So what is the correct answer?

[Megaman Z]

Recently my math class had a bit of a disagreement over what the answer to a particular multiple-choice math question was. It wasn't that half the class said one answer was right while the other half said another was, it was the fact that a good three-fourths of the class had one answer and enough work to show it was valid WHILE THE ANSWER KEY SAID ANOTHER ANSWER WAS CORRECT. So, I'm just wondering, what do you people think the answer to this problem is?

The Florida Marlins and the Montreal Expos were going to play a best 2-out-of-3 exhibitation game series. What is the probability that the series would end in two games?

A. 1/4
B. 1/3
C. 1/2
D. 2/3

(I'll put what the book says on the problem later... I wanna see what the generally accepted method of working this one out is around here)

[papillon]

I'm not posting my reasoning yet as I think you should collect some votes without my discussion swaying anyone's choice.

Of course, the part of my brain that hasn't had to do silly school multiple choice problems in a while wants to complain that it depends on how good the teams are... (And also whether the sport in question - it doesn't say which sport - allows tie games in exhibitions. Stupid detail, yes, but I *have* lost questions on tests because I did not know the rules of the SPORT they were asking about.)

[PyTom]

Zero. Since the Montreal Expos don't exist any more, there's no way they can play against the Marlins.

Seriously, this is a really poor question. The probablilty of a team winning a single baseball game depends on quite a bit of things, like how good the team is, if they have homefield advantage, weather, etc. So it's almost certainly not 50/50, and if someone knew of a perfect way to quantify all factors to get a probability, well, you could make quite a bit of money off of that.

So let's change the game a little, to something that we know quite a bit about--- coin flipping. We'll using a fair coin, one we know will come up heads 50% of the time and tails 50% of the time. To simplify the problem, we'll have one of the teams win when heads come up, and the other win when tails come up. (With a mythical fair coin, we don't have to worry about it landing on its side, or hovering in midair for all eternity, or...)

So, our series is now between the Montreal Figureheads and the Florida Fishtails. The question is, what is the chance that the series will end (by one of the teams winning twice) in two games.

And the answer is:

1/2.

There are four possible outcomes of the first two matches: HH, HT, TH, and TT. If we get HH or TT then one of the teams wins immediately, while with HT or TH we go one for another round. So we get a win 2/4 of the time, or 1/2 of the time.

I'm curious as to what the answer your class got was. Realize that in math, the correct answer isn't decided by a popularity contest. OTOH, the book can quite easily be wrong.

Okay, extra credit time, the famous Monty Hall Problem.

You're on a game show, and the host (canonically, Monty Hall) shows you three doors. Behind one of the doors is a car, and behind the other two doors are goats. He asks you to pick a door, and you do.

After you pick the door, he opens one of the other doors, to reveal a goat. (He can always do this.) He then gives you the chance to from the door you picked to the one remaining door.

Should you? Why?

[Akimaru]

^

Yea...
There's 1/3 chance that you have the right door, but when Monty picked one of the 2 remaining doors, he has to be sure there's a goat behind it and doesn't accidently pick the prize. That's why if you first picked a door with a goat in it which is 2/3 chance and switched you are certain to win.

Multiple choice math? I haven't had that since elementary......
Lol too easy.....

[papillon]

If the answer key prints only a number (or letter choice) it could easily be a misprint. Does it have any explanation of its answer? (Assuming its answer disagrees with us.)

Sometimes publishers give you prizes if you are the first to correctly report an answer key error...

And for those who don't believe the monty hall problem, this expansion makes Why It Works a little more clear.

Quoted from usenet:

Confession: Monty Hall got, er, my goat when I first
encountered the puzzle. Yes, folks, I blundered. But here's
the argument that eventually brought me to the light: imagine
that instead of three doors you'd started with one million,
one with a car and 999999 with goats. You designate a door,
and clearly your chances to this point are 1/1000000 of getting
the car. The chances are 999999/1000000 that the car is behind
one of the other doors.

Now M.H. opens 999998 of those other doors and shows you
999998 goats. (Et ab hoedis me sequestri!) Remember, the
odds are 999999/1000000 that the car is behind one of those
doors, and M.H. has kindly eliminated 999998 of them for you.
Your chosen door still has a 1/1000000 chance of concealing the
car, and all the rest of the probability has now been concentrated
in the one remaining door. Would you switch?

clarifying:

Your first choice is a choice from a huge allotment and has only a slim chance of being right. Only if your first choice was right will you win the car. If your first choice was not right, then the car was in one of the 999999 others. But now all but one of those others has been eliminated. So either you stick with your initial choice, almost certainly wrong, or you swap.

[Akimaru]

I was talking about pytom's question, which is a more harder and trickier question. I had to face that problem in a test once...but if the teacher didn't explained it I would've never figure it out....I would just say.. If 3 doors and Monty picked one out for you than you leave with 2 doors....which is 1/2 chance and regardless if you switched or not, it's still better than 1/3 chance.

[PyTom]

The thing about the Monty Hall problem is that I'll have people swearing up and down that it can't be right. While people with math or CS training tend to get the right answer without a problem, other people will argue 'till they turn blue that there's no difference between the two remaining doors.

[papillon]

That's why that expansion I posted is nice, because blown into larger numbers, it breaks through the 'common sense' feeling of "that CAN'T be right!"

[Megaman Z]

I think we have a general agreement on this...

oh, and as far as the answers...

