Chances of Smoking
Mr. Striker is walking home across the parks. It is a walk which takes exactly two cigarettes. In his cigarette case he has two craven Doodahs and a pair of El-Rashids(which he prefers).
What are the chances of his smoking two El-Rashids in such a walk if he randomly withdraws cigarettes from the case.
What are the chances of his smoking two El-Rashids in such a walk if he randomly withdraws cigarettes from the case.
Labels: mathemagic
posted by Rajesh Lal at
8:31 AM
41 Comments:
25%
Assume he pulled an El-Rashid for the first cigarette. The chances of that happening were 1/2. Then he reaches for another one. The chance of him drawing the 2nd El-Rashid are now 1/3. So the chance that he will draw both of them in two draws is 1/2 * 1/3 = 1/6.
I agree. The above answer of 1/6 (0.16666) is correct. Good explanation too...
The first answer of 25% is presumably based on there being four possible distributions of two cigarettes: DD, ED, DE and EE, where D is Doodah and E is El-Rashid.
This reasoning seems just as logical as that leading to 1/6.
So, which is right: 1/4 or 1/6?
those four possible distributions dont all have the same probability of occuring
1/6 is the right answer
No if it is random, they have exactly same amount for each cigarette, so the 25% is right because the other forgot to put the ones that start with a Doodah. Unless there is a difference in the physical properties of the cigarettes, which is not known so you must assume they're exactly the same
16.67%
Label the cigarettes D1, D2, E1 and E2. The distributions of the first two picked are:
D1 D2 D1 E1 D1 E2
D2 D1 D2 E1 D2 E2
E1 D1 E1 D2 E1 E2
E2 D1 E2 D2 E2 E1
Only two out of the twelve possibilities have El Rashids drawn first and second.
So the answer is 1/6.
Oops, spaces compressed in my previous answer.
Read the 12 distributions as:
D1 D2
D1 E1
D1 E2
D2 D1
D2 E1
D2 E2
E1 D1
E1 D2
E1 E2
E2 D1
E2 D2
E2 E1
The answer is actually 0%.
Unfortunately for Mr. Striker there is pretty much a 100% chance that upon producing any of the four items of contraband in his possession in a publicly owned and administered park he would be immediately accosted by any number of concerned comrades (er, citizens) and/or government authorities and certainly either driven from the premises if not severely injured and/or detained indefinitely.
He has no chance of successfully traversing the said park and putting women and small children at dire risk and we have the Science of Mathematics to thank for that!
-h
it just depends,
25% if he picks them at the same time
and 1/6 if he picks one then the other
um... Even if he picks them at the "same time" there are still 6 possible pairs he can pick and only one of these that matches the criteria - therefore, it is still 1/6.
{ (E1,E2),(E1,D1),(E1,D2),
(E2,D1),(E2,D2),(D1,D2) }
In fact, this is a third and possibly simpler way to look at the problem. Good insight though.
-h
The first time around he has a 50% chance of getting his favourite smoke, But after that his chances change to a 33.33% chance if you average these chances you get 41.665% chance of getting two the same
1/2
its 1/3 because these two:(D1,E1)and (D1,E2) ARE THE SAME!!! I mean how are they E1 and E2! ITS NOT LIKE THEY ARE LABLED!!
(D1,E1 = D1,E2 = D2,E1 = D2,E2)
thats one outcome
(D1,D2)
thats two outcome
(E1,E2)
thats three outcomes
I know I said the E1,E2 thing is stupid, but i only did it to make it more simple
Ben: you had the right idea. The first time is a 50% chance (1/2), the second pull is now a 33% chance (1/3). But because they are two different pulls (and the second is conditional upon the first), you must multiply them together, not average. And when you multiply those together you get 1/6 or 16.666666667% chance.
There are two ways to get to the answer. The first was with calculations like above, here is the second, and it's been pointed out already.
There are 12 different combinations of pulls he can have (6 of them are the reverse order of 6 others, but still possible combinations).
D1 + D2
D1 + E1
D1 + E2
D2 + D1
D2 + E1
D2 + E2
E1 + D1
E1 + D2
E1 + E2 *
E2 + D1
E2 + D2
E2 + E1 *
Now, out of 12 possible combinations in which he can pull cigarettes, only two of those give him the privilege of smoking two El-Rashids before a Doodah.
ur wrong adam
these were two of your outcome:
D1,E1
D1,E2
well theyre the same
as i said before they are not labled or anything its the same cigerate so theres on 1 or 2 theres only D AND E.
THESE ARE THE OUTCOMES
D,D
D,E
E,E
1/3 CHANCE (33.33334%)
yeah hes right it is 1/3, theres no E1 or E2 there exacly the same.
if you pulled one out your not going to say ,oh i pulled out el-rhashid 1/2/doodah 12 youre going to say ,oh i pulled out an el-rhashid/doodah.there are 2 different ones you can pull out,not four, because 2 are the same and the other 2 are the same.
