Nice bit of maths on the BBC

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Re: Nice bit of maths on the BBC
What often seems to happen is that you've got a rather more reliable secondary test which takes a lot more time/effort than an initial screening. So what you have to judge is the cost, both financial and potential side effects if its surgical in some way, of applying that secondary test to false positives vs that of missing some positives. Potential mental trauma too of course.
It isn't easy but there's plenty of folk in the NHS/Nice etc in charge of judging this sort of thing.
After seeing some people completely refusing to accept monty hall (the bridge version, but absolutely identical reasoning) I dunno. It clearly confuses at least some people quite badly.
It isn't easy but there's plenty of folk in the NHS/Nice etc in charge of judging this sort of thing.
After seeing some people completely refusing to accept monty hall (the bridge version, but absolutely identical reasoning) I dunno. It clearly confuses at least some people quite badly.
 Paolo Casaschi
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Re: Nice bit of maths on the BBC
No. I believe the host is not cheating because the rules say the host knows what's behind each door and always open an empty door. Fully compliant to the rules.Roger de Coverly wrote:The critical point is noting that the host is bending the rules by cheating, that is by making an initial selection to open that he knows doesn't contain the prize.
 Greg Breed
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Re: Nice bit of maths on the BBC
I freely admit that I cannot get my head around this Monty Hall problem.
If the game show host opens one door/box and reveals a goat he is effectively removing that from the equation. My brain therefore tells me the odds are still 50/50.
In deal or no deal, if we remove the banker element and just focus on the boxes, when you get to the end there are 2 left. The one you started with and the one remaining. That is a 50/50 regardless of how many went before. If they offer you to swap its still 50/50.
I can understand that showing the contents behind door 3 gives a clue but how does it improve the odds?
If the game show host opens one door/box and reveals a goat he is effectively removing that from the equation. My brain therefore tells me the odds are still 50/50.
In deal or no deal, if we remove the banker element and just focus on the boxes, when you get to the end there are 2 left. The one you started with and the one remaining. That is a 50/50 regardless of how many went before. If they offer you to swap its still 50/50.
I can understand that showing the contents behind door 3 gives a clue but how does it improve the odds?
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Re: Nice bit of maths on the BBC
The way I usually explain it is to expand it to 100 boxes. What are your chances of having picked the right box  1 in 100. So you know that at least 98 of the boxes you haven't chosen are empty. If you are shown those 98 empty boxes, does that improve your ORIGINAL odds of picking the right box? No, clearly not. Hence in this situation you should always swap as you are effectively betting against your original choice, which was 99/1 against.

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Re: Nice bit of maths on the BBC
At the start of the game, you select one box. That's a one in three chance of being a winner. The host picks one of the two remaining boxes, but not at random, so he knows there is no prize in the one he picks.Greg Breed wrote: I can understand that showing the contents behind door 3 gives a clue but how does it improve the odds?
In the remaining box, there might be a prize, there again there might not. If there isn't a prize, then you've already selected a prize winning box with a onethird probability of being correct. If you haven't selected it, then the prize must be in the second box. At probabilities add to 1, that box has a twothirds probability of being correct.
If the host made a random selection, you could win immediately with a probability of one third. If you didn't win immediately and you are allowed another try, you are then up to a 50% chance.
 Greg Breed
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Re: Nice bit of maths on the BBC
You make sense Simon, but my thick skull seems impervious to the logic.Simon Brown wrote:The way I usually explain it is to expand it to 100 boxes. What are your chances of having picked the right box  1 in 100. So you know that at least 98 of the boxes you haven't chosen are empty. If you are shown those 98 empty boxes, does that improve your ORIGINAL odds of picking the right box? No, clearly not. Hence in this situation you should always swap as you are effectively betting against your original choice, which was 99/1 against.
1. You pick 1 box out of 100 = 1 in 100.
2. 98 boxes are removed.
3. 2 boxes remain the one chosen and the other.
So the probablity stuff says that I should swap because the "other" box is 50/50 whereas the first box was chosen at 1/100?
But that doesn't change the contents. Having gotten down to the last 2 boxes, the chance of the prize being in the initial one picked is the same no? Sorry if I'm being dense
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 Greg Breed
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Re: Nice bit of maths on the BBC
I don't understand this bit, can you explain?Roger de Coverly wrote: At probabilities add to 1, that box has a twothirds probability of being correct.
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Re: Nice bit of maths on the BBC
There's only one prize, so it's in now one of two boxes. When you picked the first box you were choosing one from one hundred. If you picked correctly, you have the prize. If you didn't pick correctly, the prize is in the other box. What is the chance you were wrong on your original pick? 99 out of a hundred, therefore you swop. The key point is that the person showing you the empty boxes knows where the prize is.Greg Breed wrote: But that doesn't change the contents. Having gotten down to the last 2 boxes, the chance of the prize being in the initial one picked is the same no?

