Impeccable logic - part 2

[Observer]

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.

It seems too simple but the logic is inescapable. Show me any other starting choice by which a 'always swap' strategy would lose.

(Obviously the 19:1 probability only come into play when all the 18 other 'losing' boxes have been eliminated. We're not asked to analyse the probability of that happening.)

 

[Trazor]

The chance of the £250,000 being in either of the final 2 boxes is 50-50.

The proof is as follows....

1) The chances of the £250,000 being in any of the 22 boxes at the start of the game is 1 in 22, I,m sure we all agree on this.

2) In the first round 5 boxes are removed ( For this game we take it that the £250,000 will not be removed )
This means that the odds of ANY box containing the £250,000 is now 1 in 17, this includes the contestants box.

3) The next round removes 3 boxes, the odds are now 1 in 14, this again includes the contestants box.

4) The odds now get reduced to 1 in 11.... then 1 in 8.... then 1 in 5.... then 1 in 2 or 50-50. Each time the contestants box is included in the reducing odds.
And this was the flaw with the previous maths solution, which shows the swap as being beneficial, as the contestants box is treated as a 1 in 22 chance, and its odds never adjusted as the game progressed.

 

[Observer]

The chance of the £250,000 being in either of the final 2 boxes is 50-50.

It's a 50:50 chance if you make a random choice - toss a coin if you like.

However, if you have played the game from the start and you KNOW which box was first selected, the probability of the other box being the 250k one is 19:1, as I have proved. There will be an alternative way of calculating this probability by summing the probabilities of the single box avoiding a series of 'n' selections during the process of elimination of the other 18 boxes, but it's not necessary to go through all that. My simple and inescapable logic proves beyond contention that the 'always swap' strategy is far superior to 'random choice' and 'always stick' alternatives, and that's what the problem is really about.

[edit] The reason the first choice box has a lower probability of being the £250k box in the final 'choice of two' game is that it has been 'protected' from elimination during the preceding 'n' rounds. It started off with a 1 in 20 chance but after that took no part; so, in the final round, still has a 1 in 20 chance (contrary to your argument above). The other remaining box MUST have the balance of probability (19:1) to a person who knows which one is which.

 

[jackpot]

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.

It seems too simple but the logic is inescapable. Show me any other starting choice by which a 'always swap' strategy would lose.

(Obviously the 19:1 probability only come into play when all the 18 other 'losing' boxes have been eliminated. We're not asked to analyse the probability of that happening.)

Observer, i respect your views but have you read a previous post of mine.
There is a 250k and a £1 box left. You state that there is a 19 in 20 chance of not having the 250k box so its better to swap. So you are saying the likelyhood is that you have the £1 in pocession and better to swap.

Ok, but there is also a 19/20 box of not having the £1 box (as choosing that was also a 1/20 chance at the beginning) so you can argue that its better to swap this. So by taking the odds of the remaining box it is also likey that you have the 250k box as there is a 19/20 of not having the £1.00

Would you still swap???

hence the odds are the same for each box

 

[joe-90]

If that doesn't seal it, Jackpot - I don't know what does.

 

[blondini]

DNOD OP scenarion - I would just swap anyway.

Whatever you believe, swapping won't make your odds worse.

 

[Observer]

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.

It seems too simple but the logic is inescapable. Show me any other starting choice by which a 'always swap' strategy would lose.

(Obviously the 19:1 probability only come into play when all the 18 other 'losing' boxes have been eliminated. We're not asked to analyse the probability of that happening.)

Observer, i respect your views but have you read a previous post of mine.
There is a 250k and a £1 box left. You state that there is a 19 in 20 chance of not having the 250k box so its better to swap. So you are saying the likelyhood is that you have the £1 in pocession and better to swap.

Ok, but there is also a 19/20 box of not having the £1 box (as choosing that was also a 1/20 chance at the beginning) so you can argue that its better to swap this. So by taking the odds of the remaining box it is also likey that you have the 250k box as there is a 19/20 of not having the £1.00

Would you still swap???

hence the odds are the same for each box

Let's take the first selected box as A. There is a 1 in 20 chance it holds the prize. Conversely, there is a 19 in 20 chance that one of the other 19 boxes holds the prize. You would agree with that - no?

Box A is now out of the draw (not subject to elimination). From the remaining 19 boxes, box B is selected. The chance that it holds the prize is (1/19) x (19/20) (= 1 in 20). Box B is opened and is empty. Now there are 18 boxes left (excluding box A) from the group of 19 that had a 19/20 chance of holding the prize. Box C is selected. The chance that it holds the prize is 1/18 x 19/20 = 0.0528. The probability that box C holds the prize has increased because of the elimination of box B (that was empty).

