Lotto Probability

[macgyver]

Probability humor

Just saw this on www.theonion.com:

Losing-Powerball-Numbers Announcement Enters 17th Hour

April 20, 2005 | Issue 41•16

URBANDALE, IA—The announcement of losing Powerball numbers for Saturday's $83,000,000 jackpot entered its 17th hour Sunday. "3, 15, 17, 35, 47, and Powerball 23," said Powerball host Bill Somerford, reading from his 237-page list of losing combinations. "7, 23, 40, 46, 52, and Powerball 24. 9, 13, 27, 40, 53, and Powerball 14. 12, 15, 18, 27, 52, and Powerball 26. 1, 11, 35, 46, 53, and Powerball 36." The losing numbers will be continue to be broadcast until 10:59 EST Wednesday, after which the losing-numbers announcement for the next drawing will begin.

 

[Just thinking]

Just saw this on the ONION

Losing-Powerball-Numbers Announcement Enters 17th Hour ...

Someone really needs to get a life.

(PS: Use above link for direct access.)

 

[ynot]

Before the door is opened: You know that one of two doors is empty.
After the door is opened: You know which one of the two doors is empty,
This is a new bit of information.

Wrong! - Before the door is opened: You know that at least one of two doors is empty. The host is not showing the contestant which of the two remaining doors is empty as the other remaining door could also be empty. The host is merely confirming that one is empty. The contestant already knew this.

 

[ynot]

Yes, it does. The usefulness is demonstrated by the fact that you have a better chance of getting the prize after the door is opened than you had before.
Yes, it matters.

This would only be so if the contestant was every given the option of choosing between one of the two remaining doors (neither was revealed). As one is always revealed however, this never happens. Effectively the host is saying - “Do you want to keep the door you have chosen, or do you want to have the other two doors?”

Revealing that one of the remaining doors is empty is a subterfuge to trick the contestant in to thinking that the choice between the two doors (that are not revealed) is 50/50. Given the difficulty people have with this problem, the trick obviously works well.

 

[Just thinking]

Drkitten -- ynot is arguing the same point from the position I detailed to you earlier. It assumes the host will reveal an empty door after you make a pick from 3 doors.

 

[ynot]

Drkitten -- ynot is arguing the same point from the position I detailed to you earlier. It assumes the host will reveal an empty door after you make a pick from 3 doors.

If the host doesn't open one of the non-chosen doors to reveal that it's empty, then it's not the Monty Hall problem.

 

[illuminatedwax]

The problem with information with regard to probability is that we're not asking if we are getting new information - we're asking if that information is relevant.

It depends on your perspective. In this case, Monty hasn't asked you to consider any other possibility outside choosing a single door out of three, and then considering switching to a second door once the third is shown to be empty. In this case, the information is not relevant because your calculations have not changed. But if you were a contestant with no idea what was coming, the information would be relevant, because you were considering the possibility if Monty asked you to switch from Door 1 to Door 2 without opening Door 3. You had calculated those odds at 1/3, but now that Door 3 is open and empty, those odds jump to 2/3. But those odds could be at 2/3 without any additional information: Monty could have asked you if you wanted to instead guess the prize was behind one of the other two doors. So when we calculate probability, we are only interested in information about probabilities.

This brings up another point: P(A) = P(B) <=> A is equivalent to B.

 

[ynot]

Monty hasn't asked you to consider any other possibility outside choosing a single door out of three, and then considering switching to a second door once the third is shown to be empty. In this case, the information is not relevant because your calculations have not changed.

Yes! - This is the Monty Hall Show.

But if you were a contestant with no idea what was coming, the information would be relevant, because you were considering the possibility if Monty asked you to switch from Door 1 to Door 2 without opening Door 3.

Yes but! - This is not the Monty Hall Show.

The second scenario above is irrelevant. We are discussing the Monty Hall Show, not another possible show. I have been a contestant on several game shows and assure you that the rules and conditions of the show are fully explained before play commences.

 

[illuminatedwax]

Here's another good pedagogical example of probability:

A woman has two children. What is the probability that both are boys?

Now you are informed that one of the children is a boy. Now what is the probability that both are boys?

 

[Just thinking]

Here's another good pedagogical example of probability:

A woman has two children. What is the probability that both are boys?

I'll bite ... P = 0.25

Now you are informed that one of the children is a boy. Now what is the probability that both are boys?

P = 0.33 (at least one is a boy)
P = 0.00 (exactly one is a boy)
P = 0.50 (the older/younger one is a boy)

 

[ynot]

Here's another good pedagogical example of probability:

A woman has two children. What is the probability that both are boys?

Assuming that the statement “both are boys” is not taken to mean that at least one is a boy.

