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A simpe example. (I know its a bit different but the same principle applies)

Footall match 2 teams . One is evens to win - 2 to 1 draw and the other is 4 to 1 to win. Why do you think when a goal is scored the bookmakers change the odds. So if the 4/1 teams goes 5 nil up with 3 minutes to go are the odds going to still be 4/1. I think not

I'll say this again in case it was missed the first two or three times - in the OP,

(i) there is only one £1 box in the 20

(ii) you stipulate that the £1 box and no other losing box must be the 'other' box in the end game.

If the 'other' box can be any other losing box then the switch advantage is overwhelming.

Let's try a 'qualitative' analysis.

20 boxes. One is chosen and set aside. It is no longer subject to elimination. We're looking for the £250k box. Let's say it is in the group of 19. As successive boxes from the remaining 19 are eliminated, is it not clear that

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Look -I'm not sure whether you lot are trollingor simply do not yet get it. I'll assume the latter, withdraw my last remark and try again.

As you have only joined the argument recently and seem to quote the above, i would say it was you that it is a high probability you are trolling.

Look - I'm not sure whether you lot are trolling or simply do not yet get it. I'll assume the latter,withdraw my last remarkand try again.

As you have only joined the argument recently and seem to quote the above, i would say it was you that it is a high probability you are trolling.

Reinstated.

The probability for an event where each item in the population has an equal chance is fixed at the time the event occurs. In the DOND case and the examples I gave,the event occurs when the initial choice is made. If the probability that it has occurred is not tested , then it retains the same mathematical probability. In the DOND case, the contestant's box has 1 in 20 chance and will have that 1 in 20 chance UNLESS and UNTIL it is remerged with a population of a different number.

So wrong it becomes pointless debating, a schoolboy error.

Take a pack of cards and deal 10, face down. Each card has a 1 in 52 chance of being the Ace of Clubs. Now turn up 9 of them. None is the Ace. What is now the probability of the tenth being the Ace - still 1 in 52.

Again so obviously wrong......Its 1 in 43

According to you, in this game if you turn 51 cards over, and none are the Ace, its still 1 in 52 for the last card to be the Ace..... ( Its actually a certainty and the odds are now 1 in 1 )

Your lack of understanding about odds and probabilities, is preventing reasonable debate.

Observer, since you've just described the conditions in the OP, and said that in those circumstances the 50:50 analysis is correct, then you appear to have agreed with jackpot and trazor.Observer said:I'll say this again in case it was missed the first two or three times - in the OP,I agree the 50:50 analysis is correctIF but only IF:

(i) there is only one £1 box in the 20

(ii) you stipulate that the £1 box and no other losing box must be the 'other' box in the end game.

Ignoring (for now) the next part of your post (in which you describe a different scenario, wherein you believe the odds aren't 50:50), please can you please clarify what you and jackpot are currently in disagreement about?

Ignoring (for now) the next part of your post (in which you describe a different scenario, wherein you believe the odds aren't 50:50), please can you please clarify what you and jackpot are currently in disagreement about?

It's all in the previous posts. The assertion is that the probability of the contestant's box holding the prize, which is 1 in 20 at the time it is chosen, reduces as boxes from the other group of 19 are eliminated. That is wrong. It still has a 1 in 20 chance and the balance of probability is equally split between the number of boxes in the other group. That is why, in the MH game, switching gives a 2 in 3 probability of success. The original door is still 1 in 3 but the remaining door, after MH has revealed one which is definitely NOT hiding the car, is 2 in 3 (1 - 1/3 = 2/3). So the 'always switch' strategy wins 2 times out of 3. That is proved by theoretical analysis and by practical experimentation. The same principle applies to DOND.

I have (several times, first here) agreed that in the narrow circumstances of the OP, if it is stipulated that the £1 box and ONLY the £1 box can be the 'other' box, then 50:50 is correct. At the same time, in a real game scenario, where there is an equal probability that any of the other 'losing' boxes is the other box when it comes down to the final two, the probability that a switch secures the £250k prize is 0.95.

That is simply proved by the thought experiment I suggested, where we see that the 'always switch' strategy will always secure the £250k prize except where our first box holds the £250k prize (which will occur 1 in 20 times).

The probabilty of picking the 250k first time is 1 in 20.. Agreed?

Now what if you do pick the 250k box first time.

Are you saying that when there is only 19 boxes left the probability does not change. As this is what you are saying.

So you are also saying there is still a 1 in 20 chance of picking the 250k box on your second pick even tho there is no 250k box to pick as it has been eliminated.

If you disagree with me, then tell me why you disagree with this but not your theory as its the same thing.

As you change 1 event then the probability will change.

Perhaps so, but so is a bunch of other h*rseshit and colourful new scenarios that have nothing to do with the OP, which is why I was hoping you'd provide a brief statement that defines the point on which you and jackpot disagree.It's all in the previous posts.Ignoring (for now) the next part of your post (in which you describe a different scenario, wherein you believe the odds aren't 50:50), please can you please clarify what you and jackpot are currently in disagreement about?

