If you discuss calculating the probability of a sequence of events, in 
the first case you just multiplied independent probabilities because they
were independent. In most situations, though, the events are conditioned:
an event happening after a preceding event is conditioned by what
happened before, and we note that by noting probability of the first event
times the probability of the second event/given that the first has happened, is the 
conditioning, and so forth and so on to the last event which would be conditioned by the 
event that just preceded it.

Put another way, p(first event) * p(second event/first event) etc...

he gave an example from the top of p 6 of the handout #1 - check it out.

now, with dependency of this type, the probability is larger than if
events are independent. when you calculate, estimate, or look up
probabilities for rare events to occur, and assume that events are
independent when in fact they are not, you will calculate a smaller
probability than is actually the case. You will be UNDERESTIMATING the 
true 
probability.

Recently in clinical research this has been very important, esp with 
reference to the OJ trial. They argued in court over the probability 
calculations for the way you calculate DNA based probability. The top 
notch probabilists in the world, used by prosecution, made a mistake,which
was caught by the defense probabilist, who noted the change by a factor of 
10. What mistake did he make? THIS is the mistake he made (assuming  
independent events when in fact they are not.)

Someone remarked that she understands the numbers but the premise seems 
wrong. This seems to have confused our instructor, who remarked that he 
would have preferred it the other way around because then he could answer 
the question!

Look at page 6. For some reason people have problems with this. The
probability of collective events or the mutually exclusive system.

You're looking at one outcome, but are willing to consider more than one 
class of that outcome.

Draw one card from the deck - you can only observe one card. What is the 

probability of a Jack or a two? Well, only one can happen. If one 
happens, it excludes the other possibility. What's probability of Jack?
4/52. Same for the two. There are 52 cards in a deck - 4J and 4 twos, so
8 out of 52 cards would fulfill this.

what's probability of Q, 2 or A? 12/52
So this is pretty straightforward.

What if you need to know probability of J and 2 during two draws (but not 
in any order)? Well, it could happen two ways. J,2 or 2,J. These two events 
are of course mutually exclusive. If you see J,2 you can NOT see 2,J. But
either of these events would be a success. So by the law above, after
calculating their multiplicative probabilities, you would add them.

The mutually exclusive law has nothing to do with conditioning.

Mutual exclusivity has to do with, in a total outcome for a set of 
events, where one outcome would exclude the other - go back to the
beginning. I just said, J followed by 2 has probability of 4/52 * 4/52
with replacement. or 4/52 *3/52  w/o replacement. But, what if I said from
two draws, what is the possibility of observing a J and 2, in either
order? You can have either 2, J or J,2, but those 
events are mutually exclusive. It's no longer J-->2, it's now just J,2, 
so you can account for the probability either way that it happens. so you
multiply by the multiplicative rule for each of the separate probabilities
of J-->2 and 2--->J and then to fit the question as asked we'd add the two
separate probabilites together, under the mutually exclusive rule. 

the replacing or not replacing of the cards doesn't have anything to do 
with the order in which they happen. either could happen in either order, 
regardless of card replacement. It has NOTHING to do with replacement or 
not replacement, just am I specifying an order, or not?

Q:[unintelligible question]
A: that's true

Q:[mumblemumblecardmumble]
A:at the moment it is, and I said that.

Q: [yadaydaydayadhavetodowithmumble]
A: well, ok, there are three considerations: with or without replacement, 
that's true, the multiplicative law, but I was looking at it either way. The 
probabilites would change, but this part of the discussion would be the same.

Q:[gigglemutteermutterisn'tmumble]
A:which would negate what you just said, in other words. the calculations 

would be different numerically, which means these would be smaller, in 
either case, J first or two first. did I get to YOUR question?

Q: what i mean is, the first draw could be muttermutter, the second draw 
would have to be the opposite, so wouldn't the muttermumble 8/52?
A: no, it wouldn't suit the question

Q:right no, what i'm saying is, you're gonna get one of the two to suit 
the question, so isn't there a lower probability of the other one the 
somethingmumbleflurg?
A:where are you getting that?
Q: don't you add them together?
A: nononononono. you can't add them together. This gets calculated, this 
gets calculated, THEN you add them together. You can't logically do what you 
are saying.

Q:[hisssssssssmutter]
A: awwww, yikes. Yes, if this is 16/52squared then that would be 32 over 
52 sqared. other questions? Ok.

