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He's willing to donate to the "save a vegan" foundation if you help him.
I must be getting stupid in my old age, but the 1st question makes no sense, as there is a 50/50 chance remaining between stick and switch, and if the 2nd question is worded right, then player a is the obvious winner. :sad: :donno:
ahhh coz my young padowon, let me show you :wink: in part A the player should always switch, his chances are actually doubled in doing so. think of this, if the player does not switch it comes down to the probability that he picked the right door off the bat (1 in 3), of course one other door he didn't pick has junk in it, knowing which one it is does not help his chances if he does not switch. a more rigorous proof is:
without loss of generality say he picks the first door, there are 3 cases of where the prize is:
case I: the prize is in the first door, they open door 2 or 3 to show him it's junk, he switches to the other door and loses as the prize was in the first door.
case II: the prize is in the second door, since he chose the first door they open up the 3rd to show him it's junk, he switches to the second and he wins.
case III: the prize is in the third door, since he chose the first they open up the second to show him it's junk, he switches to the 3rd and wins.
so out of the 3 possible cases the contestant won twice by always switching (2 in 3 chances) the fact that i stuck to the contestant choosing the first door was only to help illustrate the solution, his actual initial choice is arbitrary.
as for part B, am i assuming 6 games were played and there asking for the chances player A and B won the majority of the games played? or am i just trying to find the probability player A or B won any given game.
if we're looking for the majority case, first lets find the total number of scenarios that could've occurred. player A rolled 9,9,9,9,2,2 so the total number of ways this could've happened is C(6,4)=6!/(4!(6-4)!)=15 (once we choose 4 spots for the 9's the 2's are forced). Player B rolled 3,3,3,3,11,11 and again this can happen in C(6,4)=6!/(4!(6-4)!)=15 ways. so the total number of matches that could've occurred is 15*15=15^2=225. by brute force we can see that the only way player B could've won the majority of the games is if both of his 11 rolls matched up with 2 of player A's 9 rolls, actually ties can occur as well....this is a bit deeper into combinatorics than your teacher is probably looking for, tell ya what explain the problem to me a little more before i go any further.
hehe. alrighty, before i start, i gotta letcha know that your very next reply to anything will make your post count "1234"....
k, for part B, he said that the rolls aren't neccessarily in order as they're told on the paper. so, player A coulda rolled one 9 and B coulda rolled one 3 on the first game, making A the winner. or A=2, B=11, making B the winner. Or any of the possible combinations could have occurred.
but then it says "what is the probability a/b won the game", so out of 6 games i guess it's sayin of those 6, what is the probability that either of the two won.
Ahhhh...I should've expected it. This is obviously some swarmy slick little math exercise which seems grand on the blackboard but in real life means bullshit.
The bottom line is this: According to the worded example it is truly not a choice of 1 in 3 to begin with, but 1 in 2. If the contestant knows going in that 1 of the doors will definitely be revealed and will ALWAYS be junk, then he knows that his chances going in are 50/50. The door that is revealed is thus rendered moot from the start. So, here are your TRUE cases.
case I: the contestant chooses the prize door, they open the moot door to show him it's junk, he switches to the junk door and loses as the prize was in the prize door.
case II: the contestant chooses the junk door, they open the moot door to show him it's junk, he switches to the prize door and wins as the prize was in the prize door.
case III: the contestant chooses the junk door, they open the moot door to show him it's junk, he sticks with the junk door and loses as the prize was in the prize door.
case IV: the contestant chooses the prize door, they open the moot door to show him it's junk, he sticks with the prize door and wins as the prize was in the prize door.
Don't come at the ole perfessor wit your high school bullshit!!! :flipoff:
And like I said, the wording in problem 2 is not clear enough...so get it straightened out young man, or no TV tonight! :slap: :flipoff: :rofl:
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coz is mister wizard :umm: :mrgreen:
Hey, he never missed a show of "Let's Make A Deal" with Monty Hall. Always take the curtain!Quote:
coz is mister wizard :umm: :mrgreen:
[disclaimer: Some of you may not be old enough to remember that show.]
