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The two envelopes problem
 Posted: 19 November 2012 10:00 AM [ Ignore ]   [ # 1621 ]
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OK. One step further:

Situation 1: I, the game master, have two sets of envelopes, (10,20) and (20,40). I toss a fair coin, and dependent on it, I choose one of the pairs. Now I play TEP with you with that pair. You pick 20. What could the other envelope contain? And what is your chance to gain when you switch?

Situation 2: I, the game master, come and bring you two envelopes and you may choose one. You take one envelope and it contains \$20. Now I toss a fair coin, and dependent on it I put 10 or 40 in the second envelope. What could this other contain? And what is your chance to gain when you switch?

In both situations, it is not told in advance of course what the amounts are, you have idea what I, the game master did.

Now you have all possible information to calculate the chances. Please go ahead.

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 Posted: 19 November 2012 10:28 AM [ Ignore ]   [ # 1622 ]
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GdB - 19 November 2012 10:00 AM

OK. One step further:

Situation 1: I, the game master, have two sets of envelopes, (10,20) and (20,40). I toss a fair coin, and dependent on it, I choose one of the pairs. Now I play TEP with you with that pair. You pick 20. What could the other envelope contain? And what is your chance to gain when you switch?

The other envelope could contain \$10 or \$40. I might gain \$20 and might lose \$10 with 50/50 probability. So if we imagine doing this twice and me losing \$10 and gaining \$20, I’d gain \$10 over the two goes, which is \$5 over one go.

Situation 2: I, the game master, come and bring you two envelopes and you may choose one. You take one envelope and it contains \$20. Now I toss a fair coin, and dependent on it I put 10 or 40 in the second envelope. What could this other contain? And what is your chance to gain when you switch?

In both situations, it is not told in advance of course what the amounts are, you have idea what I, the game master did.

edit: I think I misunderstood this so have deleted my comment.

Stephen

[ Edited: 19 November 2012 11:17 AM by StephenLawrence ]
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 Posted: 19 November 2012 11:51 PM [ Ignore ]   [ # 1623 ]
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StephenLawrence - 19 November 2012 10:28 AM
GdB - 19 November 2012 10:00 AM

OK. One step further:

Situation 1: I, the game master, have two sets of envelopes, (10,20) and (20,40). I toss a fair coin, and dependent on it, I choose one of the pairs. Now I play TEP with you with that pair. You pick 20. What could the other envelope contain? And what is your chance to gain when you switch?

The other envelope could contain \$10 or \$40. I might gain \$20 and might lose \$10 with 50/50 probability. So if we imagine doing this twice and me losing \$10 and gaining \$20, I’d gain \$10 over the two goes, which is \$5 over one go.

Who says that on a second time you get \$20 again?

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 Posted: 20 November 2012 12:33 AM [ Ignore ]   [ # 1624 ]
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GdB - 19 November 2012 11:51 PM
StephenLawrence - 19 November 2012 10:28 AM
GdB - 19 November 2012 10:00 AM

OK. One step further:

Situation 1: I, the game master, have two sets of envelopes, (10,20) and (20,40). I toss a fair coin, and dependent on it, I choose one of the pairs. Now I play TEP with you with that pair. You pick 20. What could the other envelope contain? And what is your chance to gain when you switch?

The other envelope could contain \$10 or \$40. I might gain \$20 and might lose \$10 with 50/50 probability. So if we imagine doing this twice and me losing \$10 and gaining \$20, I’d gain \$10 over the two goes, which is \$5 over one go.

Who says that on a second time you get \$20 again?

You know what I meant.

Fact is each time I get \$20 I should switch, which is the answer to the question.

What ever number I get if I think of the two pairs I could have and assume equal chance of having either of those two pairs I should switch.

If I switched every time regardless of what I got in your scenario I would gain nothing but that is because I would be wrong to assume double or half with 50/50 probability on every turn.

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 Posted: 20 November 2012 12:43 AM [ Ignore ]   [ # 1625 ]
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OK the answer to the TEP is if we set up an actual game and played out all the possible scenarios we would gain nothing by switching.