Important thing before we proceed, you have a multiplication rule under 
independence and a lack thereof when events are conditioned.
in each case, multiplicative.

then you have a collective rule, which means when we're asking for more 
things which will satisfy our ends than can actually happens, and in this 
case you add them.

you'll see, when we do the lotto thing. Now, my lotto thing is out of 
date. The PA lottery just had its 25th anniversary, i know you were all out 
celebrating. This is wildcard lotto, this is [mumblemumble] who knows? I [mumble] these 
things 
[mumblemumble]

who wants to be my shill? everyone's looking the other way. The correct 
answers are highlighted.

what do you usually get in a lotto game? two numbers for a dollar. Ok, if 
that's the probability - here it is

3 8 3 8 3 8 0

if that's the probability for one number and you get two numbers to win 
with, does that sound like a good chance? double it. if you double that, it's 
one out of 
919190

what is the odds of winning the 6 number game? one in [mumble]. Yes.

Q: is that accurate? what if more than one person wins?
A: yes. we're not talking about how much money you win, we're talking 
about if you get the numbers right. my luck, the day i win, the whole state of 
PA willwin.

[mumble]
they'r on the same card
oh, no no no. two sets of six. she's correct. two games. you pick two 
sets of six numbers for a dollar, which leads to the probability as said
before. this is a bit 
more obvious than the next ones. any questions? Good thing on the money

tripp: if it's over 2 mill, you should play
him: no, only if its up to 6 mill. if I win a million dollar jackpot, 
i'll be depressed, because they give you an annuity of 31K each year. wouldn't 
that upset you? not that i couldn't spend the money. but at 6 mill, if I win 
it all,  ....[tape flipped] ... they won't let you walk in and say here's
2 mill, cover the game. 

tripp: if it's 10 mill, maybe you should take that risk
A: i think this guy has contacts. when it gets up to 6 mill i start 
buying tickets. 
but not many. now.

final lotto example. win 5, miss one. you can also win money if you only 
hit 5. this problem often strikes some controversy because some people don't 
know how to multiply [voices obscuring speaker at this point].

notice [sound of chalk on board.....he's writing] this is a unique way to 
win with 5 winning #s and one losing #. but because there are six ways to hit 
five numbers, it's similar to the other thing. situation here is there are 34 
ways to lose and 6 ways to win after they've drawn the number.
so if you're gonna miss one after they've drawn the first 5 there are 34 
ways to do that, but because the losing number can be any one of the six drawn, 
you have to account for it 6 ways. so you take the individual probability 
this one here, any one of these turns out to be and notice the
denominator hasn't changed any, it's still 2 billion something, that's the
probability for any one of 
these sequences. but you'll notice the sequences - even though the losing 

position changes, you might miss any one while getting the other 5, if 
this is the joint condition probability for any one of the 6 sequences, because 
they are mutually exclusive we can add this thing up 6 times. if I do that 
assuming I can find it, it's one out of 112893.5 - ah but we forgot
someething. two number for a buck! This fraction here is this fraction reduced. we haven't 
accounted for the 6 states yet. it's any one of these is this. this is just 
reducing the fraction to the lowest terms by dividing the denominator by the 
numerator. 
for two bucks i get two of these so i multiply that by 2 and get 
1/56,446.76 - now there are 6 ways to win. this is the probability of any
one of these for a dollar.

[mumble]
it's nice to get the right answer before arguing about it. friend with 
the card, what doees it say for 5 out of 6? 
friend: [mumble]
that's an official state of PA number, there.  questions?

Q: mumblemumble
A: when they pick the number, you've already bought your tickets. unless 
you know someone. because it happens in sequence, there are 34 ways to lose 
and 6 ways to win. how many losing numbers can you get wrong? they pick 6
out of 40, you pick 6 out of 40. what's your chance of getting 5 and missing 1? 
what's your chance of missing one given you've already got 5? there's only one 
way left to win and 34 ways to lose.

Q:[mumlble]
A: this stuff here? here? did i screw up? no. the probs are the same for 
each. it doesnt' matter where the 34 goes. what this depicts here is the
order of the win. you could win, win, win,lose,win,win, or 
win,lose,win,win,win,win--this 
is not the 2nd pick, this is the last of 6 sequences, where in this case 
the loss is of this number, and here it is of this number.

