# Impeccable logic - part 2

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#### trazor

Some of your suggested swaps wouldn't be available Trazor. In three of your examples only box A would be available to swap to.

#### johnny_t

I've just been trying to think of how to reword my earlier post....

At the start of the game, you have, say twenty equally likely possibilities where you can start with a box containing £1, £2, £5 ....... £250,000

For the purposes of this problem, we are just considering the scenarios where you started with £1 and where you started with £250,000, which were both equally likely.

At the end of the game, there are two boxes left - the £1 and the £250,000.

This means that you must have either started with £1 and got £250,000 in the other, or started with £250,000 and got £1 in the other.

As both starting possibilities were equally likely, then it must be 50:50.

Can anyone say where this reasoning goes astray ?

#### johnny_t

Your probability of the £250k being in the box, is exactly the same, at all points in the game, as £1, £5 or any other uneliminated amount being in the box.

Trazor's post a couple above this explains it better - The odds of it being on one side of the fence or the other reduce over time, but at the end, its a pure 50/50...

EDIT: Or, put another way, at the beginning of the game, the chances of £1 being your box are the same as the chances of £250,000 being in your box. What do you propose happens during the game to change those odds ?

Let's try it another way. Take a game with four boxes, A, B, C and D:

Box A holds £100
Box B holds £1
Box C empty
Box D empty

I don't know which box is which.

I have to choose one box, which then remains unopened until the end of the game. Two of the remaining boxes are eliminated and I can then stick with my first choice or switch to the other remaining box. Obviously there are four possible choices for my first box so we will look at each of the four possible scenarios. In each case, box A remains as one of the last two boxes with one of either B, C or D. I adopt a strategy that I will always switch, so open the other remaining box. Let's work through the possibilities.

A1. I choose box A
A2. (say) C & D are eliminated. I switch to B and lose

B1. I choose box B
B2. C & D are eliminated. I switch to A and win

C1. I choose box C
C2. B & D are eliminated. I switch to A and win

D1. I choose box D
D2. B & C are eliminated. I switch to A and win

Four scanarios. I win 3 and lose one so my switch strategy gives me a 3 in 4 chance of success once the game has come down to a choice of two boxes and the £100 prize is still in the game.

This is a parallel situation (with fewer starting boxes) described in the OP except that the OP stipulates that, in that particular case, the £1 box (box B in my example) remained as well as the top prize. Now it's true that if you stipulate that box B and only box B can be the second remaining box, then the chance is 50:50, but that artificially forces B to remain in the game, which destroys the essence of the problem. In practice, I don't care whether my final choice is between A and B, A and C, or A and D. In the real game situation, the switch strategy gives the advantage.

Figured out what's wrong with this...

We are only talking about scenarios where you are left with £1 and £100 in your case. Therefore only two valid scenarios in your example are A & B - C & D are red herrings.

Two scenarios. One win, One lose.

#### blondini

Some of your suggested swaps wouldn't be available Trazor. In three of your examples only box A would be available to swap to.

Your 'possible choices' are what I contested.

You said.
'You can swop B for A.......B for C.......B for D'
'You can swop C for A.......C for D.......C for B'
'You can swop D for A.......D for B........D for C'

In the scenario which Observer suggested, A was the big prize so A would still be in play. Therefore if you have chosen B,C, or D initially, the only 'swap to' available would be A.

#### Observer

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.

Take 4 boxes, one contains a £10 note, the others are empty. Mix them up and choose one at random. Mark it 'A' and mark the others 'B'. What is the chance of the £10 being in the 'A' box - 1 in 4 (25%) - easy. Now open two of the B boxes at random. By chance, they are empty. What is now the chance of the £10 being in the A box. Not 50% but still 25%. The B boxes have a combined probabilty of 75% (37.5% each).

Similarly, in the DOND scenario, the chance of the first box holding the prize is 1 in 20. The probabilty event occurs when it is chosen and it is fixed with that probability unless it rejoins the population and is re-selected at random. So, while 18 losing boxes are eliminated, leaving our first choice and one other losing box, the probability of the first choice box holding the prize remains 1 in 20. At any given step, the sum of probabilities must equal unity so the combined probability of the remaining boxes holding the prize is 1 - 1/20. When we get down to the last two boxes, if the prize has not aleadt been eliminated, the probability it is NOT the first selected box is 19 in 20.

#### joe-90

I'm going like a yo-yo on this one. Each time I read the opposite argument I say "Yeah, that's it!"

#### Observer

Observer, things were easier when you were just observing.

It's not my fault this lot are so stubbornly wrong. I've given them any number of easy illustrations and not one has come up with an argument against the clear evidence that the switch strategy works in practice.

#### megawatt

Take 4 boxes, one contains a £10 note, the others are empty. Mix them up and choose one at random. Mark it 'A' and mark the others 'B'. What is the chance of the £10 being in the 'A' box - 1 in 4 (25%) - easy. Now open two of the B boxes at random. By chance, they are empty. What is now the chance of the £10 being in the A box. Not 50% but still 25%. The B boxes have a combined probabilty of 75% (37.5% each).
You're making the same fundamental error as Softus.

