Probability of 9-ball on Break

PoolBum said:
1/3 is correct. It is similar to the Monty Hall problem.
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Could you please explain this? If there are two children and one is a boy then they could both be boys or one could be a boy and one could be a girl. That's 2 possible outcomes.
 
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Drew said:
Could you please explain this?
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So, the couple has 2 kids.... that gives you 4 possible outcomes

1) BB
2) BG
3) GB
4) GG

He told you one is a boy, so he could've told you information about 1,2 or 3... so there are 3 possibilities left which are all equally likely, so a 1 in 3 chance that it's case 1.
 
skiflyer said:
So, the couple has 2 kids.... that gives you 4 possible outcomes

1) BB
2) BG
3) GB
4) GG

He told you one is a boy, so he could've told you information about 1,2 or 3... so there are 3 possibilities left which are all equally likely, so a 1 in 3 chance that it's case 1.
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In this case, the answer is slightly less the 1/3.

























There must be some non-zero chance that the other "kid" is a goat. :D
 
skiflyer said:
So, the couple has 2 kids.... that gives you 4 possible outcomes

1) BB
2) BG
3) GB
4) GG

He told you one is a boy, so he could've told you information about 1,2 or 3... so there are 3 possibilities left which are all equally likely, so a 1 in 3 chance that it's case 1.
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Incorrect...There are still only 2 possibilities as he did not require I order them. Your second and third options are identical. Either they could both be boys or one boy and one girl.
 
Drew said:
Incorrect...There are still only 2 possibilities as he did not require I order them. Your second and third options are identical. Either they could both be boys or one boy and one girl.
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Ordering is very important. Take note to how he chose his original words. He didn't tell you which kid was the guaranteed boy. #1 or #2.
 
Drew:
...There are still only 2 possibilities as he did not require I order them. Your second and third options are identical. Either they could both be boys or one boy and one girl.

briandlau:
Ordering is very important.
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Real things are "ordered". Two real kids = four real possibilities.

pj
chgo
 
Neil said:
So why aren't the odds of the 9 on the snap 50/50? It either goes in, or it doesn't. ???
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For the same reason that the odds of surviving a jump off a 40 story building are not 50/50.

(-:
 
Neil said:
So why aren't the odds of the 9 on the snap 50/50? It either goes in, or it doesn't. ???
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The odds are 100% that it goes in or it doesn't... but the distribution is another story all together.
 
Drew said:
Incorrect...There are still only 2 possibilities as he did not require I order them. Your second and third options are identical. Either they could both be boys or one boy and one girl.
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It doesn't matter if he required you ordered them or not it's about the fact that they are discrete objects which by nature implies what you are calling ordered.
 
The way that puzzle is worded is wrong...

The way that the neighbors kids puzzle is worded is too unspecific. It reminds me of many IQ questions that are too ambiguous to be valid. You can't count BG and GB as valid possible answers if the question is what are the odds that they are both boys. That is falacial logic with the wording of the question. If your question is what are the odds of them both being boys, then the only possible outcomes are two boys two girls and a boy and a girl. The wording of the question would have to include the order of the birth for the fourth possiblility to have any merit...
 
Jaden said:
The wording of the question would have to include the order of the birth for the fourth possiblility to have any merit...
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Actually, the fact that there is no ordering specified in the original question is what makes the BG and GB outcomes distinct and crucial. If I had said that the boy was born first, the answer would be 1/2. Since I did not specify which order the boy was born in, the outcomes BG and GB are distinct. To think otherwise would be to add information which is not provided in the original question.
 
Jen_Cen said:
... So, AccuStats says that the 9-ball is sunk on the break 1 out of every 35 breaks in professional matches. So, why do TV announcers and pool book authors insist on saying that the odds of it happening on the subsequent break are smaller? ...
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Because they're confused?

