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(I'll say it again) you ONLY way you lose IF you switch is if your original choice contains the £250k. The chance of that is 1 in 20 and remians 1 in 20 while all 18 losing boxes are eliminated. So the chance of the switch yielding the £250k, at that point, with 18 other losing boxes eliminated, is 19 in 20.

And this is where you keep going wrong, you can not treat the contestants box as anything other than one of the remaining boxes.

If their are 10 boxes remaining the odds of the £250,000 being in any box left is 1 in 10, INCLUDING THE CONTESTANTS BOX.

You just can not treat the contestants box as a special box, and keep its odds at 1 in 20.

Until you understand this basic flaw in your argument, you will not understand.

http://www.dond.co.uk/deal_or_no_deal_stats.php

This shows the £250,000 being in the contestants box 22 times out of 605 shows.

So it has appeared 1 in 27.5 shows in the contestants box.

By using your theory it must appear 21 times more often in the other remaining box, therefore it appeared 21 x 22 = 462 times..

So out of 605 games the £250,000 appeared 462 times in the last remaining box, ready for the contestant to swap to...Excuse me if I laugh out loud.

The premise of the OP was that only the box containing £250k and another containing £1 are known to be in play when a swap is offered.

Therefore if this scenario had occured in any of those 605 dond games, then in 22 of them swapping would have lost the big prize. In the other 583 swapping would have won.

How many of the dond games started with £1 in the contestants box? It doesn't matter does it.

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Some of you (the ones who have not yet seen the light) argue that the probability of the first choice box holding the prize changes as other boxes are eliminated. So the initial 1 in 20 chance becomes 1 in 16 when 4 boxes have been eliminated. That is wrong, although it would be correct if they were all mixed up and a new random choice was made. We are measuring the probability of an event occurring. That is fixed when it occurs, not when it is tested.

Dear oh dear.

Take a 1p, 2p, 1p, 5p, 10p, 20, 50p and £1 coin. Line them up in front of you. You have 8 seperate coins. You also have 2 x 1p pieces. Now this in respect to ration is 2:8. Now ask someone to pick a coin. The chance they will pick a penny is 2 in 8. And as all the other coins are different then the chance that either 2p, 5p, 10p, 20p, 50, or £1 is a 6 in 8 chance. And the chance of them picking the £1 is a 1 in 8 chance.

Now remove one of the pennies, 2p, 5p, 10, and 20p

You are left with 1p, 50p and £1.

Now ask them to pick a coin. Where you only have a choice of picking 3 coins are you seriously telling us that the odds of picking the £1 coin now are still 1 in 8????

I'm confused - how is this the same as the DoND example, where the offer is to swap, not to make a second 'pick'?jackpot said:.

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Now ask someone to pick a coin. The chance they will pick a penny is 2 in 8.

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Now remove one of the pennies, 2p, 5p, 10, and 20p

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Now ask them to pick a coin.

At the final hurdle there is 2 options thats have occured.

Option 1 : the 250k is in the contestants box and £1 in the (lets say with the audience).

Or option 2 where that the contestant has the £1 and audience has the 250k.

Option 1:

The chance the contestant has of picking the 250k = 1 in 20

The chance he does not pick it = 19 in 20

The chance it is somewhere in the audience = 19 in 20

The chance of one individual in the audience having it = 1 in 20

The chance the audience has picked the £1 = 19 in 20

The chance of one individual has picked the £1 = 1 in 20

The chance that the indivual left in the game picked the £1 = 1 in 20

Now option 2:

The chance the contestant has of picking the £1 = 1 in 20

The chance he does not pick it = 19 in 20

The chance it is somewhere in the audience = 19 in 20

The chance of one individual in the audience having it = 1 in 20

The chance the audience has picked the £250k = 19 in 20

The chance one individual has picked the £250k = 1 in 20

The chance that the indivual left in the game picked the 250k = 1 in 20

These odds are the same no matter what numbers are picked if £10 was left instead of the £250k then the same odds would still apply. As each contestant and box has started on an even footing at the begging of the game. People are using the odd for the whole audience to determined their outcome but it should be the odd for the certain individual left that should be used.

As these odd are the same. Then ther is no point in swapping.. Cant get simpler than that.

