MATH QUESTION (Open Face Chinese Poker)

[MadMojoMonkey]

Even if you don't know how to play, the math question is combinatorial.

If you start with 5 cards, and 3 of them are of 1 suit (say clubs), what are the odds of drawing 2 more clubs to complete a flush?. There are 8 cards to come.

Notation: (4,1) means "4 choose 1"

My guess:
(10,2)*(37,6)/(47,8) ~= 33.3%
But that only counts the chances of drawing EXACTLY 2 clubs.

So we need to take the sum of the odds of drawing 2 through 8 clubs.
SUM{n=2..8}{(10,n)*(37,8-n)/(47,8)} ~= 55.0%

Does this look right?

[spoonitnow]

With 47 cards left in the deck, 10 of which are in our suit (we'll call it clubs), the chances of getting exactly zero clubs are:

(37/47) * (36/46) * (35/45) * (34/44) * (33/43) * (32/42) * (31/41) * (30/40) = 0.1228

The chances of getting exactly one club are:

((37/47) * (36/46) * (35/45) * (34/44) * (33/43) * (32/42) * (31/41) * (10/40)) * 8 = 0.3274

The chances of getting zero or one club is 0.3274 + 0.1228 = 0.4502, so the chances of getting two or more will be 1 - 0.4502 = 54.98%.

Cliff Notes: You're right, but you're doing it the hard way.

[jackvance]

Cliff Notes: You're right, but you're doing it the hard way.

I'd do it your way but I wouldn't call it the hard way to use the more uniform statistical methods. They allow you to solve problems where it's too complex to do it by hand.

[spoonitnow]

I'd do it your way but I wouldn't call it the hard way to use the more uniform statistical methods. They allow you to solve problems where it's too complex to do it by hand.

[jackvance]

[MadMojoMonkey]

I'd do it your way but I wouldn't call it the hard way to use the more uniform statistical methods. They allow you to solve problems where it's too complex to do it by hand.

Which is my intention. Spoony got me thinking about the utility of his approach, though.

What I really want to be able to do is to get a feel for what are the odds to continue drawing to a flush vs. giving it up to pair that hand and avoid a foul.

A more useful, general formula:
GIVEN: I have w clubs, I see x "dead" clubs, there are y cards left in the deck and z to come.
LET: q = 13-w-x be the number of clubs left in the deck

SUM{n=(5-w)..MIN(q,z)}{(q,n)*(y-q,z-n)/(y,z)}

EDIT: MIN(q,z) is the smaller of the 2 values q and z, representing the maximum number of clubs that can be drawn.

[Galapogos]

I'd do it your way but I wouldn't call it the hard way to use the more uniform statistical methods. They allow you to solve problems where it's too complex to do it by hand.

Probably my favorite exchange I've seen on here in quite some time.

[spoonitnow]

My point was that "the hard way" wasn't referring to using combinations. "The hard way" is adding up the cases of 2 clubs through 8 clubs instead of just using the cases of 0 clubs + cases of 1 club.

[MadMojoMonkey]

If I was doing it by hand it would make a big difference. I'm using excel, though, so going with the combinatorial notation is my preference.

I'll keep using "clubs" as the suit here, but the odds are the same for any suit.

Using the previously stated formula:

For the first turn, you remain a slight favorite, no matter what position you are in, as long as the number of clubs played is less than or equal to the number of players who have played before you.

If you are 2nd to act and you see the person to your right has played 1 club, you are a 55.7% favorite to make your flush over 8 cards to come.
If you are 3rd to act and you see 2 clubs on the table, you are 56.5%.
If you are last to act and you see 3 clubs on the table, you are 57.7%.

Can you confirm one of these using a different approach?

EDIT: these stats are the same, no matter how many players are playing. Also, this is the break point where you are a slight favorite. If you see a single more club in any of the above circumstances, you are no longer a favorite.

[a500lbgorilla]

Enjoyable books on the topic: Unfinished Game, Drunkards Walk

[MadMojoMonkey]

If you have 4 clubs to start:
First to act, 84.4%
2nd to act and the player to your right played a club flush, you are still 58.6% to draw at least 1 club over 8 cards.
3rd to act and you see 6 clubs on the table, you are 53%
last to act and you see 6 clubs on the table, you are 59.2%

[rong]

Oh, math! I thought you said meth, was looking forward to the thread.

[jackvance]

My point was that "the hard way" wasn't referring to using combinations. "The hard way" is adding up the cases of 2 clubs through 8 clubs instead of just using the cases of 0 clubs + cases of 1 club.

[a500lbgorilla]

Faces aside spoon is clearly correct here.

[supa]

Oh, math! I thought you said meth, was looking forward to the thread.

Some years ago I actually sold meth for a living, you would not be looking forward to that thread.

[JKDS]

ama about selling meth?

[jackvance]

Looked to me like he critiqued the use of combinations when it could be done so easily by hand. It is very common to do 1 minus the odds of the cases we don't want, I learned this in high school. Simple misunderstanding it seems.

Some years ago I actually sold meth for a living, you would not be looking forward to that thread.

Did you cook it too? Because I like that show.

[rong]

Supa, you just become cool. AMA required imo.

[rong]

Some years ago I actually sold meth for a living, you would not be looking forward to that thread.

Also did you sell it or sling it?

[MadMojoMonkey]

Taking 1 minus the sum of failures doesn't eliminate the need to do a summation. A computer is doing the calculations anyway, so the notion of saving time is moot. I just want to know if I've set up the equation properly, so that I can answer that TYPE of question.

Solving the best first move is just a simplest case scenario, and really answers the question, "Should I go for a flush at all?" I'm not trying to solve what is the best first move. I'm trying to answer the question "Under what circumstances should I abandon my attempt to get a flush and go for a weaker hand?"

Can you at least wait until the subject of the thread has been addressed before you completely change the subject? Please?

[gargarvin]

Your formula is correct. Taking 1-SumOf(ProbNotHappen) is the same as SumOf(ProbHappen). The former equation however uses less summations for all time when the number of outs you need is less than half of the number of remaining draws. Note that this formula will not work for multi-card straight draws or pair draws as there are other factors to consider.

Here is an app I recently made, looking for feedback...it won't let me direct link (assuming bc I am new?)
"https://play.google.com/store/apps/details?id=com.ofcprobabilitycalculator&feature=se arch_result#?t=W251bGwsMSwyLDEsImNvbS5vZmNwcm9iYWJ pbGl0eWNhbGN1bGF0b3IiXQ"

[rong]

ama about selling meth?

This needs its own thread.

[husler777]

Hard to tell. Other apps out there already. I bought this one:

Deleted link: First post to a paid app, not happening. jyms


I hate to say good things about apps which help other players, but this one is accurate and easy to use. Paid for itself the 1st hand Maybe enough people will learn to play so open face gets mainstream enough to hit most casinos. The online app right now allows for too much cheating to use with anyone but friends.

[husler777]

Sorry about the link. Not sure of the site rules and just replied to a message with a link in it too. Can you route me to the site rules so I don't break any going forward?

[jyms]

Top of every page has " Forum rules Updated" link. It even shows on the page when doing replies

[husler777]

Yes I saw it now...thank you for that. Still I'm a little unsure how the rules are interpreted. What I see is the topic is questions about open faced chinese poker odds and I found a useful app that calculates that. Can I share that somehow if it's on topic? If not, what confuses me is several posts before someone has a direct link to a paid app which they are clearly selling (fine with me....hope they make a Million $) and that seems to be OK. But a link to another app I wanted to share gets deleted. I'm not sure I saw in the rules why one is allowed and another not. Can you please clarify? Thanks!