Your assumption is wrong:- the education I'm referring to was pre-GCE level - really quite basic. It's just that I'm averagely intelligent, have an extremely good memory, and the work I do is mostly numerate. I have no idea about your education or attributes, but you seem to be incapable of defining and analysing a problem involving statistical probability.joe-90 said:
Impeccable logic
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Is Softus really passing by the chance to prove the great Joe-90 wrong?
I can't believe that to be the truth. The first person in the history of the forum to prove the oracle known as Joe-90 wrong and what does Softus do? He declines.
Then again, maybe he recognises greatness when he sees it.
I can't believe that to be the truth. The first person in the history of the forum to prove the oracle known as Joe-90 wrong and what does Softus do? He declines.
Then again, maybe he recognises greatness when he sees it.
See the discussion of veridical paradox at your favourite web site.joe-90 said:
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The Kes problem is a non starter, if you read his post carefully his opponent discards all cards but the Ace of Spades, thus his opponent will hold the Ace 51 times out of 52 on average. hardly comparable to Deal or no Deal.
The Monty hall scenario is a red Herring, and is a totally different statistical problem, and is not comparable to Deal or no Deal.
1) Firstly Deal or no Deal uses 22 boxes not 20
2) You are given seven offers, the first after discarding 5 boxes, the remainder after discarding 3 boxes each time, until you are left with 2 boxes.
3)Your chances of HAVING the £250,000 prize after each round is as follows...
Round 1 .....17 in 22
Round 2 .....14 in 17
Round 3 .....11 in 14
Round 4 .....8 in 11
Round 5 .....5 in 8
Round 6 .....2 in 5
Round 7 .....1 in 2 ...........50:50
Multiplying the first column gives 17 x 14 x 11 x 8 x 5 x 2 x 1 =209440
Multiplying the 2nd column gives 22 x 17 x 14 x 11 x 8 x 5 x 2 =4607680
4607680 divided by 209440......gives a 1 in 22 chance of selecting the £250000 box.
Any attempt to skew the 1 in 2 chance would result in the original 1 in 22 chance being wrong, which we all know is true.
So 50:50 is correct...................
The Monty hall scenario is a red Herring, and is a totally different statistical problem, and is not comparable to Deal or no Deal.
1) Firstly Deal or no Deal uses 22 boxes not 20
2) You are given seven offers, the first after discarding 5 boxes, the remainder after discarding 3 boxes each time, until you are left with 2 boxes.
3)Your chances of HAVING the £250,000 prize after each round is as follows...
Round 1 .....17 in 22
Round 2 .....14 in 17
Round 3 .....11 in 14
Round 4 .....8 in 11
Round 5 .....5 in 8
Round 6 .....2 in 5
Round 7 .....1 in 2 ...........50:50
Multiplying the first column gives 17 x 14 x 11 x 8 x 5 x 2 x 1 =209440
Multiplying the 2nd column gives 22 x 17 x 14 x 11 x 8 x 5 x 2 =4607680
4607680 divided by 209440......gives a 1 in 22 chance of selecting the £250000 box.
Any attempt to skew the 1 in 2 chance would result in the original 1 in 22 chance being wrong, which we all know is true.
So 50:50 is correct...................
If you read the OP carefully, you'll see that it isn't at all important whether or not this scenario is (a) the same as DoND, or (b) similar to DoND.
You believing that something is not comparable is, sadly, a very different to that thing actually being not comparable.
I don't think anyone is attempting to skew anything.
What you seem to be doing is answering the wrong question. It's erroneous to attempt to derive the probability of any particular contents being in one of the two remaining boxes, with no prior knowledge of what has occurred, because the question is: Is it better to swap?
Statistically, i.e. if you repeat the attempt with a large enough sample set, then it's better to swap.
If you doubt the truth of this, then why not play it out in reality?
Very intriguing thread. I have to say I cant quite get my head around a lot of what is being discussed, but it is interesting none the less.