Possibilities = BB - GG - BG. Odds that both are boys = 1/3

Now you are informed that one of the children is a boy. Now what is the probability that both are boys?

Possibilities = BB - BG. Odds that both are boys = 1/2

It seems too obvious. What's the catch?

 

[illuminatedwax]

P = 0.33 (at least one is a boy)
P = 0.00 (exactly one is a boy)
P = 0.50 (the older/younger one is a boy)

1) You didn't answer the question!
2) P(at least one is a boy) = 1 since i told you one was a boy!!

 

[illuminatedwax]

Assuming that the statement “both are boys” is not taken to mean that at least one is a boy.

Possibilities = BB - GG - BG. Odds that both are boys = 1/3


Possibilities = BB - BG. Odds that both are boys = 1/2

It seems too obvious. What's the catch?

I should remind you that order in fact matters, so your original possibilities are BB - GG - BG - GB.

 

[Just thinking]

ynot ... are you tired?

Examine your choices (outcomes) again, please.

 

[Just thinking]

1) You didn't answer the question!

Which question? ... I answered both (with explanations).

2) P(at least one is a boy) = 1 since i told you one was a boy!!

Huh? You never asked for that probability.

 

[ynot]

I should remind you that order in fact matters, so your original possibilities are BB - GG - BG - GB.

JT must be right - maybe I’m tired (or stupid). I don’t see why order should matter. BG as far as I can see is the same as GB. Both describe that there is two children, one of each gender. Please explain why order matters and then maybe I will “get it”.

 

[illuminatedwax]

Which question? ... I answered both (with explanations).

I guess I was confused by your explanations.

P = 0.33 (at least one is a boy)
P = 0.00 (exactly one is a boy)
P = 0.50 (the older/younger one is a boy)

So, knowing that one is a boy, why is the probability that at least one is a boy 0.33? It should be 1.
Knowing that one is a boy, the probability that exactly one is a boy is certainly not zero.
Not sure what you mean by the last one, but methinks you got the problem.

 

[illuminatedwax]

JT must be right - maybe I’m tired (or stupid). I don’t see why order should matter. BG as far as I can see is the same as GB. Both describe that there is two children, one of each gender. Please explain why order matters and then maybe I will “get it”.

It's no different than a coin toss. Mathematically, this is selection with replacement.

 

[Just thinking]

To all ...

Just as there are four outcomes to flipping a coin twice (HH - HT - TH - TT) there are four outcomes to having two children (BB - BG - GB - GG) where the order of births matches the labeled ordered pair.

Then there is the detail of what might be meant when one states something like -- "one is a boy". It could mean at least one is a boy or exactly one is a boy. So I cleared that up a bit.

OK -- if we simply ask what are the chances of both being boys, we see that it is only one outcome out of a possible four, or 25%. Just like flipping two heads in a row. If we ask what are the chances if at least one is a boy that both are boys, we eliminate the GG outcome entirely. Now, only one of three is the desired outcome, or 33%. (Of course, if exactly one is a boy, then P = 0 that both are boys -- self-explanatory.)

If, we state that either the older or younger one is a boy, then it becomes 50% -- since it's BB out of BB - BG or BB out of BB - GB.

That's it.

 

[illuminatedwax]

To all ...

Just as there are four outcomes to flipping a coin twice (HH - HT - TH - TT) there are four outcomes to having two children (BB - BG - GB - GG) where the order of births matches the labeled ordered pair.

Then there is the detail of what might be meant when one states something like -- "one is a boy". It could mean at least one is a boy or exactly one is a boy. So I cleared that up a bit.

OK -- if we simply ask what are the chances of both being boys, we see that it is only one outcome out of a possible four, or 25%. Just like flipping two heads in a row. If we ask what are the chances if at least one is a boy that both are boys, we eliminate the GG outcome entirely. Now, only one of three is the desired outcome, or 33%. (Of course, if exactly one is a boy, then P = 0 that both are boys -- self-explanatory.)

If, we state that either the older or younger one is a boy, then it becomes 50% -- since it's BB out of BB - BG or BB out of BB - GB.

That's it.

Pretty much. The point of the problem is to demonstrate independence and can trip people up. You could view the problem like this: having the two children is an independent event. P(B,B) = P(B) * P(B) = 1/2 * 1/2 = 1/4.

But if you know one is a boy, then it simply becomes P(B) = 1/2, since the two events are independent.

The other way is enumerating the options: BB, GG, BG, GB. In knowing that one of them is a boy, we have to assign one child that role. Most people, expecting a trick, would say "Now our options are reduced to BB, BG, GB, so the probability is 1/3" but really, GB isn't an option.

You people are too smart!