Impeccable logic - part 2

[Softus]

Understand MH and be enlightened.

I fully understand.

I don't think you do.

Lets play with 3 boxes again and the top prize can be anywhere. If the contestant picks a box (1 of 2) as he cant touch is own. He hits the 250k and that box is eliminated. Game over he is left with another prize.

This isn't the scenario that Kes' stipulated in his first post on the first topic.

If you eliminate the £250K, then you're not following Kes' post. What you've done is describe a different scenario - one that doesn't match the Monty Hall scenario - then use that as the basis for saying that Kes' scenario is not the same as MH.

 

[Softus]

But there is also a probablility of 1:20 that the 250k can be in box B before any box has been opened, and the same for your second point. The probability of the 250k in box B will never change.

Really? I thought that all along you'd been insisting that the probability of it being in Box B reduced to 1/2.

 

[jackpot]

But there is also a probablility of 1:20 that the 250k can be in box B before any box has been opened, and the same for your second point. The probability of the 250k in box B will never change.

Really? I thought that all along you'd been insisting that the probability of it being in Box B reduced to 1/2.

Yes when the 2 boxes are left remaining a 1:2 chance. At the beginning a 1 in 20 for 1 single box.

As i said if you read carefully what you have quoted me. You said a 1 in 20 for box A and that probablity will never change.
I said the same thing for Box B.

Which concludes as they are the same probabilities then hey a 1 in 2 chance.
My my other previous post before these and compare the difference between the 2 games.

 

[megawatt]

Softus Wrote:

I don't know what megawatt wrote, but I can assure you that it's wrong.

Typical arrogant nonsense from someone who will never accept that they are wrong despite factual evidence having been provided.

Then he wrote:

2. The probability of the £250K being in Box A never changes, just like in the Monty Hall scenario.

This is total nonsense which any schoolboy could confirm.

As has been explained by a number of previous posts by both myself and others, the probability of the £250K being in Box A changes with each box opened and this is the basic flaw in all of Softus's posts on this subject.

At the time only 2 boxes remain the odds are 50:50.

Either Softus has no grasp whatsoever of basic probability or he is being deliberately obtuse ... I know where I'd put my money

I will repost the earlier maths but someone else will need to explain the following to the obtuse one as he doesn't read my posts fortunately.

At the time of choosing box A from n boxes
Odds of holding the prize = 1 in n
After the first box is opened, odds of holding the prize = 1 in (n-1)
.
.
After (n-2) boxes have been opened, odds of holding the prize = 1 in (n-(n-2))

= 50:50

If Softus wishes to attempt to discredit the above explanation it should prove interesting though its unlikely he will because he knows it's correct.

MW

 

[jackpot]

I fully understand. Its a different experiment for a number of reasons.

1. With 3 boxes left the contestant can only choice to get rid of 1 of 2 boxes. Where as Monty he is picking from 3 boxes.

2. Let play 22 boxes on Montys game. The contestant has a 21 in 22 chance of not have the box. 20 boxes are discarded knowing that the top prize is not in any of the boxes. So 100% of the time this will happen. As thats the game. Its impossible for the top prize to be in 1 of these 20 discarded boxes.

However in the deal or no deal. Its not impossible as proved in the show on many occasions. The top prize can be eliminated at any time but on this occassion the top prize is in one of the remaining 2 boxes.

3. Prity much the same as above which proves its a different experiment

Lets play with 3 boxes again and the top prize can be anywhere. If the contestant picks a box (1 of 2) as he cant touch is own. He hits the 250k and that box is eliminated. Game over he is left with another prize.

Now in Montys he picks 1 of 3 boxes as above but he can pick his own box now. And picks the same box. A goat is revealed in one of the other boxes but the contestant still does not know where the top prize is. Then he gets a choice to swap.

The person hardly has the same choice of options does he? The experiment is hardly under the same circumstances. Therefore you can not compare the 2 experiments together as they are different experiments for different scenarios.[/quote]

So are you saying this is not how deal or no deal is played?
Are you saying with 3 boxes left the contestant can pick 1 of all 3 boxes?
Are you saying the Op does not refer to a game of deal or no deal?
Are you saying that with regard to Monty game and dond, the person choosing has completely the same amount of options?

As for your last post about eliminating the 250k. It never happened in the Ops scenario but as i game of dond the op refers to it can happen as it often does on the game. Its impossible for this to happen on Montys.

 

[Softus]

How do you reconcile the following two statements:

But there is also a probablility of 1:20 that the 250k can be in box B before any box has been opened, and the same for your second point. The probability of the 250k in box B will never change.

