midconnectors prob of hitting a straight (challenging)

[deanmarcus]

I have been trying to find the odds of this for ages:

If i hold 89 off suit preflop, what is the chance, in percent, that by the river i hit a straight? (not including a straight on the board that doesn't involve my cards)


HELP!!!

[lwestatbus]

This doesn't seem right but here's what I have:

You have five chances to draw one of eight cards (from 50) that are immediately adjacent to your connectors 5 x 8 / 50 = 80%

You have four chances (one of them was needed for the first card) to draw one of eight cards (from 49) that are immediately adjacent to the three card sequence you have built 4 x 8 / 49 = 65.3%

You have three chances to draw one of eight cards (from 48) to complete the straight 3 x 8 / 48 = 50%

The chance of all three of those things happening is the product of the probabilities = 26.1%

I'm not sure that I treated the denominators (50, 49, 48) properly. I also treated the order in which the cards are drawn as immaterial. However if you hold 8-9 and the first card drawn is a Queen then you have reduced the value of the draw to 5-6-7. Note also that this is only good preflop. Once the flop appears the probabilities change overwhelmingly. Also, this analysis assumes that you are using both of your cards. Drawing T-J-Q-K gives you the straight but one without much value.

Edit: In thinking about this it seems that you need a decision tree approach where you calculate the probabilities of each of the four straights you can build using both of your cards (5-6-7-8-9, 6-7-8-9-T, 7-8-9-T-J, 8-9-T-J-Q) and sum them. Now my head hurts.

Anyone else?

[deanmarcus]

Thanks for replying, that's an interesting approach but the solution is way too high.

I don't think the chance of hitting a straight with 89 is more than 20% higher than the chance of hitting a flush with suited cards.

SOMEONE HELP

[deanmarcus]

I'm thinking something like finding all of the possible community card combinations that will give you a straight if u have 89.

i.e. 5 6 7 x x
6 7 10 x x
7 10 J x x
10 J Q x x
obviously any order any suit.

where x is any card. including one of your hole cards or even a card that gives a higher straight draw, as long as there is no straight or flush on the board.

if a number can be found for the amount of possible community cards to make a straight, you could divide the total number of possible community cards (50*49*48*47*46/(5!)) = 2118760

I don't know how to work out possible number of successful community cards. Someone please work on this!!!!!

[snoogins47]

You've hit on probably the simplest way to think about it.

Given that, you seem smart enough to do the calculations yourself. Why aren't you working on it? ;P

It would probably be way, way easier by the way (this is a gut feeling, and I've had a lot to drink, so be warned) to isolate situations using both of your cards, and situations using one of your cards.

So, for instance, find out how many boards are
567xx
679xx
79Txx
9TJxx
5679xx
and...so on.

Then there's plenty of really tedious ways to figure it out. If I was still working 8 hours a week at a shitty motel job where five of those hours were downtime, I'd probably do it for you. Too bad for you.

(The tree thing is one way. For case 1: how many times is the first card a 5? 46 no, 4 yes... when it's yes, how many times is the next a 6? when it's no, how many times is... etc. etc. Very tedious. Did I say tedious? I meant exciting)

I think somewhere I worked out numbers of how often you flop different draws (OESD, GSSD, etc) with connectors, one gappers, etc. If I can manage to find those you could probably do this pretty easily, but that was years ago, so I don't hold out much hope. Do the work yourself (it's FUN! heh), or, better idea, go to a stats or computer coding forum and ask the question in the form of a "challenge." Much better chance of getting the answer from a bunch of dorks with no life outside of that sort of thing, than trying to get a bunch of dorks with no life outside poker to figure it out. Us poker folk are lazy.

[suitedaces84]

Combinatorials are the way to do this. There are a total of 50c5 boards given your hole cards. You need to figure out how many of those combinations will make a straight. For example, there are 4c1*4c1*4c1*43c2 ways to make a straight 5 to 9.

[deanmarcus]

that all sounds good, I've never tried to work out odds in hold'em, i just find the odds on sites like these and use them when i play. If i had 5 spare hours to work this out i probably could but i'm in my final year at school and am struggling.

I'm sure that one of you pro prob people can do this in around half an hour and come up with an answer expressed as a percent.

thou hast to be in it to win it, but to win properly thou must play properly

[davepoker]

well mike caro on his site had some tables

with QJ in the hole, you flop an oesd 6.04% of the time.
http://www.poker1.com/mcu/tables/Table21.asp

then, since you are about 32% to fill up, the odds should be around 2%, unless i am missing something.
so does 1/50 seem to be correct?

[MrDarling]

The way I see it from the link you provided you got :
you've got about 2% chance flopping a str + 6% chance flopping an OESD (which will hit about 16% by the river). To this you have to add the times you flop pair+gut draw. Then you're going to have to add back door draws..

So you're not there yet Sir.

[suitedaces84]

4c1*4c1*4c1*43c2 ways to make a staight 5-9
4c1*4c1*4c1*43c2 ways to make a staight 6-T
4c1*4c1*4c1*43c2 ways to make a staight 7-J
4c1*4c1*4c1*43c2 ways to make a staight 8-Q
4c1*4c1*4c1*4c1*42c1 ways to make a staight 9-K

Add all these up and divide by 50c5 and you'll have your answer. This number will include the times there's a flush on the board so your staight doesn't play, the times there's an 8 and 9 on the board and the board plays, the times your hand is very weak because the board is scary, etc.

[deanmarcus]

I think i understand what you have done and it sounds good. After deducting when there is a flush on the board and, the 9-K straight (because that one isn't cool) and when the board has 9-Q on it, i got 10.71%.

That number feels too high even though the calculations seem accurate. To verify, did you include 43c2 (and 42c1 for the last) in the calculations to eliminate the higher card which would then make that straight part of another one? (e.g. for 5-9 straight, you assumed that a 10 couldn't come out as well as you would then have a 6-10)

Anyway, thanks for replying, it was really helpful.

Dean Marcus.

[LeeG]

According to the calculations here, the chances of small unsuited connectors flopping a straight or a straight draw is 2.64%

.33% of flopping a straight, and 2.31% of flopping a draw.

www.pokerstove.com/analysis/unsuited.php

[deanmarcus]

i looked at that page and had absolutely no idea what was going on, i prefer suitedaces84's hypergeometric approach; the most simple and effective in my opinion.

[suitedaces84]

Yes. To make a straight 5 to 9 you can pick 1 of 4 fives, 1 of 4 sixes, 1 of 4 sevens, then you pick 2 more cards that cannot be Ts. If you take out the 5, 6, 7, 8 and 9 which have already been used as well as the 4 Ts which you are not allowed to pick you are left with 43 cards to choose those 2 from.

The straight 6-T, 7-J and 8-Q work the same way.

[holdemhelpem]

The odds for getting a straight with 89 is that uses at least one of your hold cards 9.035%. This does not account for flushes. The best cards to have for a straight are TJ with 9.119%.

The worst odds for a straight of 0 to 3 gap hole cards is A2 with 2.558%. I have a complete list on my website.