What do you think the answer to that math riddle is? - Page 2 - GoFuckYourself.com

Damian_Maxcash · 12-17-2004, 05:37 PM

Quote:

Originally posted by SilverTab
let's say you pick A.....

DAMIAN - I have picked "A", its not in "A", so that is out, I have 2 answers left.. B or C

if you STAY with your choice, you will only win if the car is behind door A...

It cant be "A", it went in the last round! It no longer exists!

if you switch, you will win no matter if the car is behind B or C...TRY IT damn it

WTF>>>> I AM LOST>>>>> I WANT THE ORIGIONAL QUESTION POSTED HERE!!

SilverTab · 12-17-2004, 05:41 PM

here's the exact question

Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat. He says to you, "Do you want to pick door number 2?" Is it to your advantage to switch your choice of doors?

The Machine · 12-17-2004, 05:41 PM

which came first - the chicken or the egg?

alexg · 12-17-2004, 05:43 PM

Quote:

Originally posted by SilverTab
here's the exact question

Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat. He says to you, "Do you want to pick door number 2?" Is it to your advantage to switch your choice of doors?

well, that's just what I asked in other words

psili · 12-17-2004, 05:46 PM

Trust the numbers man. I still don't quite get how the numbers come out like they do, but "always switching" always results in higher net wins than "never switching".

Damian_Maxcash · 12-17-2004, 05:48 PM

Quote:

Originally posted by SilverTab
here's the exact question

Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat. He says to you, "Do you want to pick door number 2?" Is it to your advantage to switch your choice of doors?

OK I get it....

U assume the host is crooked

I am a trusting idiot and I assume people are honest

very good

ytcracker · 12-17-2004, 05:50 PM

Quote:

Originally posted by psili
Trust the numbers man. I still don't quite get how the numbers come out like they do, but "always switching" always results in higher net wins than "never switching".

silvertab drew an awesome picture explaining it

SilverTab · 12-17-2004, 05:56 PM

Quote:

Originally posted by damian2001
OK I get it....

U assume the host is crooked

I am a trusting idiot and I assume people are honest

very good

hahaha well that was the question in the first place ;)

Wolfy · 12-17-2004, 06:12 PM

you guys are either really patient or very bored.

it's pretty simple, really. 66%.

Wolfy · 12-17-2004, 06:16 PM

Quote:

Originally posted by psili
Trust the numbers man. I still don't quite get how the numbers come out like they do...

when picking randomly out of 3 boxes, you ALWAYS have a 1:3 chance of winning = which means you are going to be wrong 2:3. Eliminate 1 of the wrong answers AFTER you have taken, with 66% probability, the other wrong answer - and you have your explanation.

Libertine · 12-17-2004, 06:29 PM

Quote:

Originally posted by Kevsh
Yes, it does. Because 1 of the 2 boxes left has a beer in it (unless you chugged it ) ....

But if I were predisposed to throwing away an empty box, me throwing away a single box would simply mean that _if it's one of my two boxes, it isn't that one_. That doesn't lower the chance of it being one of my two boxes though.

Quote:

Originally posted by Kevsh
People get a grip and forget the math, that's what confusing everyone. Think in practical terms, not percentages and raised or lowered chances.

Think about if you were in a beauty pageant (a real stretch for most of us, I know). There are 12 contestants left. One by one they are told by the host they didn't win until there are 2 left.

If you and a friend were betting on who would become Ms.Whatever, and did so when there were 12, logic says that each of you, on his own, would have a 1/12 chance of guessing who the winner would be.

Now, there are 2 contestants left. Each of you were brilliant: You each have 1 of the remaining contestants, but of course only 1 of you will win...

What are the odds that you choose the right one?
Would you think it's logical / mathematically / practical to change your pick?

One should not forget the math, because math yields truth.

Your comparison, on the other hand, does not. You fail to account for the fact that the box the contestant chose has immunity from being revealed because you chose it. Thus, the fact that it goes on to the next round is meaningless.

Let's compare it to the models again:
10 models get voted for in a live tv show - all at once - and afterwards you get to pick one of them as your personal favorite.

Now, after the vote, one by one, the models with the least amount of votes are sent off. However, the model you chose as your favorite can stay until the final two no matter how many votes she got.

Clearly, in the final round, she'll have way less chance of winning than the other model, simply because there's an 8/10 chance that your model is only there because you chose her. The fact that you having chosen her gave her immunity until the last round has no impact whatsoever on the amount of votes she got.

So, by default, the 8/10 chance that she's just there because of the immunity you gave her goes to the other girl, and only within the remaining 2/10 chance does she have 1/2 chance - so, what she actually has is 1/10 chance.

In the case of the 3 boxes, the immunity acounts for 1/3 (which by default goes to the box which did not have immunity), and the initial decision has 1/2 of the remaining 2/3 - or a total of exactly 1/3 chance.

chodadog · 12-17-2004, 10:37 PM

Quote:

Originally posted by SilverTab
indeed

See, this is where i think the problem is. You say "The host reveals B or C"

How can two seperate events be counted as one?

Nobody has adequately explained to me why that is.

I've gone through every possible combination of events in this problem and i wind up with a 50% chance of winning.