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Re: Nice bit of maths on the BBC
I don't understand this bit, can you explain?[/quote]Greg Breed wrote:
(Probability of A) plus (Probability of not A) adds up to 1. So if the Probability of not A is easy to figure out, then work out what that is and derive Probability of A by subtraction. Say with two dice, you want the probability of not scoring a given number. The easiest solution is to work out the probability of throwing that number. The probability you want is 1 minus it.
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Re: Nice bit of maths on the BBC
So those playing deal or no deal  if they decide not to accept the banker's offers and try to pick the "winning" box should always swap?Roger de Coverly wrote:There's only one prize, so it's in now one of two boxes. When you picked the first box you were choosing one from one hundred. If you picked correctly, you have the prize. If you didn't pick correctly, the prize is in the other box. What is the chance you were wrong on your original pick? 99 out of a hundred, therefore you swop. The key point is that the person showing you the empty boxes knows where the prize is.Greg Breed wrote: But that doesn't change the contents. Having gotten down to the last 2 boxes, the chance of the prize being in the initial one picked is the same no?
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Re: Nice bit of maths on the BBC
It depends what rules the bank are playing to.Greg Breed wrote: So those playing deal or no deal  if they decide not to accept the banker's offers and try to pick the "winning" box should always swap?

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Re: Nice bit of maths on the BBC
I said it was hard Really the best thing to do is to accept that your intuition sucks at this  a very, very common state to be in  and work out the cases by hand. Quite managable here.
The thing about the behaviour of the host is crucial. Take boxes A,B,C. I'll put * by the one with the thing in it. Presume that you always pick box A to start with. Three base cases of course: A*, B, C; A, B*, C; A, B, C*.
Consider a position where you've chosen A and they've opened B. Then yes there are two cases for the boxes: A*, B(o), C or A, B(o), C*. Also the two base cases of A*,B,C and A,B,C* are equally probable, thus the inituitive suggestion of 50/50. However, the normal assumption is that the host would have chosen between two possible empty doors with equal probability.
In that case, if the prize was in the box you chose, they would have opened door B half the time and C the other half. However, if the prize was in box C they had to open B. So the chance that you're in case A*,B,C once the host opens a specific door is actually cut in half yielding the 21 odds on switching.
(This is I think the fundamental way to explain but sadly far from intuitive.).
For a lot of host behaviours you do actually get 50/50 on switching. The most obvious if they open an empty box before you make your choice. Or if they wait until you've chosen and then open a fixed box, C say. Then you've either got a 50/50 choice between A and B or a rather easier problem if the prize was actually in C
It gets slightly more complex if they'll always open B except when it contains the prize. In that case you only get 50/50 on switching when they open B, but absolute certainty on switching when they open C.
This actually looks rather simple in terms of strategy as I don't think you can ever lose by switching under normal assumptions. This sort of thing can get very much messier in bridge and hugely so in game theory.
The thing about the behaviour of the host is crucial. Take boxes A,B,C. I'll put * by the one with the thing in it. Presume that you always pick box A to start with. Three base cases of course: A*, B, C; A, B*, C; A, B, C*.
Consider a position where you've chosen A and they've opened B. Then yes there are two cases for the boxes: A*, B(o), C or A, B(o), C*. Also the two base cases of A*,B,C and A,B,C* are equally probable, thus the inituitive suggestion of 50/50. However, the normal assumption is that the host would have chosen between two possible empty doors with equal probability.
In that case, if the prize was in the box you chose, they would have opened door B half the time and C the other half. However, if the prize was in box C they had to open B. So the chance that you're in case A*,B,C once the host opens a specific door is actually cut in half yielding the 21 odds on switching.
(This is I think the fundamental way to explain but sadly far from intuitive.).
For a lot of host behaviours you do actually get 50/50 on switching. The most obvious if they open an empty box before you make your choice. Or if they wait until you've chosen and then open a fixed box, C say. Then you've either got a 50/50 choice between A and B or a rather easier problem if the prize was actually in C
It gets slightly more complex if they'll always open B except when it contains the prize. In that case you only get 50/50 on switching when they open B, but absolute certainty on switching when they open C.
This actually looks rather simple in terms of strategy as I don't think you can ever lose by switching under normal assumptions. This sort of thing can get very much messier in bridge and hugely so in game theory.
Re: Nice bit of maths on the BBC
In Deal or No Deal, the banker does not know the contents of any box, so your chances are 50/50 and swapping will not affect the odds.Roger de Coverly wrote:It depends what rules the bank are playing to.Greg Breed wrote: So those playing deal or no deal  if they decide not to accept the banker's offers and try to pick the "winning" box should always swap?

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Re: Nice bit of maths on the BBC
In front of you are a simple set of scales and a large number of sealed cardboard boxes. The boxes appear identical, but actually all contain different things. You pick one at random as your box. You pick three more, and find two are heavier than your box, and one is lighter. What is the probability that the next box you pick will be heavier than yours?

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Re: Nice bit of maths on the BBC
Now there's a hard question I can actually remember doing a course about trying to address that sort of thing as sanely as possible.... Which isn't very!