Re: Deal or no Deal

Siggy wrote:
"Ed Chilada" <nospam@xxxxxxxxxx> wrote in message
news:1r860259vh9h4ojgqr317i3qmkvpjc84eq@xxxxxxxxxx
On Fri, 24 Feb 2006 15:50:09 -0000, "John Dean"
<john-dean@xxxxxxxxxxxxxxxx> wrote:

michael adams wrote:
"zcx" <zcx@xxxxxxxxxxxxxxxxxx> wrote in message
news:UtDLf.1361$Ru5.1112@xxxxxxxxxxxxxxxxxxxxxxx
I have a question which has been bugging me.
OK did anyone see Deal or No Deal last night?

The guy had 5 boxes left i.e. his own plus 4 others.

4 were blue and 1 was red.

He picked 3, and they were *all* blue.

There is now 1 red and 1 blue box. - one is his and one is still
left.

Obviously this implies that there is a 50/50 chance of it being
red or blue.

However, does the fact that he picked 3 boxes and they were all
blue mean that it is *more* likely that the blue box is the one
left (i.e. surely it's easier to pick 3 blue boxes if there are 4
left out there?)

No

starting off there are 5 boxes which can be numbered 1 - 5

No. Starting off there are 22 boxes which can be (and actually are)
numbered 1 - 22
It don't signify what happens during the game, the probability of a
red number in your box starts at 0.5 and stays at 0.5 all the way
through.

The probability of you having initially chosen a red stays at 0.5
throughout the game, I agree.

However, the probability/likelihood of it turning out to be red when
you open the box at the end changes depending on what happens in the
game and how many of each colour you reveal.

When you're at the stage of there being four blues and one red, it
quite frankly defies any common sense or logic that the odds of you
having red are 0.5.

Which is why we're not relying on common sense or logic but on probability theory. I'm sure someone on-line is anal enough to keep a record of DOND games in one or more countries. If you went through them and identified all the games where, at the five box stage, there were four of one colour and one of the other, you would find that the final box was red half the time and blue half the time. The point you're missing is that the final box was chosen at a time when the probability of a given colour was 0.5. Hence, half of all games end with a red box and half of all games end with a blue box. Always. (subject to the usual small variation. If you check 10,000 games, you would see something like 5,011 red last box and 4,989 blue. Percentages 50.1 and 49.9. Just like at school when you tossed a coin a hundred times and got 52 heads and 48 tails, that confirms a probability of 0.5). It doesn't matter what sequence the other boxes are opened.



It least that's something we agree on the probability changes
everytime you make a decision. It's like in horse racing. If one of
the horses pulls out before the start of the race, do the other
horses odds change? No.

The difference here is that the final box has been chosen. You just don't know what's in it. It's like seeing horses fall in the Grand National after one horse has already passed the winning post. It doesn't matter. If Noel opened a box *before* the contestant was chosen, *that* would make a difference.
<Just when I thought that I was out they pull me back in>
If you think the probability changes every time a box is opened, you're saying that a change in information changes the probability. Consider the other way round - that after the contestant has sat down with his "money shot" box, you *add* half a dozen blue boxes. The probability that the money shot box was blue was 0.5 because it was one of 11 red and 11 blue boxes. Has that probability changed because you've added 6 blue boxes, that there are now 17 blue and 11 red?
Obviously not. Your gut reaction tells you it's absurd. But probability theory doesn't have to rely on gut reaction. It's simple to work out why adding six boxes doesn't change the probability of the final box. It's the same reason that taking six blue boxes away doesn't change the money shot probability.
Let's consider a few sporting wagers. Imagine we watch the contest and at certain points when there's an imbalance of colours, I offer you a bet and I ask for odds.
Let's say there are 15 boxes left and 10 are red and 5 are blue. I bet you 100 UKP that the money shot box is blue. Since there are now twice as many red boxes as blue, I'd like odds of two to one please. While that bet is being held in safe hands, the boxes reduce to 10. Now there are 6 red and 4 blue. Again, I want to bet 100 UKP that the money shot box is blue. I'd like odds of six to four. You presumably accept that the probability has changed again.
So at the end of that game, if the money shot is red you win 200 UKP and if it's blue I win 350 UKP.
Fair odds?
We make these bets on every game from now till hell freezes over or Noel stops using hairspray. On every occasion, I bet the underdog and you give me odds according to the spread of colours remaining.
If your theory is correct, we should break even as the Winter Olympics begin in Hades. If my theory is correct, I clean up. I'm expecting that, no matter what the situation looks like at any point in the game, half of all the games will have a blue money shot and half a red.
In fact, here's one you can make at home. Assume I want to bet after the first box is opened. At that point there are 10 boxes left of the same colour as the first box opened, and 11 left of the other colour. I always want to bet that the money shot is the same colour as the underdog, the first box. So if a contestant opens a blue box, and we see 10 blue and 11 red left in the game, I want to bet 100 UKP that the box in front of the contestant is blue. And I'd like odds of 11 to 10 please, because there are 11 boxes on which I lose and only 10 on which I win.
Keep track of the games on your telly from now on to see how many times the contestant's box is the same colour as the first box. I'll tell you now it will happen almost exactly half the time. And it will be the opposite colour almost exactly half the time too.
So if we watch 2,000 games, I lose 1,000 games for a loss of 100,000 UKP. And I win 1,000 games for a gain, at those odds, of 110,000 UKP. Net gain to me 10,000 UKP. And all because I believe that the probability of the contestant's box being a given colour is 0.5 and that the probability remains the same even when a box of the same colour is removed.
If you want to dispute the fact that the probability remains at 0,5, you have to find an explanation for the fact that half the games end with a blue box on the table and half with a red. What probability gives *that* outcome?
--
John Dean
Oxford

.

Relevant Pages

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