D,D
D,E(E,D)WHICH IS THE SAME
E,E
1/3 chance
my previous text had an error
the(el-rashid 1/2/doodah 12 youre)part is ment to be shown as this(el-rashid 1/el-rashid 2/doodah 1/ doodah 2, youre)
sorry for the mistake
im only 13 thougth
50/50 of course. Because the two El Rachidos and the two Doodahs are the same, so it would be as though there were only 2 cigarettes.
It's not highly mathmatical, people!
you are saying that the two el rachidos and two doodahs are interchangeable but consider this; when rolling two dice there are three ways to roll a seven, 1-6 2-5 3-4, and there is one way to roll a two, 1-1, if you assume that the sevens are interchangeable then the chances of rolling a two is 1/11 and the chance of rolling a seven is 1/11. as we all know its harder to roll snake eyes than it is to roll a seven. basically what im saying is 1/6 is right
Yes I know that E1 and E2 are identical cigarettes. Now try and follow me so I can show you how much that makes since, many posters above me get it.
Pull 1: You have four cigarettes, and two out of four (or 1/2) chance of pulling a El-Rashid. Thus you can either pull the first one E1, or the second one E2. It doesn't matter. Or if you don't understand that... just know that on the first pull you have a 1/2 chance of getting an E.
Pull 2: Assuming you get the El-Rashid on the first pull, you now have one E left, and two Doodahs left. So you can now either pull the second E (E2) or the one of the Doodahs (D1) or the other Doodah (D2). Following me yet? Yes they are the same, but you are twice as likely to pull a Doodah now, because you have twice as many. So your chance of getting that second El-Rashid is 1/3.
Because getting the second El-Rashid is conditional upon getting the first, you must multiply: 1/3 * 1/2 = 1/6 = 16.667%
If you are still making the mistake and coming up with 1/3 (or even 50% or 25% like many others) I propose another, exactly similar, situation.
I have 4 sticks in a box, one is red, one is blue, one is orange, and the last is purple. What are the odds that you will pick the two primary colors with only two picks? (primary colors are red and blue)
Here are the possible outcomes of two draws:
red, blue *
red, purple
red, orange
blue, red *
blue, purple
blue, orange
purple, red
purple, blue
purple, orange
orange, red
orange, blue
orange, purple
This is exactly the same as the cigarette question. Yes orange and purple are both complimentary colors so red,purple and red,orange is the same right? No, because they are two different sticks, they must both be taken account of. So the chance of getting both primary colors is 2/12 = 1/6
Just like the cigarettes, yes the two doodahs are the same, but you must take into account both of them, because on the second draw (if you got an E on the first) you have two Doodahs and only one El-Rashid.
If you don't understand this, then let's just hope you avoid any casino, because with your inability to understand odds, they would steal all of your money.
ok adam,
to prove its 1/6
wat are the six outcomes?
Ok, if you still need more proof, here you go.
Although it is true there are only three possible outcomes, the likelihood of each of those outcomes occurring is not equal.
Outcome #1 (EE)
only one possible combination
Outcome #2 (DD)
only one possible combination
Outcome #3 (ED)
because there are two E cigarettes available to use, and two D cigarettes available to use, there are 4 possible combinations (2 x 2)
So with three different outcomes, there is now 6 possible combinations (1+1+4). Although it is true that 1/3 of the outcomes is EE, the odds show that EE will not occur as often as ED will. The likeliness he will actually draw EE (one of six different combinations) is 1/6.
If with three different explanations (this explanation, and the color stick explanation, the simple 1/2 x 1/3 = 1/6 explanation) you still don't understand, than I don't know how else to explain it to you.
Matt - maybe when you get a little older than 13, and actually get to high school and take a math class that involves more than adding and multiplying you'll understand.
To further my point, I'll elaborate on what Tommy Boy was saying with the dice.
There are 11 different outcomes you can have with dice (2 through 12). However, the chance of rolling snake eyes (1-1) is 1/36 (1/6 x 1/6). The chance of rolling a 7 (any of the combinations) is 6/36 (1/6), ask any gambler. This is because there are 36 possible outcomes (6 numbers on die #1 x 6 numbers on die #2).
Now you are saying because something has the same outcome, it doesn't matter how many combinations there are. If this was true than getting snake eyes and getting a 7 have equal odds, 1/11. But they don't, because you aren't accounting for all possible combinations.
on your so called outcome 3, you said there 4 outcomes for it.
but really they are all the same.
they all end up with one el-rashid and one doodah so thay are all the same, and that thing about getting to highschool, im in high school and im the best at it in my whole year group i get level 7 or higher on every test. i asked my maths teacher the question and he agrees that is is 1/3, so there!
Well, if your math teacher doesn't understand, than I don't expect you to. Did you English teacher also teach you how to spell? "thay" "wat" "youre"
Yes on outcome #3 (ED), there are 4 possible combinations. They are the same outcome, but there is four possible ways to get there (see dice explanation, and color stick comparison). Having 4 combinations to get to the same outcome makes it much more likely to occur (4 times to be exact). This results in 6 possible combinations total. Thus EE is one combination, out of a possible 6 combinations.