I'm confused - how is this the same as the DoND example, where the offer is to swap, not to make a second 'pick'?jackpot said:.

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Now ask someone to pick a coin. The chance they will pick a penny is 2 in 8.

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Now remove one of the pennies, 2p, 5p, 10, and 20p

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Now ask them to pick a coin.

Sorry Softus i was not comparing it to dond. It was Observers stateent about 4 boxes being removed that there still remains a 1 in 20 chance of the box being amongst the 16 left. It was an experiment for him to show that is impossible to have a 1:20 chance when there is only 1 item amongst the 16 boxes he had left.

Hence at the end of mine. i Had 3 coins left so it could not remain a 1 in 8 chance.

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,

Similarly with the boxes.

Here's another parallel.

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.

Now deal 1 more and turn it over. What's the chance of it being the Ace. NOT 1 in 43. It is (1-1/52)/42 (2.33% or 1 in 42.8 ) [edited to fix formula]. Note - if you replace the tenth face down card and shuffle the pack, then each card will then have a 1 in 43 chnace of being the Ace but until youi do that, the original mathematical probability is fixed.

Your argument is that when the first card of the 10 is turned over and found not to be the Ace, the probabilty of any of the remaining 9 becomes 1 in 51, then 1 in 50 and so on. Wrong. Completely wrong. To prove this, imagine all ten cards are turned over simultaneously. Obviously each has the same probability from the time it's selected from the population to the time it's re-selected from a different population.

And still none of the unbelievers has tried to explain why the switch strategy (in the 4 boxes illustration) succeeds 3 times out of 4. Or, for that matter, why it works in the MH game here. You have empirical evidence of the mathematical principle I'm explaining. Why do you refuse to accept it?

I agree that what you say in your 8 coins example is correct, but the coins aren't hidden and more importantly you haven't picked out and isolated one coin first so I don't believe it is analogous to the problem which is being debated.

The logic in your next post is very convincing, but have you actually run through the possible scenarios? It doesn't have to be twenty boxes or eight to experiment with. Four will do, and Observer has listed the scenarios for this.

If you do run through the possible scenarios then it shows that swapping wins 3/4. It is very convincing.

I think the key to the solution lies with the fact that one box is isolated.

You have a pack of cards. There are 4 jacks. The probability of getting jack on your first pick is 4 in 52. If a jack is not picked then there is a new event happening and the next pick is a probability of 1 in 51.

You are altering the event of the game which reduced the amount of options which results in the change on odds or probability.

Lets say you have a different type of cards. 10 in total. (8 jacks and 2 different cards). The odds of getting a jack first pick are 8 in 10.

Now lets say by pure luck 7 jacks are picked, 1 pick after another and you are left with 1 jack, and 2 other cards.

So are you saying you still have an 8 in 10 chance (80%) of getting a jack if you picked another card????

OMG.

Oh sooooooo wrong. Probability of the Ace being on the table is now 1 in the number of cards still undetermined (43).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.

Now deal 1 more and turn it over. What's the chance of it being the Ace. NOT 1 in 43

... Errrrmmmmm,

Let's try your logic from a different perspective and see how it holds up ...

Grand National starts with (say) 30 runners one of whom we have backed. Assuming no individual odds each has a 1 in 30 chance of winning the race. After the first fence 3 other horses fall, what are the odds now of our horse winning the race 1 in 27 or 1 in 30?

If by the time we get to the last fence only two riders remain ours plus one other what is the chance of our's winning 1 in 30 or 1 in 2?

I'd hope everyone will see a 50:50 odds at the end but I won't hold my breath.

Taking a practical example of this, the bookies lay the odds at the start of the race. If two runners pull out prior to the race starting do you think you'd still get the same odds on our horse winning? If you don't get this example please don't attempt to win anything on the horses ... You'd lose your coat

If you get this right try applying the same logic to deal or No Deal and you'll see (as many of us do) that the final outcome is 50:50 swap or no swap.

MW

This thread has gone nowhere since Softus, who was making little headway, started posting as Observer.

I assure you I am not Softus and have no idea who s/he is. Anyway, you're probably right - as they say, "you can't fix stupid".

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