Keep on working through and you will see that when we are down to the last box (in the group of 19) all other boxes having been eliminated (and proved to be empty), the probability that it holds the prize is 1/1 x 19/20 = 0.95. The probability that box A holds the prize remains at 1 in 20.

 

[joe-90]

What if the box with a quid in it also had the keys to a new house worth £1m. Which is then the big prize? The big prize doesn't come into it. Any of the boxes are the 'big prize'.

 

[Trazor]

Let's take the first selected box as A. There is a 1 in 20 chance it holds the prize. Conversely, there is a 19 in 20 chance that one of the other 19 boxes holds the prize. You would agree with that - no?

Box A is now out of the draw (not subject to elimination). From the remaining 19 boxes, box B is selected. The chance that it holds the prize is (1/19) x (19/20) (= 1 in 20). Box B is opened and is empty. Now there are 18 boxes left (excluding box A) from the group of 19 that had a 19/20 chance of holding the prize. Box C is selected. The chance that it holds the prize is 1/18 x 19/20 = 0.0528. The probability that box C holds the prize has increased because of the elimination of box B (that was empty).

Keep on working through and you will see that when we are down to the last box (in the group of 19) all other boxes having been eliminated (and proved to be empty), the probability that it holds the prize is 1/1 x 19/20 = 0.95. The probability that box A holds the prize remains at 1 in 20.

Sorry but your thinking is wrong, yes based on 20 boxes the original odds are 1 in 20 for the £250,000 to be in the contestant,s box.
But where you make the mistake is in removing that box from further calculations.

Lets say you remove 5 boxes to begin with, leaving 15 boxes, the odds now of the £250,000 being in ANY box left, is now 1 in 15. AND INCLUDES THE CONTESTANTS BOX
You can not remove the contestants box from your calculations.
The contestants box DOES NOT REMAIN AT 1 in 20.....

See my above post for the full explanation.

 

[blondini]

my rethink was sparked by a valid observation that jackpot made.

Softus, if that was the same 'observation' which I am thinking of, it certainly shook my confidence. Jackpot seemed to have hit the nail on the head, but if jackpot was right I still couldn't see what the flaw was in my previous conclusions and more thought was needed.

 

[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 ?

 

[megawatt]

Observer Wrote:

My simple and inescapable logic proves beyond contention that the 'always swap' strategy is far superior to 'random choice' and 'always stick' alternatives.

I'd say that Trazor has escaped your simple logic.

 

[Observer]

No - I disagree with that. If you remove 5 boxes to begin with, open them all and all are empty, then the chance of any one of the remaining 15 holding the top prize is 1 in 15. However, if you remove five boxes and open 4 (which are empty) but the fifth remains unopened, the chance of the fifth holding the prize is 1 in 20 because that is the original probability. If that has a 1 in 20 probability, the probability of any of the other 15 holding the prize must be 1/15 x 19/20 = 0.0633. I don't remove the fifth box from the probability calculation but the probability of it holding the top prize doesn't change. The probabilty of the prize remaining in the other group is also unchanged but the probability of it being in any one of the other group increases as the other boxes are opened and found to be empty.

Consider - if we play the game 20 times, then (with perfect distribution of probability), 1 time in 20 our first choice box will hold the top prize. If, playing the game 20 times, the top prize box is not eliminated, then, when the game comes down to two boxes, one must be the top prize box and the other must be empty (or a lesser prize). If we always switch, then 1 time out of 20, we will lose. The other 19 times, when we originally chose a box other than the top prize, we will win.

That's all we need to know.

The alternatives strategies are:

(i) we make a random choice (eg: flip a coin and call switch if heads and stick if tails). This would be a 50:50 strategy

(ii) we always stick. This would lose 19 times out of 20.

But we're concerned with winning the top prize. The OP stipulates that that the £1 box was also not eliminated. But we wouldn't care if the £1 box had been eliminated on the way down to the last two or not. It is not a necessary condition (for us to win the top prize) that the £1 box remains, all that matters is that the top prize box remains.

The fundamental and unarguable fact is that the 'always switch' strategy ONLY loses if our first choice box holds the top prize. In any other case, whether the other remaining box is the £1 box or any other one, we win the top prize.

 

[megawatt]

I've sussed this ... Observer is Softus

 

[megawatt]

The fundamental and unarguable fact is that the 'always switch' strategy ONLY loses if our first choice box holds the top prize.

I think you'll find that its actually VERY arguable and there'll be a few along I'm sure to prove the point