We are wrong to assume half or double with 50/50 probability.

As in GdB’s example although the probability was 50/50 if I selected \$20, it was different if I selected other possible amounts.

This makes sense but still the question arises if I’m wrong to assume 50/50 probability what probability should I assume?

The answer to that, I guess, is I shouldn’t assume any probability at all, I just don’t know. But this doesn’t sit well with I know there is nothing to gain by switching

So I still need to fit these two together in my mind to say I know the solution.

Stephen

[ Edited: 20 November 2012 12:47 AM by StephenLawrence ]
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 Posted: 20 November 2012 12:53 AM [ Ignore ]   [ # 1626 ]
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StephenLawrence - 20 November 2012 12:33 AM

You know what I meant.

Fact is each time I get \$20 I should switch, which is the answer to the question.

No. Fact is that with 2 pairs (10,20) and (20,40) you have a chance of getting 20. You do as if the chance for getting 20 is 100%. It isn’t. Same with completely unknown pairs. You treat the amount you find as its chance was 100% anyway. That is wrong. With one pair (10,20) your chance of getting 20 is 50%. After you have seen it the chance still was 50%.

Therefore the correct treatment with two pairs is saying that the chance for 10 was 25%, for 40 it was 25%, and for 20 it was 50%.

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 Posted: 20 November 2012 01:09 AM [ Ignore ]   [ # 1627 ]
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GdB - 20 November 2012 12:53 AM
StephenLawrence - 20 November 2012 12:33 AM

You know what I meant.

Fact is each time I get \$20 I should switch, which is the answer to the question.

No.

Sorry GdB but yes. In your set up I should switch.

Fact is that with 2 pairs (10,20) and (20,40) you have a chance of getting 20. You do as if the chance for getting 20 is 100%. It isn’t.

Nope. Sorry but you are just wrong. I’m doing it as if there is equal chance of the pairs being (10,20) and (20,40) and I have \$20. That was the correct answer to your question.

Same with completely unknown pairs. You treat the amount you find as its chance was 100% anyway.

No this is wrong. We are agreeing to treat the chance of the two possible pairs as equal, if we do that we should switch. The probability of me picking 20 has nothing to do with it. Each time I do pick \$20 I should switch.

Therefore the correct treatment with two pairs is saying that the chance for 10 was 25%, for 40 it was 25%, and for 20 it was 50%.

As you know the relevant probability is 100% once I know what my number is. As you know I should switch each time I get \$20 in your scenario.

The correct treatment is saying If I get \$10 I should switch if I get \$20 I should switch and if I get \$40 I should not switch, in your scenario.

What you are doing makes no sense.

Stephen

[ Edited: 20 November 2012 01:12 AM by StephenLawrence ]
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 Posted: 20 November 2012 01:18 AM [ Ignore ]   [ # 1628 ]
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GdB - 20 November 2012 12:53 AM
StephenLawrence - 20 November 2012 12:33 AM

You know what I meant.

Fact is each time I get \$20 I should switch, which is the answer to the question.

No. Fact is that with 2 pairs (10,20) and (20,40) you have a chance of getting 20. You do as if the chance for getting 20 is 100%.

This is actually funny, sorry again.

Imagine I am given two envelopes one with \$10 in it and one with \$50 in it. I open one and get the \$50. Now I’m offered the chance to switch that for \$38.

Should I do it as if the chance of my having \$50 is only 50% and switch!

Stephen

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 Posted: 20 November 2012 01:36 AM [ Ignore ]   [ # 1629 ]
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GdB - 20 November 2012 12:53 AM
StephenLawrence - 20 November 2012 12:33 AM

You know what I meant.

Fact is each time I get \$20 I should switch, which is the answer to the question.

No. Fact is that with 2 pairs (10,20) and (20,40) you have a chance of getting 20. You do as if the chance for getting 20 is 100%. It isn’t. Same with completely unknown pairs. You treat the amount you find as its chance was 100% anyway. That is wrong. With one pair (10,20) your chance of getting 20 is 50%. After you have seen it the chance still was 50%.

