Impeccable logic - part 2

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Lets put it another way now as its time that sense was sought.

Let’s say that at the end of the game you’re left with 2 boxes, the top prize and the bottom prize. You can, of course, apply the reasoning that the odds were 1/22 for you to pick the top prize, having opened all the boxes but the remaining box you can say that it is therefore 21/22 that the other box contains the 250k.

Lets flip it the other way around. At the beginning it was also 1/22 that you chose the bottom prize (£1), and when you reach the end of the game, according to the exact same conditional probability softus is saying, it’s also a chance of 21/22 that the last remaining box contains the £1 also.

The exact same probabilities as if you had the 250K or again any other box.
Lets imagine you get to the final two boxes again, 1 box contains the 250k and 1 contains another prize. There are 2 ways you could have got here, either by choosing the jackpot as your box, or by choosing to open every box but the jackpot.

The odds of choosing the jackpot are 1/22
The odds of leaving the jackpot until last are 21/22 * 20/21….*2/1 = 1/22

Hence once again a 50:50 chance
 
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jackpot said:
So having 19 boxes and only 1 prize amongst them is not 1 in 19 :eek:
It's certainly a 1 in 19 choice, but you're never offered that choice in DoND.
 
Jackpot:

I don't believe Softus is actually so stupid as to believe what he has been posting and simply think he is goading you now ... This is what Softus does, trust me ;)

Any year 12 maths student knows that probability is based on the circumstances presented and isn't influenced by prior events so there's little point arguing it ... It's a fact!

For Softus to say that there is a 1 in 20 chance with 20 boxes and still a 1 in 20 chance with 19 simply tells me he is winding you up :LOL:

Ignore him with any luck he'll go away.

MW
 
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Megawatt your right. I have proved my point analytically and statistically on a number of occasions and there is no argument left. I would like to see him argue with my last two statements that prove this as he wont be arguing with me but the whole concept of Mathematics.

Its been very entertaining tho
 
You have indeed as have I and others but we have all, I fear, been debating in vain with the Soft fella ;)

Take a look back over some of the Softus diatribes (search for his posts) ... Some of them are stunning :LOL:

To be fair and balanced though he does post useful stuff also and certainly makes me laugh at times.

When he gets like this though you really are wasting your life ;)

There should be a Softus health warning in the forum information section to protect OP's :LOL:

MW
 
Can someone recap for me please as I've lost the plot. :oops:

In Deal-no deal it is 50-50 as that particular game comes up every x number of games - so swapping makes no difference. Yes or no?


In the Monty Hall thing swapping increases the odds to 2/3? (even though i still can't see how). Yes or no?
 
DoND 50:50 ... No idea about the MH one as I haven't given it any thought.

Others can comment but reading back through prior posts I think the general consensus seems to be 2/3 as you say.

MW
 
Yup, at the start 1:n of the top prize, n:1 of a dud ... When down to the last 2 boxes though 50:50 on either ;)

Did you manage your paper exercise Blondini?
 
I have to congratulate Softus and a few others who have patiently tried to explain why their analysis of the Monty Hall and DOND problems are correct. I believe they are and I have been searching for the simple and succinct explanation that will convince the 50:50 sceptics. I think I've found it.

Taking Monty Hall first, the simple question is whether an 'always swap' strategy is better than a random 'stick or swap' choice. It is certainly true that the random 'stick or swap' is a 50:50 chance but the 'always swap' does give a 2 in 3 or 66% chance of success. Here's why. The initial choice of door, obviously, gives a 1 in 3 chance of picking the car. An 'always swap' strategy will ONLY lose if the initial choice is the 'correct' one (with the car). With any other first choice, bearing in mind that Monty will always expose another 'losing' door, we will win. If the losing chance is 1 in 3, the winning chance must be 2 in 3. So the ONLY losing probability is determined by the initial probability of choosing the 'correct' door and the winning probabilty is what remains. This is indisputable.

Similarly with the DOND problem. If an 'always swap' strategy is adopted, we will ONLY lose if our initial choice was the £250k box. If any other box was chosen and all the other 'losing' boxes are eliminated (it doesn't matter how), then 'always swap' will win the £250k prize. If we start off with 20 boxes, it is clear we have a 1 in 20 chance of picking the £250k box with our initial choice. So an 'always swap' strategy has a 19:1 probability of winning. It must be so. Again, a random 'stick or swap' startegy gives a 50:50 chance, but that's not smart.

I hope everyone will now see that Softus et al were quite correct.
 
Observer, somewhat bizarrely, I tentatively disagree with you regarding DoND. I agree with your analysis of the MH scenario, and your analysis of the DoND scenario certainly matches mine on this topic, but my rethink was sparked by a valid observation that jackpot made.

I've been thinking about this over the last few days, and have nearly come to a conclusion, which I hope to put into words and post here.

Pending that, I'd like to take this opportunity to thank you for your kind words, which are in stark contrast to the gloating and crowing displayed by those who think they've been in an argument and that they've won something.
 
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