Statistics 101

The probabilities we're talking about, though, aren't conditional.When you say "what are the odds that you did win the lottery", you're not implying a conditional probability. You're just changing the way you indicated the subjunctive mood.Chiro is right that, in this case, it isn't about probability, it's about the English language. Obviously if you restrict the sample space to only one outcome, that outcome occurs 1/1. If you say "what are the odds that you've won the lottery given that you've won the lottery", well, congratulations, you're just repeating a statement that I originally addressed a hundred posts ago.But that's not the situation that's being talked about. Since we're using the subjunctive mood we're indicating that the subject is the odds of winning amongst all the outcomes, regardless of what actually wound up happening.That's the argument I advanced several pages ago. Nobody's addressed it, to my knowledge. Maybe you can be the first?

Actually that's backwards. The probabilities YOU are talking about aren't conditional. Everybody else is talking about conditional probabilities. In my first post the condition was that you had won. Thus P(A|A) which is equal to 1.If you can find a probability text that says P(A|A) ≠ 1 then we have something interesting to discuss. No, what we are saying is 'what are the odds that you have won the lottery given that x' where x is 'you picked all the numbers that were drawn'. Which is what I have been saying all this time. 1 is a valid probability. Though not in the real world. In the real world our justified tentativity means we say almost 1. It's amazing really. When you first brought this up in our dialogue I told you that was what I was saying, yet you still continued to argue the point. I never said it wasn't trivial. I only invested a single sentence to it in my original post, spending more time on it after you disputed it. I stressed then, that the problem was likely due to a misunderstanding, but you insisted that that was not the case and instead you believed the issue was that I was wrong. It is not the only situation being discussed. It is the issue you have been getting confused about this thread though. See my Message 21:

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If I hold a ticket in my hand that has the winning number on it as confirmed by me and my partner. I go to the lottery office and they confirm it is the winning number for the correct date. They run double checks through their authentication systems and it agrees that it is a winning ticket.

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What are the odds that the ticket I have is a winning ticket?

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To reword...Given that I have won the lottery, what are the odds I have won the lottery. You're answer was 'That's still 1 in 146 million'. Since you want to discuss how there is a problem with the English language here, might I might to the first word in the above quote? IF. Usually comes just before a conditional statement. That's probably what you'd expect to see in an Englishified conditional probability argument. I beg to differ, I addressed this in Message 25: S'ok Crash, I don't think you're stupid or ignorant or anything. I just think you got the wrong end of the stick. Hopefully you'll turn the stick around and we'll be reading from the same page while we shoo others out of the kitchen.In all sincerity, can you think of a good, non clumsy, non ambiguous way of saying P(A|A) = 1 using non-jargon English?Edited by Modulous, : No reason given.Edited by Modulous, : No reason given.

When the ball doesn't go where you wanted it to go.

I admit it, I was a little unsure of myself posting that, I really didn't think it through. It was more of a joke.I wish I could remember how to program, I would like to generate a program that flips the coins, and see what result we get.

No, those statements aren't synonymous. The statement I quoted was a statement in the subjunctive mood. This statement is a completely different statement.In a sense, the subjunctive mood is used to refer to alternate outcomes, for instance when one is formulating a hypothesis. That's why"what are the odds that you will win the lottery"
"what are the odds that you would have won the lottery"
"what are the odds that you did win the lottery"all have the same answer, because they're three different tenses in the same mood.That's been my point all along. Reference P(A|A) all you like; it doesn't change the fact that that's not the situation expressed in the English statement "what are the odds that you did win the lottery", and to try to conflate them is to be purposefully confusing. And I can't see why anyone would want to do that. No, my answer was "your question is nonsense". Because it is. It's the degenerate example. "What are the odds of a certain outcome, given that the sample space contains only that outcome?" That's how I'd put it. I certainly wouldn't try to do it in pluperfect subjunctive simply for the purpose of obfuscation.