Therefore the correct treatment with two pairs is saying that the chance for 10 was 25%, for 40 it was 25%, and for 20 it was 50%.

What if I got \$10? I suppose I’d be wrong to switch then too because I might have had \$40.

Stephen

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 Posted: 20 November 2012 01:51 AM [ Ignore ]   [ # 1630 ]
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Interestingly GdB, what you did was set up a scenario in which looking made a difference. If I didn’t look my expected gain was 0 but that changed on looking, just as it would have if I’d found a \$10.

So perhaps that’s part of the solution, looking does make a difference but in the TEP we don’t know what that difference is.

Stephen

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 Posted: 20 November 2012 02:06 AM [ Ignore ]   [ # 1631 ]
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StephenLawrence - 20 November 2012 01:09 AM

Sorry GdB but yes. In your set up I should switch.

No. You forget I said you don’t know what the amounts are that the game master is selecting. You would be correct when you know that the pairs ar (10,20) and (20,40), but you don’t.

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 Posted: 20 November 2012 01:03 PM [ Ignore ]   [ # 1632 ]
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GdB - 20 November 2012 02:06 AM
StephenLawrence - 20 November 2012 01:09 AM

Sorry GdB but yes. In your set up I should switch.

No. You forget I said you don’t know what the amounts are that the game master is selecting. You would be correct when you know that the pairs ar (10,20) and (20,40), but you don’t.

Let me check what I know. Are you saying I don’t know that there is either half or double what is in my envelope in the other envelope? And are you saying I don’t know that you tossed a coin to decide which pair of envelopes to play with?

Stephen

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 Posted: 20 November 2012 06:54 PM [ Ignore ]   [ # 1633 ]
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StephenLawrence - 20 November 2012 01:03 PM

Let me check what I know. Are you saying I don’t know that there is either half or double what is in my envelope in the other envelope? And are you saying I don’t know that you tossed a coin to decide which pair of envelopes to play with?

You know the procedure, that there were two pairs of envelopes, (X,X/2) and (X,2X), that I tossed a coin to decide which pair I would take, but you do not know what X is. The pairs could have been ((3,6),(6,12)), ((5,10),(10,20)), ((10,20),(20,40)), ((20,40),(40,80)), ((111,222),(222,444)), or whatever.

You open you envelope and it contains 20. What was the chance for having 20? And for getting 10 or 40 by switching?

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 Posted: 21 November 2012 12:05 AM [ Ignore ]   [ # 1634 ]
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GdB - 20 November 2012 06:54 PM
StephenLawrence - 20 November 2012 01:03 PM

Let me check what I know. Are you saying I don’t know that there is either half or double what is in my envelope in the other envelope? And are you saying I don’t know that you tossed a coin to decide which pair of envelopes to play with?

You know the procedure, that there were two pairs of envelopes, (X,X/2) and (X,2X), that I tossed a coin to decide which pair I would take, but you do not know what X is. The pairs could have been ((3,6),(6,12)), ((5,10),(10,20)), ((10,20),(20,40)), ((20,40),(40,80)), ((111,222),(222,444)), or whatever.

You open you envelope and it contains 20. What was the chance for having 20? And for getting 10 or 40 by switching?

OK, so given what I have the two pairs could have been A, (5,10),(10,20)  or B,(10,20),(20,40) or C, (20,40),(40,80)

I can’t use the chance of getting \$20 to do the calculation since I have it. If that’s wrong you do need to explain why.

Assuming a probability 1/3 that I have either one of these three pairs, I should switch.

Clearly I shouldn’t assume that probability as clearly this is wrong because it means I should switch whatever number I see.

Stephen

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 Posted: 21 November 2012 12:15 AM [ Ignore ]   [ # 1635 ]
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StephenLawrence - 21 November 2012 12:05 AM

I can’t use the chance of getting \$20 to do the calculation since I have it. If that’s wrong you do need to explain why.

You throw a die. What is the chance that you throw a 3?
You have thrown a die. It was 3. What was the chance on throwing a 3?

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