Here is a little homework for you. Grab the four aces out of a deck of cards. Mix them up until you don't know which is which anymore, and pick two. Then repeat about 60-100 times. Record all the results. You'll see what the odds should really be.
I did that and got Black Black(BB) 7 times (out of 60), Red Black (RB) 21 times (out of 60), Black Red (BR) 21 times (out of 60), and Red Red (RR) 11 times (out of 60). For your 1/3 rule to get BB held true you'd expect 20 (I got 7 for BB and 11 for RR). That is much closer to my 1/6 (expected 10 for each BB and RR). I would also expected 2/3 for one card of each color, and I got 42/60 = 21/30. That's pretty close. If you would actually try this (honestly without cheating), and repeat it enough times 60-100, then you'll get results like that.
So after giving your actual numbers, if you don't see it now, then you should just quit before you make yourself look more ridiculous. I guess I didn't need my year of calculus, plus two advanced calculus courses in college, statistics, trigonometry, and 2 years of algebra to solve this after all.
adam
youre a pompus snob and you think you know everything!
if i didnt know better i say u were a pervert!
and i might be wrong but i still think its 1/3
and i have never spelt they with an "a". and "youre" is a word, i only forgot the apostrophy-so ur wrong!!
Ok, so if the 2 geniuses have stopped their below-the-belt antics, can we have the correct answer? my answer a few days back was 16.67%; which coincides with Adam's 1/6 theory (makes me happy or what).
I'm sorry if you think i'm pompous, but I was just trying to explain the right answer to you. I didn't start getting smart until you were rude to me, and refused to accept the correct answer. Also, why would you think i'm a pervert? Talk about an immature attitude. And you did spell it thay in your 8:10 post, and I was pointing out your lack of an apostrophe.
I'm guessing you finally understand that 1/6 is the right answer. (Just try the experiment yourself if you aren't sure). I've been proven wrong about many things and learned a lot of lessons the hard way, don't feel bad, just learn for future reference.
Rahul, I know I wasn't the first to say 16.67%. I was just trying to explain it better for those who didn't understand.
Adam.... I don't care if I was the first one to answer or the last one to answer. We all solve things in our own way using our own brains. My point was that we should stop this unnecessary bickering and come to the final answer. Whoever may be right. It doesn't matter. We have this forum to improve our thinking skills with new puzzles and riddles. If I may speak on everyone's behalf, I think we all are sick and bored of this immature fight. I don't think any of us care who started it or who is wrong or right. Let's solve issues and not make enemies on this forum.
Now, if the person who created this quiz can please give us the right answer, we all would appreciate it.
ANSWER
--------------------
His chances are One in Six,
But don't you guys already know that !
i was wrong adam
and im sorry for calling you a, you know. if you forgive me i would like to make ammends(if thats who you spell "ammends")
and im sorry for my immoture behaivour. me and my maths teacher were wrong and you were wright. and i agree with you on "deaf spy" question. sorry
I'm not mad or wasn't trying to be rude. I was just trying to explain the correct answer to you in a variety of ways.
His chances are 100 percent. It doesn't say he has too randomly withdraw only two cigarettes, just that he randomly withdraws cigarettes. He prefers El-Rashids. If he pulls a Craven Doodah out, he simply puts it back and keeps choosing at random until he gets the cigarette he wants.
Dude I think its a
nCr (p^r)(q^n-r)
problem
n=total number of times your going to do the activity
r= total number of sucesses your looking for
p= probability of a sucess in one activity
q= probability of failure in any one activity
and you would only have to do one equation because your looking for him to get EXACTLY 2 El-Rashids, not at least or at most.(In which case you would have to use the equation twice.)
so
the equation would look like this:
2C2((2/4)^2)(2/4)^2-2)
which when put in a graphing calculator equals:
-7/16
The probability of him smoking two El-Rashids in such a walk is -7/16
That is my final answer :D
Arg just kidding I think I did that wrong...
okay so
nCr(p^r)(q^n-r)
and
1C1((2/4)^1)((2/4)^1-1)
+
1C1((1/3)^1)((2/3)^1-1)
in which case it equals -13/36
but that really doesnt make sense either?
and now that I look at other peoples comments i agree it should be 1/6
But thats just how my brain thought to do it.
I guess I'm making it way harder then it really is, and maybe its not actually a probability question?
it is 1/2 because there are 2 cigarets it say 2 pairs of cigarets witch means there are 2
@ matt and the anonymous above him, they ARE the same, but they are two different outcomes.
picking E, THEN D is different than
picking D, then E.
Say that youre playing a game, and if you pick D at all, you lose. If you survive the first pick, you get 1 point. if both are E, then you get 5 points.
If you pick E, Then D, you get one point, but if you pick D, Then E, you dont get any. they are the same cigarettes, but they are definately different outcomes
~GR655
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