After all, if you can't even agree what it is that you disagree about, what hope is there is resolving that difference and coming to any agreement? None - that's what I say, and at the moment you're just trading mildly condescending posts without any party attempting to see anyone else's point of view.

You say it's wrong, but in one of your very recent posts you appeared to agree with it. Still, at least now we know that you think it's wrong.The assertion is that the probability of the contestant's box holding the prize, which is 1 in 20 at the time it is chosen, reduces as boxes from the other group of 19 are eliminated. That is wrong.

IMHO, this is the whole crux of the argument. There are those who believe that same principle applies, and those who don't. And yet I haven't seen anyone who thinks it's different give a reasoned, dispassionate,....in the MH game, switching gives a 2 in 3 probability of success.

.

.

.

The same principle applies to DOND.

I wish I'd read this before typing some of the above. Now I'm completely lost as to whether or not you agree with the OP.I have (several times, first here) agreed that in the narrow circumstances of the OP, if it is stipulated that the £1 box and ONLY the £1 box can be the 'other' box, then 50:50 is correct.

Why are you even writing about a real game scenario, when the OP is so clearly not about that?!At the same time, in a real game scenario, where there is an equal probability that any of the other 'losing' boxes is the other box when it comes down to the final two, the probability that a switch secures the £250k prize is 0.95.

Oh well, if you've proved something then it must be right.That is simply proved by the thought experiment I suggested, where we see that the 'always switch' strategy will always secure the £250k prize except where our first box holds the £250k prize (which will occur 1 in 20 times).

Frankly, the only two people are showing the least shred of honesty and humility on this topic, at the moment, are joe-90 and blondini. Both of those posters have had the guts to say that they're unsure of the answer, and are not attempting to browbeat anyone who disagrees with them.

I've love to debate the prevailing uncertainty with those people who are willing, but that's impossible at the moment because this topic is now overwhelmed by opinionated ramblings from posters who have no intention whatsoever of trying to find the middle ground.

Here - let me simplify things and save a log of time for you and trazor and jackpot:

Someone: It's 50:50.

Someone else: No it isn't. You're wrong.

Someone: No.

Someone else: How laughable it is that you think you're right.

Someone: I'm the one who's laughing, because you're wrong.

Someone else: I'm incredulous that you can think that.

Someone: Well only a stupid person would be incredulous when they're wrong.

Someone else: You're wrong.

Someone: Am not.

Someone else: Are too.

[

Like agreeing with his point of viewI'd love to debate the prevailing uncertainty with those people who are willing, but that's impossible at the moment because this topic is now overwhelmed by opinionated ramblings from posterswho have no intention whatsoever of trying to find the middle ground.

Why don't you debate it with Observer then Softus ... That should be interesting to watch.

MW

Take a look at some of those ridiculous posts that you have quoted and look underneath at the statements you make.

I have proved it mathmatically and i have proved it logically.

OK Observer. If the probability does not chance then take 20 boxes.

The probabilty of picking the 250k first time is 1 in 20.. Agreed?

Agreed.

Now what if you do pick the 250k box first time.

Are you saying that when there is only 19 boxes left the probability does not change. As this is what you are saying.

I am saying that the probability that you had picked the £250k box is still 1 in 20.

So you are also saying there is still a 1 in 20 chance of picking the 250k box on your second pick even tho there is no 250k box to pick as it has been eliminated.

If the £250k box has actually been eliminated there is obviously zero chance of choosing it from the reduced group. The probability that you had picked the £250k box first time is still 1 in 20 (but has been tested and shown not have materialised by virtue of the fact that it has been eliminated). If it has not been eliminated (which I think is an essential condition for this debate to have any purpose) the probability of it being any one of the remaining group of n boxes (say 18 ), is (1 - 1/20 )/18 = 0.0528 or 1 in 18.95 - i.e. there is a slightly better probability that any one of the remaining 18 is the £250k box than the contestant's original box. As further eliminations occur, if the £250k box remains in the game, the probability that it is in the 'group' increases. When it comes down to the last two, the probability is 19 in 20 (because only 1 time in 20 will it be the contestant's first chosen box).

If you disagree with me, then tell me why you disagree with this but not your theory as its the same thing.

I think I have. I have set out illustrations starting with smaller numbers as well as the mathematical theory. The MH game starts with three boxes (or doors). The principles are exactly the same. In the MH game, Monty knows where the car is it must stay in the game. In DOND, the OP stipulation is that the £250k box stays in the game.

As you change 1 event then the probability will change.

The contestant's box cannot be eliminated. It is sitting there, having been chosen from a population of 20, with the same probability that it is the £250k box (or any other specific box) as it had when it was selected.

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