In John's game today one of the ladies said with 8 blue and 4 red showing (something like that) that John had a 50-50 of having a red.

She was ridiculed by Noel and everyone else. Yet she was telling the truth.

When John picked his box at the beginning there was 11 red and 11 blue. Ergo he had a 50-50 chance of getting one or the other when he picked his box.

This odd clearly does not change throughout the game nothwitshtanding the selction of boxes.

I think Noel should apolgise tomorrow and go to maths school.

There were 5 blues and 3 reds so it's not 50-50.

But when he picked the box it was 50-50!!!

That odd does not change.....ever.

To prove it just look at how many reds were at the end compared to how many blues.

I bet its close to 50-50.

Why can't people get this?

The way that lady was ridiculed was shameful

He wasn't deciding whether to deal at the beginning of the game though (and frankly I don't think any player would take offence to Noel's ribbing - he's had a go at people for far less than not able to work out rudimentary probability)

This answers it nicely.

Still think he has a 50% chance of having a red in his box?

Or imagine a scenario where the player has 7 blues and £250,000. You think it's a 50% chance he has £250,000 in his box? Clearly it isn't.

psychokiller - you are missing the point.

When the all blue guy picked his box he had a 50-50 chance of picking a red or a blue. If you go to the end of the game
the odds of having one or the other is 50-50.

Using your scenario where the player has 7 blues and £250,000 it is still 50-50 he has a red or a blue.

Forget the numbers. At the beginning of each game the odds of each player having a red or blue is 50-50.

That maths statistic cannot change with the opening of random boxes.

But that is exactly what does happen as the boxes are opened!
Yes, at the beginning of the game, before any boxes are opened, the chances of a red being in the box are 50-50. That is, there are 11 blues and 11 reds, so the probability of a red being in there is 11/22 = 1/2 = 0.5 (or 50%).

Now suppose that a box is opened and it is a red. We now have new information which modifies the probability of a red being in the player's box. There are now 21 boxes left, of which 10 contain reds (as one red has been revealed), so, with the additional information that we have gained as a result of the box opening, the probability of the player's box containing a red is now 10/21 = 0.476 (or 47.6%).

Now suppose that a second red is revealed. We now have further information; as there are 20 unopened boxes left, of which 9 are red, the probability of a red in the player's box is 9/20 = 0.45 (or 45%).

If a blue is now revealed, we now have 19 unopened boxes, of which 9 are red, so the probability has become 9/19 = 0.474 (or 47.4%).

And so on. As the above screenshots so graphically illustrate, in this case, we have gained sufficient information through the box openings to allow us to deduce that there cannot possibly be a red in the player's box. With 9 boxes remaining, and none of them reds, the probability is 0/9 = 0.

Take it to the extreme: suppose there were 10 million boxes; 5 million blue and 5 million red. Suppose 4,999,999 boxes were opened, all revealing a red, leaving 5 million blues and 1 red. Do you really still assert that the probability of there being a red in your box is 50% ?

David22 - Your maths is so very wrong.

We don't care how the odds change as the game progresses. At the time you pick your box you have a 50-50 chance of a red.

Taking it to the other extreme lets say we have 22 boxes and you pick one. Then we say lets not bother opening the other boxes and just say we will stick with our box and open it. We might as well open it now and stop the game coz I am not playing it.

What are the odds then of you having a red?

50-50!!!!!!!!!!!!!!!!!!!! Cos there are 11 blue and 11 red boxes.

Ergo when you finish the game you still have a 50-50 chance you have a red unless as you correctly state you can demonstrate that you have in fact picked a blue due to random box openings (which was a 50-50 chance).

Using your 10 million example I can then throw it back at you and say what are the chances of me now picking that lone red amongst your 5 million blues!! Think about it.

Er, contradiction?? I thought you said that the statistic cannot change...


Yes, absolutely. Agreed.



NO!! Only if there are equal numbers of reds and blues remaining. If the numbers are not equal, then the chance is not 50-50.





Another apparent contradiction....

1/5,000,001.



Three words: Pot, kettle, black.

Dave22 - To be honest you can forget about any contradictions and your clumsy response. I am not writing a maths paper for my university.

Only thing which counts is you agreed with my main point.

At the start of the game when you pick a box you have a 50-50 chance its a red. When you take one box away you still have a 50-50 chance its a red. The only time that odd changes when you know for certain there are no more reds!

As I said I am sure there is some person on here who can tell us how many red and blue boxes there have been at the end. I bet its about 50-50.

To be honest if I was to play the game I would just pick my box and say forget playing and open it right now. I am not going to deal.

I would then have 36% chance of hitting a red of £10k or above (8/22)....but then I suppose Dave22 you would think
those odds change as well.............

See above.

Sorry posted reply too early!!

Question is this:

I show you three ring boxes with one diamond ring and two empty. I close boxes and shuffle them and ask you to pick one but not look at it (I know which box ring is in). I then open a box and reveal its empty. I then ask you whether you would like to swap your ring box with the last one remaining.

Going by your logic you would say its 50-50 so doesnt matter. Correct? Or should you swap and if so why?

Sorry cfd missed out the beginning of the post!! Your maths is wrong. Are you saying man had a 12.5% chance of having a red when he started game at the beginning?

Look at the question I just posted. On your logic you would say you would hold onto your box as no point swapping as 50-50

This is the Monty Hall problem and is different for one key reason:

The host knows where the prize is.

So yes, in a Monty Hall problem you are best switching.

DOND is different. No-one (apart from the IA but that's got no bearing on this argument) knows where the prizes are. The boxes are opened randomly. In the Monty Hall problem they are opened in a very specific way (i.e. to leave the big prize as the only one left (if it is there))


Answer it this way. Let's imagine we're using the same 7 blue/1 red example. Now let's assume the prizes have no colours attached to them. Do you still think it's a 50% chance the player has £250,000?

The observations and beliefs posted on this thread by sarlat are nothing less than a glimpse into the jaws of insanity.

Good job that he appears to have done a runner as otherwise I would have got just a bit frightened by his presence on here. An unpleasant 'you're being followed' sensation that one would associate with investigating a disused mental hospital, or perhaps visiting the toilets in a restaurant so rough that it no longer allows food orders to be placed, would have plagued this forum had he not taken his leave.

Well done to cfd and David22 (and in part me !) for combating the defective 'LogicBot 2012'. Normal service appears to be in place again.

Give him some credit, though.

He wasn't afraid to speak his truth.

But his truth was slightly distorted from the actual truth, and therein lay his problem...

Any particular reason why you've bumped this topic up from last October?

Er, no particular reason at all, to be honest.

It's an interesting display of one man/woman's dogged determination to prove several others wrong, however. Good reading.

Ahh I see. Somebody obviously scrubbed replying to this thread off your to-do list from last October and you've only just remembered! I'll let you off then.