I had 1/2, almost all the class had 1/2, the key said it was 1/4. I'm still trying to find some believable logic to their answer, though the popular theory is that they were, in reality, asking about one of the teams specifically.

the accuracy of it doesn't surprise me, seeing as this question was on a practice sheet with questions from a standardized test (the "FCAT", if anyone's wondering...).

oh, well... at least the class didn't get marked wrong for putting "C"...

Zero. Since the Montreal Expos don't exist any more, there's no way they can play against the Marlins.

about how long ago did the Expos go away? they may have still been around when the question was written (which was a few years back...)

So let's change the game a little, to something that we know quite a bit about--- coin flipping. We'll using a fair coin, one we know will come up heads 50% of the time and tails 50% of the time. To simplify the problem, we'll have one of the teams win when heads come up, and the other win when tails come up. (With a mythical fair coin, we don't have to worry about it landing on its side, or hovering in midair for all eternity, or...)

okay, sense of humor aside, this is EXACTLY what I think everyone did.

I'm curious as to what the answer your class got was. Realize that in math, the correct answer isn't decided by a popularity contest. OTOH, the book can quite easily be wrong.

see spoiler box above.

If the answer key prints only a number (or letter choice) it could easily be a misprint. Does it have any explanation of its answer? (Assuming its answer disagrees with us.)

Sometimes publishers give you prizes if you are the first to correctly report an answer key error...

sadly, no. I don't think I can yell at the guys who wrote the question. though, at this point, I wonder if I'm getting extra credit for this... (wishful thinking...)

all that said, I think someone will have found a normal way to unlock Memories on DDREXTREME US PS2 before we get a full agreement on what the solution should be.

*(Memories is an extremely elusive unlock that, so far, has only been unlocked via hacking either the disc or the save data.)

[ShiraiJunichi]

You're wrong Py'Tom! For the Monty Hall problem, you definitely shouldn't switch doors. He picked a door that didn't have the prize, which means your door probably has the prize- so you should never ever switch. Anyone who thinks you should switch is dumb, and needs to go back to elementary school and relearn their arithmetic.

LOL

[Guest]

You should definately switch doors.

Example: suppose we have three doors A, B and C. Where A = goat, B = goat, C = car.

There are 3 cases.

case 1: you choose door A. Monty Hall opens door B. By switching to C you get the car.
case 2: you choose door B. Monty Hall opens door A. By switching to C you get the car.
case 3: you choose door C. Monty Hall opens either A or B. By switching you lose the car.

By the above case, if you switch you have a 2/3 chance of getting the car.

[Beta]

the best answer I can come up with for the original problem (the baseball one) is 1/3. here's my reasoning...

let's assume that both teams have a 50/50 probability of winning. let us also assume that the series ends ONLY WHEN ONE TEAM ACCUMULATES TWO VICTORIES, and there are no draws. so it becomes a coin-flipping situation.

H=Heads
T=Tails
X=Not applicable (game never happened)

1)HHX; 2 games
2)HTH; 3 games
3)HTT; 3 games
4)THH; 3 games
5)THT; 3 games
6)TTX; 2 games

based on the above, there are TWO ways (out of six) that the series could be played out and end after exactly two games.

*reads the rest of the thread, including spoiler boxes*

...and it seems like nobody remembered that there would be a third game or couldn't figure out how it would affect the answer.

[PyTom]

While you're doing a case analysis, which is good, you're doing it wrong, which is bad. You're assuming that all of the six outcomes that you've listed have the same possibility, which isn't the case. Let's consider a sport which has a best-of-three championship series, but which, for whatever reason, demands that they play all three games, even if the outcome is preordained. We'd get the following possible outcomes:

HHH *
HHT *
HTH
HTT
THH
THT
TTH *
TTT *

The lines I've starred are the ones in which the games are won in two flips. All of the lines have the same probability, so the chance of someone winning in 2 games is 4/8 == 1/2.

Now, the thing to realize is, these probabilities don't change, even if we don't bother playing the third game when it's unnecessary. Nothing we do in the third game affects if the same team has won the first and second games... so it really doesn't matter if we play it or not. That's why we all gave analyses that only considered the first two games.

To better illustrate where you went wrong, say you have a lottery ticket. There are two possible outcomes: You win, or you lose. But you don't have a equal chance of getting those two outcomes, so your chance of winning is much less that 1/2.

Today's hard problem: Come up with a method for coin flipping over the phone, that's better than:

Tom: Call it.
You: Heads!
Tom: Sorry, it was tails. Better luck next time!

Your method should work without needing a disinterested party to do the coin-flipping. People who have read "Applied Cryptography" are disqualified.

[Counter Arts]

Yeah, the baseball 1/2 sounds about right.

The thing about the 1/3 arguement is that it just assumes all outcomes have an equal possiblitly of happening.

All possible variations for three coin tosses...

TTT (1/8 )
TTH (1/8 )
THT (1/8 )
THH (1/8 )
HTT (1/8 )
HTH (1/8 )
HHT (1/8 )
HHH (1/8 )

They all have an equal chance of happening. Merge the two wins first cases and you get...

TTX (2/8 = 1/4)
THT (1/8 )
THH (1/8 )
HTT (1/8 )
HTH (1/8 )
HHX (2/8 = 1/4)

Add the probablitly of getting those two cases together you get 1/4 + 1/4 =1/2 chance of the game ending in two games.

This supports PyTom's arguement.

Yeah... I'm definately in CS.

EDIT: Darn... PyTom beat me to the punch.

[PyTom]

That's cause I'm a Master of CS!* Also, I have no life.**

*Literally. I have a Master of Science in Computer Science.

** That is, I spend my days at a computer.