Q: [silence]
A: if this number is a loser, there are 34 ways to lose. you picked it. 
34 ways to lose out of 40. so forth and so on. don't be confused by the ordering. 
what it's really saying here is because you only have 5 winning numbers
and 6 were picked as winning #s you could win 5 different ways, because
you exclude one of the 6 and have 5 left. that's what this depicts, 6 ways.

Q: if you [mumble]
A: NO
Q: OH! [mumble]
A: i'm just depicting it that way. it doesn't matter. that's all. 
Q: if the number's something something, why something?
A: LOOK, there's more than one way to do this. this is the simplest. 
don't worry about the sequencing. the important thing is there are 6 ways
to win by getting 5 #s if they pick six. because you can - if you're watching them 
do it, you could make the calculations in order, but you don't have to. 

Q:how would you calculate it if you had to get [mumbl]
A: it would be one of the sequences.

Q: [mumble]
A: speak louder
Q: mumlbe
A: that's what he just said. no, the order doesn't matter.

ok, i'm going to instead of taking a break now, i wanna do the craps 
example first because you probably won't have quesetions because youwon't 
understand. this isn't in the first handout. 

bringing together all these rules, you may or may not know this, but in 
craps, the guy with the dice for each game has just under a 50% chance of 
winning. .4929 in fact.  but even at 50-50 you'd still lose all your
money so it doesn't matter. the bank/casino makes money off each actual
game. on the main part of the game, the bank gets about a penny/dollar
bet. the craps table line bet is  for probability afficionados.  All the
other games on the table pay WORSE than the main game. the best bet on
the table is the socalled pass or don't pass line. 

table. bunch of people all around. people on other side with sticks, 
helpers on each side. very colorful. if they're not smoking. ex smokers are the 
worst. :)

guy comes out. man with dice starts new game. people make bets. you take 
chips on the pass line or don't pass line. there are tables in las vegas 
where you can bet like 25,000 bucks. once a college professor tried to bet 
$250000 on one roll...and he WON.

first roll:  7 or 11 you win. game ends.

you could roll a 4 or a 5 or a 6 or an 8 or 9 or 10. these are called 
POINTS. game continues - roll again. if you roll your point again before
a 7 you win. that's the game. 

2, 3, or 12 on first roll LOSES - game over. 

Q [mutter]
A: you put your money on this line. you roll. if 7 or 11 comes up, you 
win, 
everyone's happy.

[laughter]
Q: [mutter]how does [mutter]
A: the guy with the dice is determining the game, he keeps rolling. 
everyone else bets with him or against him. most people bet that he will WIN. but 
half the time he will lose. 

the game is determined by what I said. if you bet with him and he wins 
you win, if he loses and you bet with him you lose. if you bet against him 
and he loses, you WIN. 

what's that bar 12? there's slightly more than 50-50 chance of him 
losing, so if you bet "wrong way" eg that he will lose, and he rolls 12,
you don't win or lose. 
you get money back. that's how they even the odds for wrong way betters. 
but we're not talking about them.

say we're all betting on the pass line. new game. dice are rolled. that's 
how game is determined. now if he doesn't roll a 7 or 11 or 2, 3, 12 he may 
continue to roll for half an hour or so, making neither the point nor the 7. then, 
the people betting on the roller will start making side bets on the other 
numbers. 

[story about gambling trip long ago, men in tuxes, women in gowns, come 
to dollar bet table in 1972. the guy puts $500 chips everywhere. the dice 
guy rolls a bunch of numbers, makes 3-4000 bucks, and leaves. 

Q:[mumblemumble]
A: you win. you lose. no. it's still the first roll. 

[someone near the recorder is talking]

Q: [silence]
A: Once a point is established, all that matters is a seven or that 
number. All other numbers are irrelvant.

Q. 
A. You lose. Before the seven, yes. First roll: win, lose, or establish a 
point. Then, try to get point again. If seven comes up, you lose.

do not try to remove money from the table. that is a bad idea. you're 
dealing with the wrong people.

Outcomes:

could get 2,3,4,5,6,7,8,9,10,11,12
with two dice. no doubt about it. How many ways could that happen?
6 ways....5,4,3,2,1. Now, we know this by enumeration. How many ways can 
you roll a 2? one way. a three? 2 ways. etc.