Take 4 boxes, one contains a £10 note, the others are empty. Mix them up and choose one at random. Mark it 'A' and mark the others 'B'. What is the chance of the £10 being in the 'A' box - 1 in 4 (25%) - easy. Now open two of the B boxes at random. By chance, they are empty.

Agree with above

What is now the chance of the £10 being in the A box.

Not 25%. This isn't about the A or the B boxes its about the probability based on the total possible outcomes remaining and with only 2 boxes left 1 must contain the £10 note.

The B boxes have a combined probabilty of 75% (37.5% each) ... Totally irrelevant as we are only interested in the probability for the remaining unopened boxes.

MW

#### megawatt

It's not my fault this lot are so stubbornly wrong. I've given them any number of easy illustrations and not one has come up with an argument against the clear evidence that the switch strategy works in practice.
Hmmmnnnn ... Such arrogance in someone so new ... Unless you are, as I suspected, really the Soft one.

There have been numerous posts with clear detailed explanations of the 50:50 argument but you seem too closed minded to me to recognise your own shortcomings.

Stick to observing would be my advice cos you ain't cutting it as a participator

MW

#### Observer

johnny_t";p="795402 said:
At the beginning of the game, you choose a box. That has an equal chance of containing any amount on offer as a prize, whether it is £1, £2, £50, £100 or whatever number of prizes are on offer. So, at the outset, it is equally likely that your box has £1 or £250,000 in.[\quote]

Agreed.

Now, at the end, say you have two boxes. Your original one, and one other, and you know that one contains £1 and one contains £250,000. Nothing has altered the possibility of your original box containing either that £1 or £250,000, which was 50:50.

No not 50:50. The box you chose has an equal (1 in 20) chance of holding either £250k or £1.

Therefore, the chances of the other box containing £250,000 or £1 must be 50:50 too, no ?

We KNOW the other box holds either £250k OR £1 because that's stipulated.

If there was ONLY one £1 box and you insist that ONLY that box can be left at the end (as well as the £250k box) then you have a 50:50 chnace of winning. But that's not the real point of the OP. The question is whether it is better to switch, stick, or makes no difference. Switch is the best strategy because (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.

There are only really two, equally likely possibilities for how the game has panned out if you ended up with £250,000 and £1 in the boxes...

Option 1) You had the £250,000 box to start with, and have eliminated everything else to make the other box have £1 in.

Option 2) You had the £1 box to start with, and have eliminated everything els to make the other box have £250,000 in.

and both are equally likely.

As above - ONLY if you insist that the the other box must be the £1 box and not any other 'losing' box.

I've tried explaining it in simple terms, then with more complicated illustrations, then simple terms again. Nothing seems to get through.

#### megawatt

Have you considered the fact that it may be you who is wrong?

#### Observer

Take 4 boxes, one contains a £10 note, the others are empty. Mix them up and choose one at random. Mark it 'A' and mark the others 'B'. What is the chance of the £10 being in the 'A' box - 1 in 4 (25%) - easy. Now open two of the B boxes at random. By chance, they are empty. What is now the chance of the £10 being in the A box. Not 50% but still 25%. The B boxes have a combined probabilty of 75% (37.5% each).
You're making the same fundamental error as Softus.

Take 4 boxes, one contains a £10 note, the others are empty. Mix them up and choose one at random. Mark it 'A' and mark the others 'B'. What is the chance of the £10 being in the 'A' box - 1 in 4 (25%) - easy. Now open two of the B boxes at random. By chance, they are empty.

Agree with above

What is now the chance of the £10 being in the A box.

Not 25%. This isn't about the A or the B boxes its about the probability based on the total possible outcomes remaining and with only 2 boxes left 1 must contain the £10 note.

The B boxes have a combined probabilty of 75% (37.5% each) ... Totally irrelevant as we are only interested in the probability for the remaining unopened boxes.

MW

Sorry - I made a silly mistake on the illustration. The probability of the ONE remaing B box holding £10 is 75%. I repeat, the probability event occurs when we split the boxes between A and B. The initial probability is fixed until a new event occurs. Until then, the A box holds the 25% probability and the remaining B boxes hold a combined 75% probability (as they originally had). Merely opening a box which is already in a pre-selected group to test the contents is NOT a new event.

#### Observer

Have you considered the fact that it may be you who is wrong?

Yes - I've considered it and would be willing to retract if just ONE person can produce a cogent argument to demonstrate that the switch strategy does not work. Nobody has done so. Give it a try. I'll repeat the hypothesis.

If I ALWAYS switch then the ONLY time I lose is IF my first choice box was the winning box. If the chance of that was 1 in n, then the probability of the switch strategy winning must be 1 - 1/n. That's a very simple to understand hypothesis. Disprove it if you can.

#### blondini

Have you considered the fact that it may be you who is wrong?

Have you?

If you asked me that I would say yes.

#### Observer

Figured out what's wrong with this...

We are only talking about scenarios where you are left with £1 and £100 in your case. Therefore only two valid scenarios in your example are A & B - C & D are red herrings.

Two scenarios. One win, One lose.

Yes - as I said it IS a 50:50 chance IF you insist that ONLY the £1 box and no other losing box can be the 'other' box of the final two.

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