At one of the Reno Opens the players were "racking their own" and on one of the tables the nine ball was consistently moving towards the pocket. I recall a player who was genuinely apologetic for racking and making the nine twice in a row. His opponent checked the rack both times. The table was funny. Here's a related story: http://forums.azbilliards.com/showpost.php?p=976743&postcount=3

With a really tight rack, the nine ball doesn't move unless kicked by another ball. The Accu-stats stats are not for tight racks, they are for the slightly loose racks you see in the typical US tournament.
 
Jen_Cen said:
Before anyone excoriates me, I did do an honest search on this topic, and I didn't find anything.

Anyway, if you flip a coin 100 times, and it comes up heads 100 times, what is the probability that it will come up tails on the 101st flip? The answer, of course, is there is still a 50% chance that it will come up tails. The coin does not know or care or remember what has happened previously.

So, AccuStats says that the 9-ball is sunk on the break 1 out of every 35 breaks in professional matches. So, why do TV announcers and pool book authors insist on saying that the odds of it happening on the subsequent break are smaller?

It's still a 1 in 35 chance each time they break. Even if they did it 2 or 3 times in a row. On that 4th break, the odds are still 1 in 35!!
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That is correct,

The probability that event B occurs, given that event A has already occurred is

P(B|A) = P(A and B) / P(A)

Example 1: P(101H|100H) = ((1/2)^101)/((1/2)^100) = 1/2
Example 2: P(4-9B|3-9B) = ((1/35)^4)/((1/35)^3) = 1/35
 
Jen_Cen said:
Before anyone excoriates me, I did do an honest search on this topic, and I didn't find anything.

Anyway, if you flip a coin 100 times, and it comes up heads 100 times, what is the probability that it will come up tails on the 101st flip? The answer, of course, is there is still a 50% chance that it will come up tails. The coin does not know or care or remember what has happened previously.

So, AccuStats says that the 9-ball is sunk on the break 1 out of every 35 breaks in professional matches. So, why do TV announcers and pool book authors insist on saying that the odds of it happening on the subsequent break are smaller?

It's still a 1 in 35 chance each time they break. Even if they did it 2 or 3 times in a row. On that 4th break, the odds are still 1 in 35!!
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I knew this was going to go into outer space...:confused:
 
Jen_Cen said:
... Anyway, if you flip a coin 100 times, and it comes up heads 100 times, what is the probability that it will come up tails on the 101st flip? The answer, of course, is there is still a 50% chance that it will come up tails. ...
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If it's a theoretical coin, I'd agree with you. If it's a real coin, then I would bet on heads because it is a two-headed coin. Phony cheater-coins are far more common than a fair coin landing heads 100 times in a row. I have two of them myself. Similarly, if I see the nine ball fly straight into the corner pocket three times in a row, I'm not thinking to myself that it only has a 3% chance on the next break. I'm thinking the table is funny.

Trivia question: In his million-dollar run of 11 racks of nine ball, how many times did Earl Strickland make the nine on the break?
 
Bob Jewett said:
Trivia question: In his million-dollar run of 11 racks of nine ball, how many times did Earl Strickland make the nine on the break?
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Was it five?
 
PoolBum said:
Suppose you know that your neighbors have two kids, and you know that at least one of them is a boy. If the odds for having a boy are exactly 50-50 for any given birth, gievn what you know about your neighbors what are the odds that both of their children are boys?
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There is no interpretation here, 1/3 is absolutely false. The question states the odds that both children are boys. We already know that one of them is a boy. The other child only has 2 options. The question never stated whether the child was born first or second nor did it ask what gender the first or second born was. Therefore, the order of birth is irrelevant. All you can answer is whether or not both children are boys.

At least one child is a boy. What are the odds that the second born child is a boy and both children are boys?

This has now become a completely different question and changes the probability. Now the possibilities look something like this.

First born is the given boy second born is also a boy.
First born is the given boy and second born is a girl.
Second born is the given boy and first born is a girl.

Now the odds are 1/3. But unfortunately the purpose of the exercise would be lost if the question were worded in that particular way.
 
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