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Old 12-11-2002, 07:07 PM
Fly1357 Fly1357 is offline
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Default I need the math, 4 cards to a flush and 3 left to draw.

My first 4 cards are 4 cards to a flush, I have 3 left to draw, what is the math to show me the odds of filling out the flush?
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  #2  
Old 12-11-2002, 10:39 PM
PseudoPserious PseudoPserious is offline
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Default Re: I need the math, 4 cards to a flush and 3 left to draw.

Hi Fly,

Here's the easy way:

Calculate how likely it is NOT to fill out your full. Call this q. Then the probability p of filling your flush is

p = 1 - q

To find q, you need to multiply the probability of not drawing your flush suit on each of your draws. We'll call those three probabilities q1, q2, and q3, one for each of the draws.

q = q1*q2*q3.

To find each qx, divide the number of non-flush cards (mx) by the total number of cards available for that draw (nx).

qx = mx/nx

So,

p = 1 - (m1/n1)*(m2/n2)*(m3/n3)

As an example, let's say that on your first draw there are 8 flush cards out of 38 available, on your second there are 7 of 33, and on your last draw there are 7 of 30. Then the chances of you hitting your flush are:

p = 1 - (30/38)*(26/33)*(23/30) = about 50%

Hope this helps...if you have any questions, just ask irchans. He's much better at this than I am.

PP
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Old 12-12-2002, 01:21 AM
BruceZ BruceZ is offline
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Default Re: I need the math, 4 cards to a flush and 3 left to draw.

As an example, let's say that on your first draw there are 8 flush cards out of 38 available, on your second there are 7 of 33, and on your last draw there are 7 of 30. Then the chances of you hitting your flush are:

p = 1 - (30/38)*(26/33)*(23/30) = about 50%



I only read this because I was going to say I thought you were sandbagging. Since you answered vastly more difficult problems correctly, how could anyone be "much better" at this elementary thing? But the above is no good!. Unless I'm missing something or you're psychic, how can you know on 4th street how many flush cards will be available on each future round? The correct probability of making your flush by 7th street if there are 8 unseen flush cards out of 38 unseen cards (meaning you see 8 of your opponent's cards and there is 1 flush card out in addition to your own, is

p = 1 - (30/38)*(29/37)*(28/36) = 51.9%
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Old 12-12-2002, 03:53 AM
PseudoPserious PseudoPserious is offline
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Default Re: I need the math, 4 cards to a flush and 3 left to draw.

Well, I didn't say it was practical example, just that that's how the method works -- IF you knew the information given, THEN that's how you'd calculate the probability.

Why bog yourself down worrying about actual numbers, especially when none were given in the original problem? The poster didn't specify that the game is 7-stud, after all...it could be 5-card triple-draw, solitaire, or 7-card draw and he's only looked at his first four cards.

When giving an example, you can choose to make a reasonable-sounding assumption that may or may not apply -- this sometimes makes the example easier to follow, but you run the risk of the example answer being taken as gospel even though it might not apply.

On the other hand, you could also use random numbers that would be rather silly in a practical situation -- if the silliness doesn't muddy up the method, then the example is just as clear and it's more obvious that the answer does not apply to the actual question being asked.

In this particular case, the numbers I used in my example were probably confusing, since I didn't explain why I picked them, or how you'd figure them out in a practical situation. I just assumed that he'd be able to figure them out...my bad [img]/forums/images/icons/frown.gif[/img]

Sorry for the confusion -- I in no way meant to imply that I'm psychic, although that would be a rather useful skill at the poker table. ("I fold, and can you pay me back that $500 I loaned you? Let's just say you shouldn't make any plans after next Thursday...")

Cheers,
PP


P.S. BruceZ -- I must admit that this statement confuses me:

I thought you were sandbagging

The guy needed help, I tried to help him. How is that sandbagging?
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  #5  
Old 12-12-2002, 09:17 AM
BruceZ BruceZ is offline
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Default Re: I need the math, 4 cards to a flush and 3 left to draw.

The "sandbagging" was meant as an abstract compliment. You said irchans could do this much better, but it's an elementary problem which I believed you to be capable of doing perfectly well, so it would be impossible for anyone to do it better. That would be like saying someone else is a much better tic-tac-toe player than you. You either know how to play or you don't, and if you do then nobody can be much better. See? Maybe sandbagging was the wrong word. Nevermind, I was off my meds [img]/forums/images/icons/smile.gif[/img]

But I strongly disagree that your formula adresses the poster's question, even for the games you mention or any other that I can imagine. I'm not arguing about the specific numbers in the example; I'm saying the concept behind the formula you used is fallacious. The poster asked the probability of making a flush by drawing 3 cards to a 4 flush. It is extremely important to realize and point out that no matter what game we are playing or how these cards will be drawn; the number of unseen cards and the number of remaining flush cards after each card is drawn is, for the purpose of computing this probability, known before ANY of the cards are drawn. In other words, the formula should be:

p = 1 - (m1/n1)*(m1-1/n1-1)*(m1-2/n1-2)

There is no m2,n2,m3,n3 involved in computing this probability. We don't know what those numbers will be yet (hence the psychic comment) and we don't need those numbers to calculate the probability in question. Unseen cards are unseen cards, no matter if certain ones become seen later on.


PS: OK. I suppose you can argue that my formula is a special case of yours where n1=n2=n3 and m1=m2=m3. But this is the only special case we need to answer the type of question I'm sure the poster is asking. Your general formula applies to a contrived situation, and makes it look like we have to know how the number of unseen cards changes or we can't compute the probability. This would confuse just about everyone who didn't already know how to do the problem, and not really show them how to compute what they are interested in.
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  #6  
Old 12-12-2002, 10:46 AM
PseudoPserious PseudoPserious is offline
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Default Re: I need the math, 4 cards to a flush and 3 left to draw.

All very good points, Bruce...I agree completely. I shouldn't have assumed that Fly would know how to find m1, m2, etc. That was pretty confusing of me <bonks self on head with a hammer>.

I won't even mention how I'm on a 7-game losing streak at tic-tac-toe.

PP
"Confusing innocent bystanders since 10/08/02"

P.S. Instead of m1=m2=m3, the 'special case' is m2=m1-1 and m3=m1-2 (likewise for n). Unless you're drawing with replacement, that is -- in that case m1=m2=m3 and n1=n2=n3. However, if you're drawing with replacement and the dealer says he's going to tear up a flush card after each draw, well, then m1=m2=m3, n2=n1-1, and n3=n1-2.

P.P.S. Actually, I do think irchans is better at this than I am. Not only is he better at math, but he's also very good at explaining things clearly (which I pretty much suck at). [img]/forums/images/icons/smile.gif[/img]
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