Impeccable logic - part 2

[jackpot]

If the top prize has not been eliminated how can it still be 1/20 when there are only 16 boxes left. The probablity can not stay the same as you have altered the number of boxes. Its a 1/16.

And as the £1 box is still amongst all that, then its also a 1 in 16 chance the contestant will have the £1.00 box or any other prize that still remains.

 

[Softus]

Please do not use the quote button excessively.
Thanks.

What would you say is that probability of that happening?

 

[jackpot]

Please do not use the quote button excessively.
Thanks.

What would you say is that probability of that happening?

about a 100:1

 

[Thermo]

Please do not use the quote button excessively.
Thanks.

What would you say is that probability of that happening?

odds on where you are concerened

 

[Softus]

But what if a poster has the choice of

and

buttons?

If the poster initially chooses

, but is then offered the option of swapping, does that poster increase the odds of receiving a comment from a Moderator if he swaps?

 

[Trazor]

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.

You pick a random box, the fifth one, which you do not open and immediately fix its odds at 1 in 20, you can not do this.

The moment you discarded your 4 boxes then the odds of ANY remaining box containing the £250,000 must be 1 in 16

You are giving the contestants box a position in the game it does not merit, it is purely one box amongst others, and at any point in the game its odds must be calculated in exactly the same way as any other box.

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.

The above statement is wrong, and shows a lack of understanding of probabilities. You can only state the above if your original premise that each game favours swapping is correct, and it is not.

I have posted the mathematical proof for showing that the odds of any game are 50-50, I,m getting perplexed that such a simple mathematical problem can be wrongly interpretated by so many posters.

 

[megawatt]

Me too

 

[megawatt]

Me three

 

[megawatt]

We must remember that not everyone is good at maths.

 

[Observer]

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.

 

[blondini]

We must remember that not everyone is good at maths.

And not everyone is good at comprehension.

 

[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.

I can't quite get my head round your example just at the moment, but I think your list of possibilities are incomplete. Either way, you are overcomplicating the problem......

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.


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.

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


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.

 

[Trazor]

Observer......Your new game has 12 possible choices, not 4...

You can swap A for B.......A for C.......A for D
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

If you now work out what you win and lose you will find that you will win 5 times lose 5 times and come out even twice.

You will win a total of £301 and also lose £301 over the 12 games, and in my book that is 50-50.....

 

[blondini]

Observer......Your new game has 12 possible choices, not 4...

You can swap A for B.......A for C.......A for D
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

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

 

[Softus]

Observer, things were easier when you were just observing.