Irrelevant. I told you exactly what we have been saying all along. Yes, all fo theose can have the same answer. None of those are what we are talking about. What we are talking about is what are the odds that you did win the lottery, given the information that you did win the lottery. The setup in the OP details a scenario where one has won the lottery, given this information, what are the odds that you won the lottery? See how, the words 'given' are used? That is a strong indicator that we are using a conditional statement. It's essentially the same as saying "What is P(A|A)?" It is not nonsense, it's just trivial. Are you saying that P(A|A) is nonsense? Incidentally, that was not your original answer. So you're original answer was, that P(A|A) doesn't change P(A) which is not really answering the question 'what is P(A|A)?'. That's what I thought you were saying. However, it can easily be interpreted as P(A|A)=P(A) which is nonsense. emphasis mine. Clumsy and jargon filled I'm afraid. I certainly wouldn't put it in anyway for the purposes of obfuscation. You (and possibly rR) are the only people who have been confused about this concept for the past 125 posts.What I do find completely amazing is that we are still talking about it, when I removed the English from the statement entirely and said it in maths in Message 105. I'm not sure how you might have thought that my mood was obfuscative then.I am fairly sure I have apologized for the ambiguity and any misunderstanding that I had any part in on several occasions in this post. I have tried wording it explicitly, I've tried breaking it down, I've tried giving it in purely mathematical languange. Perhaps there is another, better way, to put this rather trivial point? If I didn't find it, please accept my apologies for that as well.I will point out, however, that the people in this thread who are familiar with Maths - > English and English -> Math translations had little to no trouble discerning what I said.Edited by Modulous, : No reason given.

Sorry Crash, but everyone with a decent understanding of probability theory understood what Modulous meant, after his second post explaining the first - which is why there were so many people disagreeing with you.I really think you need to look back and try to understand how you could get things so badly wrong.

This thread, has reminded me of a classic scene in Red Dwarf. They are travelling faster than light, and Rimmer has just witnessed a future event in which Lister is killed: Not entirely on topic, but I thought we could do with a small lifting of the mood

Just to change the subject - I'm playing a game where I roll a ten-sided die, with a special rule called "spiking", where, if you roll a 10, you roll the die again and add the result to the first roll (so, 10 + the new result.) If that roll is a 10, you can do it again; theoretically, then, a 10-sided die (a d10) under this system is capable of generating any number, as long as you roll tens all along the way.Because the die generates conceivably any number, I'm not quite sure how to figure out the probabilities. What are the odds of rolling higher than 5? 10? 15?Further, in this game you can roll multiple dice and keep the highest result. What are the odds of beating 10 and 15 if you roll 2 dice and keep 1? 4 dice and keep 1?

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Because the die generates conceivably any number, I'm not quite sure how to figure out the probabilities. What are the odds of rolling higher than 5? 10? 15?

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This part is fairly easy. To roll 1 through 9, you cannot roll a 10; hence, you roll only once. The probability of 1 through 9 is 1/10.To roll 10 is impossible: if you get a 10, you roll again; if your roll less than a 10 you don't get to roll again.11-19, you have to roll a 10 and then 1-9, so the probability for each of these is 1/10 * 1/10 = 1/100.20, or any multiple of 10, is impossible if you think about it.To roll 10*n+k (where k is 1 through 9) will be (1/10)^(n+1).I'll have to think about the other question; class is coming up in a couple of minutes (Intro to Probability, ironically).

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Actually, if their god makes better pancakes, I'm totally switching sides. -- Charley the Australopithecine

The probability of rolling higher than 5 is 0.5 (you need to roll 6,7,8,9,10)The probability of rolling higher than 10 is 0.1 (you will beat 10 if and only if you roll a 10)The probability of rolling higher than 15 is 0.1 * 0.5 = 0.05 (You need to roll a 10 and then roll higher than 5)To get from the probability of beating a number with 1 die(p) to beating it with n dice use:1 - (1-p)^n(The probability that one die fails to beat the target is 1-p
The probability that n dice fail to beat the target is (1-p)^n
The probability that at least one die beats the target is 1- (1-p)^n)So for 2 dice the probability of beating 5 is 0.75
The probability of beating 10 is 0.19
The probability of beating 15 is 0.0975

Bugger, what system is this???? I remmeber it but i can't for the life of me remember which games use it. I was asked this same question then, and I seem to remeber drawing up some rather nice charts. But this was 10-12 years ago...

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Further, in this game you can roll multiple dice and keep the highest result. What are the odds of beating 10 and 15 if you roll 2 dice and keep 1? 4 dice and keep 1?

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Oops. I just noticed that PaulK already answered this one.Damn, these are good test questions! I gave this problem to my colleague who is teaching the statistics course, and he might make use of it, too.

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Actually, if their god makes better pancakes, I'm totally switching sides. -- Charley the Australopithecine

The Legends of the Five Rings roleplaying game uses this system*, plus other combinations of dice rolled/dice kept, indicated by k notation, where something like "4k2" means roll four dice, keep the results of two and add them together.*well, they did in first edition. Second edition used a different system, and then the property was turned over to Wizards and redone for D20 System.

Its probably one of the White Wolf roleplaying games. My guess is Vampire: The Masquerade.