumption that we both discard two in the first draw and one in the second. 5 discards in all as far as we are aware the stub has 52-5=47 live cards. Of course 5 of them are in the villains hand and he is about to discard another so from the villains point of view he is looking at a 41 card stub. Should we adjust that for the cards that are in our hand? It is very difficult because we are not sure if we hold any of the villains outs.

I will assume that we hold 2 of the villains outs and reduce the stub to 36.


We will assume the villain has 4 really good cards (any 4 of 2,3,4,5,6,7)

When the villain draws his last card he has the following probabilities

He makes a a good hand 10/36 = 28%
He gets a medium high card (J,T,9,8) 14/36 = 30% 2 assumed to be discarded
He gets a big card (A,K,Q) 9/36 = 25% here I assume 3 of these big cards have been discarded by the villain or hero.
He gets a bad card pairs straights and the like(the rest of the time) =17% one of which was discarded (in the case of a pair)

Now to try and work out the EV of betting when we are first to act.
Firstly if we have a weak hand. We will fold to the raise and loose to a call.

Will the villain bluff? Definitely sometimes, say 5% of his weak holdings are bluffed.
this means in total we will loose

Now I will assume he will raise 80% of the time and call 20% of the time with a good to nut hand.
He will call 72% of the time with a medium hand raise 3%
23% of the time when he has a high card raise 2%
only 10% of the time with a pair or worse (for many people this would be zero)

.28+0.75*30%+.25*25%+.10*.17%= 58% of the time.

Expected value calculation is now

E(Bet)=Pot-0.58*bet

So the expected value will be positive as long as the pot is bigger than .59*bet. (roughly need the pot to be bigger than 1 bet) Which it almost always is.

So this suggests if we are first to go, or we are checked to we have positive EV if we bet fold all hands.

Should we call the raise?
Villain is raising 80% of good hands, 5% of all other hands. This means when he raises there is a 94% chance he has a good to nut hand. Without a good to nut hand we should probably fold. It is unlikely that we are getting the odds on our weak hands.

E(without a made hand)= .06*pot-.94*bet

EV is positive when .06*pot>.94*bet
So pot >15.7*bet.
Those are big odds to get.

Do Flushes and straight count?

F;ushs and str8s count, best hand is 23457 when not all are the same suite
A count as high only

I try to keep my maths simple at the table.

I play this game limit, so I don't really care too much about pot odds, because they will always be excellent with a strong hand multi-way after the first draw, which more than makes up for any "bad call" you might make first round of betting.

Taking this into account, I don't work out pot odds, and instead concentrate on the action in front of me, trying to decide how strong the others are. I only apply pot odds if there was no raise to start, and nothing but checks after the first draw.

On the button, I'll call any 27 or 23 to no raise. If the others draw lots, then I'll probably draw one and start firing, standing pat with whatever crap I have, unless someone suddenly stands pat. Otherwise, I'll draw the three and fold if I don't improve.

Otherwise, I'll play 3 card draws with a duece, and preferably not a six, pretty much to any action first round. After the first draw, if I don't improve, I'll use the info I have from other draws to decide if it's wise to continue. If there's more than one person drawing only 1, or anyone standing pat, I'll get away. Otherwise, I might tag along for 1 bet.

If I do improve, and I'm drawing on a seven, I'm in 'til the end, and calling an eight better at the finale.

If I'm just drawing on an eight, I'll stay in unless someone's standing pat. I might keep drawing on a perfect eight.

Nines are not worth bothering with. I see too many people standing with nines (and sometimes tens) from the beginning. I'm pretty certain that if your opponent is standing from the off with a nine, and you have a 1-card seven draw, you're favourite. And if he mucked the nine first draw, he's roughly 50-50 to improve or get the nine back.

Maybe a nine or ten is worth calling at the final if you hit it last card and he drew one as well. The last bet is the round where a bluff is most likely to occur.

Most people at low levels play rougher than this, that's why I'm making money at this game.


Finally...

Only today, I got dealt JJTTT out of pos. I raised, reraised, stood pat, the lot. He drew until the end and missed, folding maybe jack high or pair of threes to a 20c bet into a $1.20+ pot. I showed my beautiful hand. I was lucky he missed, otherwise I'd look a right fool. Heh.