Outcome number 1:

Prize is in Box A.
Contestant chooses Box A.
Box C is removed.
Stays = Win
Switches = Lose

Outcome number 2:

Prize is in Box A.
Contestant chooses Box A.
Box B is removed.
Stays = Win
Switches = Lose

Outcome number 3:

Prize is in Box A.
Contestant chooses Box B.
Box C is removed.
Stays = Loses
Switches = Win

Outcome number 4:

Prize is in Box A.
Contestant chooses Box C.
Box B is removed.
Stays = Loses
Switches = Win

Nobody has explained to me why the first two outcomes together are somehow only as likely as one of the other outcomes?

Equinox kept posting the results from that C program that someone had written. I can't read programming for shit. My suspicion is that the person has also lumped the first 2 outcomes into a 1 in 3 chance instead of a 2 in 4 chance.

In the end, there are 4 possible outcomes regardless of what box is initially chosen. And out of those 4 outcomes, 2 will win when you stay, and 2 will win when you switch.

chodadog · 12-17-2004, 10:48 PM

Okay, i've thought about it some more. Initially, you have a 2 in 3 chance of choosing a situation that will lead to switching winning, whereas you only have a 1 in 3 chance of choosing a situation where staying will win.

I'M CONVERTED!

So what punkworld said about the two staying choices being a subset is correct. They each become a 1 in 6 chance.

In my defence, it was 6 am when we were discussing this earlier.

haha.

SilverTab · 12-18-2004, 01:02 AM

Quote:

Originally posted by chodadog
Okay, i've thought about it some more. Initially, you have a 2 in 3 chance of choosing a situation that will lead to switching winning, whereas you only have a 1 in 3 chance of choosing a situation where staying will win.

I'M CONVERTED!

So what punkworld said about the two staying choices being a subset is correct. They each become a 1 in 6 chance.

In my defence, it was 6 am when we were discussing this earlier.

haha.

Nice! all this work was not in vain! ;)

12-18-2004, 01:11 AM

Quote:

Originally posted by Kevsh
There is a roulette theory about betting black (or red) that sounds great on paper:

You start by betting $5 on black.
If you win, fine, you pocket $5 profit.
If you lose, you double the bet ($10).

Since it pays out 100% of the original bet, a win on the 2nd try will net you $5
(You bet $5, lost, then $10, won. $15 spend but you won $20 = $5 net)

If you lose the 2nd (and so on) you keep doubling your bet. No matter how long it goes red it will eventually go black and you will $5 no matter what the bet ($5,$10,$20,$40 ... $512, $1024)

A friend had this idea our first trip to Vegas. The thinking was, what are the odds it will go black 9 times in a row? (Which would have meant you gambled $2555 total so far)

Well, mathematically, 1/512 chance. Fairly small right? Wrong. As the above post points out, past history has no effect so really its 50/50 each time (1/2).

As my friend found out $2555 later - and imagine having that much on a roulette wheel? If he last the next bet would have been $2560 (and over $5k total) which he didn't have.

We won, luckily, but sh1t his pants along the way. Lesson learned.

I do not like this

as you will only be winning 5$ per winning session...

unless you have alot of money, and alot of time, to be begging 50, 100, 200, 400, 800 and such

View Poll Results: What chance would you have to win if you switch?
50% - It doesn't matter if I switch or not...	13	50.00%
66.666% - I should definitely switch...	11	42.31%
I have another answer...	1	3.85%
I have no idea what you're talking about...	1	3.85%
Voters: 26. You may not vote on this poll

12-17-2004, 05:41 PM	#52
SilverTab Confirmed User Join Date: Nov 2001 Location: MTL Posts: 5,060	here's the exact question Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat. He says to you, "Do you want to pick door number 2?" Is it to your advantage to switch your choice of doors? __________________ mmm my sig was too big... no more cool animation but hey still! need php? ICQ: 94586959

12-17-2004, 05:41 PM	#53
The Machine Confirmed User Join Date: May 2002 Posts: 859	which came first - the chicken or the egg? __________________ Complex, large-scale, multi-platform custom programming. icq # 106 769 997

12-17-2004, 05:46 PM	#55
psili Confirmed User Join Date: Apr 2003 Location: Loveland, CO Posts: 5,526	Trust the numbers man. I still don't quite get how the numbers come out like they do, but "always switching" always results in higher net wins than "never switching". __________________ Your post count means nothing.

12-17-2004, 10:48 PM	#63
chodadog Confirmed User Join Date: Apr 2002 Posts: 9,736	Okay, i've thought about it some more. Initially, you have a 2 in 3 chance of choosing a situation that will lead to switching winning, whereas you only have a 1 in 3 chance of choosing a situation where staying will win. I'M CONVERTED! So what punkworld said about the two staying choices being a subset is correct. They each become a 1 in 6 chance. In my defence, it was 6 am when we were discussing this earlier. haha. __________________ 26 + 6 = 1

12-17-2004, 06:12 PM	#59
Wolfy Confirmed User Industry Role: Join Date: Dec 2003 Location: Wisconsin Posts: 3,574	you guys are either really patient or very bored. it's pretty simple, really. 66%.