:wink:
Yeah, it's a damm shame, but I actually liked that crap! :rofl:
hehe, i knew coz would try and defend himself but sorry always staying with the chosen door yields only a 1 in 3 chance of winning and always changing doubles you're chances, your cases are neither mutually exclusive nor collectively exhaustive..........this is actually a popular problem with introductory statistics classes, i'm sure it's around on the web if ya want a proof different from mine, but if you really wanna stick to you're 50/50 belief and want it to mean something in real life i say we each get a grand simulate the experiment with $10 a game and see who has more dough at the end, me thinks you'd believe me then when ya lost 6 hundred someodd at the end :thumup:
and yeah i still don't understand what the second question is really looking for, won any given game or won more games?
I'm sorry FrEky, but bullshit!!!
Either the example is worded wrong, or someone is tryin to pull a fast 1.
And remember, I'm the gambler here.
Perhaps you need to challenge a professor or 2, 'cause it sure sounds like funky stuff to me...heheheh.
I'm gonna come up with a mathematical formula for ya...you always believe that shit!
:rofl:
I guess the assumption here is that when you choose the first door, regardless if you chose the expensive prize or one of the cheap ones, would they ALWAYS reveal the other cheap prize? If that is true, I believe what Coz said. If they will sometimes show you the expensive prize, then you are screwed anyway.
Sometimes when I watched "Deal" it was hard to tell when they showed you the one whether that prize was one of the booby prizes or not.
My 2 cents.
:umm:
HAHAHA, i knew coz aint gonna take this one lying down.......ok, lemme give ya this one then (it follows directly from this)
let's say there were 100 doors (99 with crap in em), you pick one and they open up 98 "junk" doors and ask you if you want to stay or switch to the one remaining door........still think the chances are 50/50? the chances of winning if you stay are the same if they hadn't shown you which doors were junk: 1%, but should you switch your chances are 99% and again i'll come to your house tonight if ya wanna put a grand each toward this one........ me thinks the 10 dollars or so left in your hand would definitely convince you!
again if you were to just pick one door out of 100 and they just opened it right up you obviously have a 1% chance of winning. and it's also obvious that at least 98 of the doors contain crap in em because there's only one winning door, if they just show you which doors are crappy and you don't switch it don't help you out none, you already know that 98 of em contain crap don't make no difference if you know which ones they are if you don't switch, you still left with the chance that you had picked the right one, still 1%...........switching is like being able to choose the whole other 99 doors at once with a safety net that if just one of em is the winning one you still win (the opening of the junk doors being the safety net)........think about it long enough, you'll get it....and the 3 doors is the same exact thing, 1/3 chance vs, 2/3 chance.......... 1,000,000 doors = 1/1,000,000 chance not switching, 999,999/1,000,000 or 99.9999% of winning always switching...........
yeah it's highschool bullshit, shouldn't be too hard for "the ole perfessor" to nail down :poke: :rofl:
Shit!!! Ihate being wrong!!! :sad: :slap:
:bowdown: :bowdown: :bowdown: :bowdown: :bowdown: :bowdown: :bowdown: :bowdown: :bowdown: :bowdown:
thats ok cozzie.......i was born all wrong and i am the personification of wrong :weird: :weird: :weird: :rofl: :rofl: :thumup: :thumup: :wave: :wave: :haha: :haha: :mrgreen: :mrgreen:
yeah took me a minute to get it when i first saw this one too, cuz it's still stuck in your head "it always comes down to the fact that i'm left with 2 doors and i gotta stay with one or go with the other.....gotta be 50/50" and it's hard to see past this, the 100 door one made it clear for me too though. ok got another one for youz, i might have done this one here before, i can't remember. but anyway..... (note C^n means C raised to the nth power)
define 2 variables such that:
1. A = B
2. AB =B^2 multiply both sides by B
3. AB - A^2 = B^2 - A^2 subtract A squared from both sides
4. A(B - A) = (B + A)(B - A) factor (B - A) from both sides
5. A = B + A divide through by (B - A)
6. A = A + A substitue A in for B (see step 1)
7. 1 = 1 + 1 divide through by A
8. 1 = 2
so, does one indeed equal 2? (using our normal real one dimensional number system) or did i do something tricky? where'd i pull off the trick if so? (what numbered step)