This is my take on it.... apologies if it seems quite simplistic, but it would be interesting to have explained to me what I am missing... here goes..... now.......
Lets say we are at the point where the contestant has the last two boxes, and is offered the chance to swap. Now it is argued that swapping would be statistically favourable thing to do. However......
Lets pretend the show was some how 'put on pause', and a bloke from the street was wheeled in and asked to choose a box instead. His chance of picking the jack pot would be 50 \ 50, right??? So why would the first chaps be say, 55 \ 45 ( I am picking numbers out of the air here, but you get my drift )
This is my take on it.... apologies if it seems quite simplistic, but it would be interesting to have explained to me what I am missing... here goes..... now.......
Lets say we are at the point where the contestant has the last two boxes, and is offered the chance to swap. Now it is argued that swapping would be statistically favourable thing to do. However......
Lets pretend the show was some how 'put on pause', and a bloke from the street was wheeled in and asked to choose a box instead. His chance of picking the jack pot would be 50 \ 50, right??? So why would the first chaps be say, 55 \ 45 ( I am picking numbers out of the air here, but you get my drift )
Well explained Trazor and I agree 100% with your analysis.
Softus ... Why don't you, for once, stop doing your usual "I'm not prepared to accept I'm wrong so I'll f@nny around with big words and daft statements" routine and hold your hands up.
Now if I were Joe-90 I'd say (he won't) but, as I'm not, I won't.
MW
Softus ... Why don't you, for once, stop doing your usual "I'm not prepared to accept I'm wrong so I'll f@nny around with big words and daft statements" routine and hold your hands up.
Unprovable nonsense straight from the Softus classic archives I'm afraid
Now if I were Joe-90 I'd say (he won't) but, as I'm not, I won't.
MW
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Grrinc. It's not really a paradox at all. When the original player faces Monty Hall there are 2 goats and 1 car involved and as MH knows where the car is - he can then influence the game.
When the guy on the street enters the show there is only 1 goat and 1 car involved.
What this means is you are comparing cabbages and cauliflowers or mathematically 1/3 compared to 1/2.
If MH didn't know where the car was then he would often reveal the car when he opened the first door - but in this example he NEVER DOES - and that is where the odds are distorted.
You got it right. Well done.
When the guy on the street enters the show there is only 1 goat and 1 car involved.
What this means is you are comparing cabbages and cauliflowers or mathematically 1/3 compared to 1/2.
If MH didn't know where the car was then he would often reveal the car when he opened the first door - but in this example he NEVER DOES - and that is where the odds are distorted.
You got it right. Well done.
Wrong.grrinc said:
Even if that were the scenario, and it isn't, the probability would be 0.5 only if there were two boxes originally, i.e. such that the top prize could only ever have been in one of two boxes.
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Here's another one...
Two people disagree. The first maintains an open mind, is prepared to politely accept an opposing point of view, and is happy to agree to disagree. The second belligerently and sarcastically insists that he's right and that anyone who disagrees is wrong and is talking "piffle". What's the probability that the second person is joe-90?
90 to 1
Following
Here's another one ...
Two people disagree. The first tries to convince everyone that he actually wears his underpants outside his trousers in an attempt to mask the fact that either:
(a) He hasn't got a scooby's about what is being discussed.
or
(b) He knows he's wrong but is simply too infantile to accept it and move on.
Whilst the second, who is generally a pain in the arris but on this occasion is 100% correct.
What is the probability that the first is Softus and he's sat somewhere sucking on his thumb?
Let me see ... Errrmmm ... Answer = 1
MW
Here's another one ...
Two people disagree. The first tries to convince everyone that he actually wears his underpants outside his trousers in an attempt to mask the fact that either:
(a) He hasn't got a scooby's about what is being discussed.
or
(b) He knows he's wrong but is simply too infantile to accept it and move on.
Whilst the second, who is generally a pain in the arris but on this occasion is 100% correct.
What is the probability that the first is Softus and he's sat somewhere sucking on his thumb?
Let me see ... Errrmmm ... Answer = 1
MW
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