odds question

[fishyak]

So far, our issues are how to frame the math before we get to the issue of folding Kings. I don't know how to do the Excel spreadsheet stats thing. I have a minor in statistics, however, Excel had not yet been invented! Yes, I am a geezer. But I am interested in the question - does the # of limpers matter? Since 16 other cards were dealt out, each one of those cards had an equal chance of being a King. It appears that makes the worst case scenario odds closer to 50/50 that one King or more is out against you.

Then we would get to the non-random elements, folding. We know the some people would fold some hands PF if you raise, even if they have a King. That may be tough to quantify.

We may be left with the math on your worst case scenario as the only solid math. From that point forward, it becomes "guesstimates" that are not statistically valid.

[One Outer]

This thread is making my head hurt. I'm still eager to see a consensus result.

[Frond]

Count I'll get out my Abacus and get you the exact figures later.

[fishyak]

[ QUOTE ]
This thread is making my head hurt. I'm still eager to see a consensus result.

[/ QUOTE ]

There are two reasons I am taking this one on today. 1) The correct calculation of these odds is important to our game. And 2) I have my helicopter beanie screwed on real tight today!

[fishyak]

[ QUOTE ]
You are correct if we make the assumption that nobody would fold with a King in their hand. We could then say with certainty that there are only 39 unseen cards. My assumption that any of the five opponents will play any King actually has nothing to do with my calculation, now that I look at it.

1 - c(37,10)/c(39,10) = 1 - 348330136/635745396 =~ 0.452


I think that making the assumption that nobody would fold with a King is incorrect since there are many unplayable hands that include a King, even at 3/6. I understand your point though. It is certainly more likely that the hands that folded did not contain a King than hands that did contain a King. I have seen these points debated in the Probability forum and I'll see if i can dig some up.

[/ QUOTE ]

I don't think we even need to make that assumption. The fact is that 16 cards were dealt out, each with an even chance of being a King with a maximum limit of two kings being dealt out in the 16 slots.

From there, the events are NOT RANDOM. Folding is NOT a random event. I agree that some people would fold some hands with Kings in it. Further, I believe that people are making an analytical mistake when they say that the fewer limpers they face, the less likely it is that they will face a King. Inherent in that statement is the assumption that non-King hands will be folded at an equal rate to King hands, which clearly cannot be true.

I submit the best we can say is that there is a 45% chance that at least one King got dealt out against you. From there, I don't believe stat geeks can truly help because events from that point forward in the problem presented are not random.

I am interested in seeing if I am right or wrong in how I have structured the math in this issue.

[One Outer]

[ QUOTE ]
[ QUOTE ]
This thread is making my head hurt. I'm still eager to see a consensus result.

[/ QUOTE ]

There are two reasons I am taking this one on today. 1) The correct calculation of these odds is important to our game. And 2) I have my helicopter beanie screwed on real tight today!

[/ QUOTE ]

lol

We have something loosely in common. You minored in stats.
I dropped stats sophomore year and completely changed my political science concentration just so I didn't have to do it.

[Ricks]

I seemed to have missed your main question.

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Ricks, assumining a 9 seat table, how does the fact that all 8 opponents were given two cards that could have been Kings impact the appropriate formula?

[/ QUOTE ]

If we are unaware of their cards and no one has acted, thus we make no assumptions, it has nothing to do with the formula except that we have to consider the number of cards dealt to them.

After someone has acted you can make assumptions, which certainly supports your assertion that events are no longer random. I don't disagree with this.

Edit: added "except that we have to consider the number of cards dealt to them."

[Ricks]

[ QUOTE ]
I get about a 57% chance that a king was dealt to one of the eight people. Here is the calculation for the odds of no one getting dealt a king:

0.434 =(1-2/47)*(1-2/46)*(1-2/45)*(1-2/44)*(1-2/43)*(1-2/42)*(1-2/41)*(1-2/40)*(1-2/39)*(1-2/38)*(1-2/37)*(1-2/37)*(1-2/36)*(1-2/35)*(1-2/34)*(1-2/33)

Have I made an error somewhere?

[/ QUOTE ]

You were correct.

1 - c(45,16)/c(47,16) = 1 - 6.46626423E11/1.503232609E12 =~ 0.57

[Ricks]

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From there, I don't believe stat geeks can truly help because events from that point forward in the problem presented are not random.

[/ QUOTE ]

Perhaps binomial distribution . [img]/images/graemlins/smile.gif[/img]

[threeducks]

Look at this page

The absence of aces might be the same problem as the absence of Kings?