Probability Question: What are my odds of picking the right prize?

yes, you will always have a 1/3 chance, regardless of when you pick.

If you are picking second and the first guy picks your desired prize you have no chance of getting it, but if he does not pick your desired prize, your chance of getting it increases by the exact amount to counteract the first situation. => 1/3 = (1/3)*0 + (2/3)*(1/2)

Similar math/idea holds for picking third.

With three people, it's pretty easy to enumerate all the situations.

Call the people P1, P2, P3. Call the prizes A1, A2, A3, so that the numbers of the people correspond to the prizes they initially put into the envelope.

Then they could get the prizes back in the following ways:

P1 - A1

P2 - A2

P3 - A3

(everyone gets the same prize -- if I read it right, this doesn't count)

P1 - A1

P2 - A3

P3 - A2

(One person gets their prize.)

P1 - A2

P2 - A1

P3 - A3

(One person gets their prize.)

P1 - A2

P2 - A3

P3 - A1

(No one gets their prize.)

P1 - A3

P2 - A2

P3 - A1

(One person gets their prize.)

P1 - A3

P2 - A1

P3 - A2

(No one gets their prize.)

All these situations are equally likely, so the probability that exactly one person gets their prize back is 1/2.

Looks like we each interpreted the question differently -- your answer is correct for your interpretation, since there are two situations out of six in which P1 gets A1, etc.

This is why I LOVE letsrun. Thanks!

Your order doesn't affect your odds of picking your own prize.

Timmy Jenkins wrote:

yes, you will always have a 1/3 chance, regardless of when you pick.

That's wrong timmy. Prob Theory has it right. 50%

Conditional probabilities are the summation of the independant probabilities.

The probability of the first player picking his own prize is 0.333 and the next player not picking his prize (0.50) is

0.333(0.50) = .167

The probability of the first player NOT picking his own prize is 0.333 and the next player picking his prize (0.50) is

0.667(0.50) = 0.333

The chance that either of these scenarios will happen is

0.167 + 0.333 - 0.50

proof?

The only other possibilities eveyone wins their own prize or that no one wins their own prize.

Everyone chooses own = 0.333(0.50) = 0.167

No one chooses own = 0..667(.50) = 0..333

0.167 + 0.333 = 0.50

malmo wrote:

The probability of the first player picking his own prize is 0.333 and the next player not picking his prize (0.50) is

0.333(0.50) = .167

The probability of the first player NOT picking his own prize is 0.667 and the next player picking his prize (0.50) is

0.667(0.50) = 0.333

The chance that either of these scenarios will happen is

0.167 + 0.333 - 0.50

proof?

The only other possibilities eveyone wins their own prize or that no one wins their own prize.

Everyone chooses own = 0.333(0.50) = 0.167

No one chooses own = 0..667(.50) = 0..333

0.167 + 0.333 = 0.50

Typo

Malmo, you misread the question.

The OP asked the odds that a single person picks the prize they originally put in the envelope. The OP says, "Obviously, the person who picks 1st has a 1/3 chance, but what about the 2nd and 3rd person?"

The 2nd and 3rd person also have a 1/3 chance.

ppawee wrote:

Malmo, you misread the question.

The OP asked the odds that a single person picks the prize they originally put in the envelope. The OP says, "Obviously, the person who picks 1st has a 1/3 chance, but what about the 2nd and 3rd person?"

The 2nd and 3rd person also have a 1/3 chance.

.

I did not misread the question and neither did you. I answered the correction correctly.

You are answering a question that wasn't asked. You are simply answering "what is the probability that any participant picks his own prize". That is different from "what is the probability of ONLY ONE of the participants picking his own prize." Understand?

The question is: “what are the odds that a single person gets the prize they originally put into the envelope?” [The answer is 50%. Though to answer his question correctly in the context that was asked the odds would be 1:1]

Obviously, the person who picks 1st has a 1/3 chance, but what about the 2nd and 3rd person? [ Each independent probability is 33.33%. The odds are 1:2]

Do we have to also consider the odds of me picking first (1/3), so my initial odds of getting the prize I want, before anyone picks would be 1/3 x 1/3 = 1/9? [This is the answer to an unasked an unrelated question to the original. This question begs to ask, “what is the probability that any specific individual picks his prize twice in two consecutive trials?”]

You guys should really let go of arguing with semantics.

"A single person," as it was stated in the question, could either be written as "exactly one person" or "someone," which formally in mathematics would be stated "at least one person" - depending on your interpretation.

I know you guys can each argue why your interpretation was right/better, but the fact of the matter is mathematics avoids use of ambiguous terms such as "a single person."

malmo, in case you're thinking about arguing your case, consider the following:

A: When I showed up at the locker room, there wasn't a single person there!

Do we think A is negating

1) There was not at least one person there?

2) There was not exactly one person there?

Would he really have said that if there had been two people in there?

*my bad*

1) There was at least one person there?

2) There was exactly one person there?

Kartelite, you're the one arguing semantics. If there was not a single person in the lockerroom then no one was there. If a single person was there, than one person was there.

Geez...

Putting "not" in English generally negates the statement. I'm sure you can appreciate formal logic.

A: There was a single person there.

Then,

~A = There was not a single person there.

This, according to you, means:

No one was there.

So,

~~A = There was not no one there (can't think of a better grammatically correct way to phrase it).

We can interpret this as:

~~A = Someone was there.

Now, we all know ~~A = A, so

There was a single person there = Someone was there.

Of course I appreciate formal logic, however, this is straightforward.

I answered the OP question correctly and I answered your lockerroom logic question correctly.

You keep moving to goalpost until you can frame the question into something that wasn't asked by the OP.

"The question is: what are the odds that a single person gets the prize they originally put into the envelope?"

A single person is ONE no matter how many times you move the goalpost.

You're welcome.

I'm not arguing the general use of the language. What I'm saying is that it IS ambiguous enough, when dissected formally, as to not be a properly framed question. Anytime you can easily twist the language to say ~~A != A, that's gotta be the case. So you would never see "a single person" stated on a math/logic exam, you would see "exactly one person."

The use of "or" presents a similar dilemma, actually.

Well, malmo you are changing your definition of a single person from "one and only one person" to "at least one person". In the case of this question that changes the answer and you have to admit that the question could be read both way. There's really no need to attack people for answering the alternative reading.

The chances that only one person gets their prize back is 50%. (I think this is the intended answer)

For an individual, there is a 1/3 chance of getting their own prize back.

The odds that at least one person gets there prize back is 67%.

clearly a single person gets the prize could be exactly one ((A and not (B v C) v (B and not (A v C) v (C and not (A v B)), or a single person gets the prize could mean at least one person gets the prize (A v B v C).

jrm wrote:

Well, malmo you are changing your definition of a single person from "one and only one person" to "at least one person". I

1) What the hell are you talking about? I haven' changed any definiton.

2) Who am I "attacking"?