Yes when the 2 boxes are left remaining a 1:2 chance. At the beginning a 1 in 20 for 1 single box.

______

As i said if you read carefully what you have quoted me. You said a 1 in 20 for box A and that probablity will never change.
I said the same thing for Box B.

You're being very evasive. Does the above mean that you agree with me about the probability for Box A, and that it never changes?

 

[megawatt]

As i said if you read carefully what you have quoted me. You said a 1 in 20 for box A and that probablity will never change.
I said the same thing for Box B.

Both of the above statements are wrong.

 

[Softus]

jackpot, I've finally seen where you're getting confused.

Probability isn't the same as choice.

In Kes' game, it's true to say that there are only two boxes left, and that the contestant has only two boxes to choose from, but you can't derive the probability of winning the prize from the number of choices that the contestant has.

This is also where people get confused about Monty Hall - the contestant can only ever choose to swap between two doors, and yet the probability of the car being behind door B is 2/3.

joe-90 - would I be right in thinking that this is the apparent contradiction that you've found hard to resolve?

 

[jackpot]

No matter what box you pick you have a 1 in 20 at the beginning. This can not change at this moment. As soon as you have manipulated the game in some way, i.e got rid of a box thats not 250k those odds will start decreasing 1:19. But from the start you had a 19 in 20 of not having it. Its the same odds for every box.
Hence its not like the monty game, because the options you have are different.

 

[megawatt]

Softus wrote:

Jackpot, I've finally seen where you're getting confused.

See Jackpot, its you who's getting confused ... Everything's crystal clear in Softus World

Softus wrote:

You can't derive the probability of winning the prize from the number of choices that the contestant has.

Even by Softus's posting standards this one is priceless

How exactly does Softus think probability is derived?

If you start with two boxes and one prize would anyone on this forum argue that the probability of choosing the prize is 50:50?

How then does this change if you start with 1000 boxes ... After 998 have been opened you are still faced with the same scenario ... Two boxes and a single prize.

Bluster on Softus you're not fooling anyone here.

MW

 

[jackpot]

Softus answer me this if you would.

Having 3 boxes. Why in Montys game cab the contestant pick one of 3 boxes but in deal or no deal he can only choose from 2?

 

[Softus]

As soon as you have manipulated the game in some way, i.e got rid of a box thats not 250k those odds will start decreasing 1:19.

No - there being only 19 boxes left doesn't affect the probability.

If you select 1 from 20 boxes enough times, then you would to win £250K roughly 1/20th of the number of times you make the selection. This is why 1/20 is the probability of £250K being in the box the contestant originally chose.

This probability doesn't change when you open one of the 19 other boxes and discover that it isn't £250K.

Since you claimed to agree with me 100% about Monty Hall, how do you explain the fact that the probability doesn't change in the Monty Hall scenario?

 

[jackpot]

This might be a bit technical for old Softy boy


Let K be the event a 250K is in the original box
Let S be the event we are offered a swap

For MONTY, we are always offered a swap:
P(K|S) = P(K and S) / P(S) = (1/3) / 1 = 1/3

For DOND with N boxes, contestant always keeps
choosing boxes until the 250K is revealed or only two boxes left.
P(K|S) = P(K and S) / P(S)

Clearly,
P(K and S) = 1/N
Since if the 250K is chosen, a swap will always be offered

And
P(S) = 1/N + (N-1/N)*(1/N-1) = 2/N
i.e. the probability of a swap is the probability you chose the 250K plus
the probability you didn't choose it and didn't open it in the random choices.

So,
P(K|S) = (1/N) / (2/N) = 1/2
So in the case of DOND the swap doesn’t matter.

 

[jackpot]

As soon as you have manipulated the game in some way, i.e got rid of a box thats not 250k those odds will start decreasing 1:19.

No - there being only 19 boxes left doesn't affect the probability.

So having 19 boxes and only 1 prize amongst them is not 1 in 19

 

[Softus]

Softus answer me this if you would.

Having 3 boxes. Why in Montys game cab the contestant pick one of 3 boxes but in deal or no deal he can only choose from 2?

That's just part of the game. Remember that at the point of swapping it's two doors in Monty Hall and two boxes in DoND, but it's the initial nomination that's important.

However, as I said earlier, the number of choices, boxes, doors, or goats at the moment of swapping, doesn't affect the calculation of the probability, and it's the probability that determines whether or not a swap is the right decision.