In these days there's a dispute, in Italy, extended by TV show "Striscia la Notizia" and by newspaper "Libero", it tells that, during the TV Show "Affari Tuoi" ("Deal or not deal"), broadcasted by Rai 1, it happens a strange fact: the boxes with higher prizes would resist TOO OFTEN till the very end of the game; this, for some people, is made on purpose, to rise the suspense and the share. But is it against the probability?
I plunge myself into the close observation of the fact, analyzing by the probability, and - of course! - realizing a nice simulator.
Well now, directly from the website striscialanotizia.it, here's the sequence of the episodes under examination:


MAR 03.10.2006: 250.000 € in the last three boxes
MER 04.10.2006: 250.000 € in the last two boxes
GIO 05.10.2006: 500.000 in the last six boxes
VEN 06.10.2006: 500.000 in the last four boxes
SAB 07.10.2006: Affari Tuoi didn't broadcast
DOM 08.10.2006: "Fiorello" broadcast instead of Affari Tuoi
MAR 10.10.2006: 250.000 € in the last two boxes
MER 11.10.2006: Affari Tuoi didn't broadcast
VEN 13.10.2006: 500.000 € in the last six boxes
SAB 14.10.2006: 500.000 € in the last three boxes
LUN 16.10.2006: 250.000 € in the last five boxes
MAR 17.10.2006: 500.000 € in the last two boxes
MER 18.10.2006: Affari Tuoi didn't broadcast
GIO 19.10.2006: 500.000 € in the last five boxes; 250.000 € in the last two boxes
VEN 20.10.2006: 500.000 € in the last two boxes
SAB 21.10.2006: 500.000 € in the last two boxes
DOM 22.10.2006: Affari Tuoi didn't broadcast
LUN 23.10.2006: 500.000 € in the last two boxes
VEN 27.10.2006: 500.000 € in the last two boxes
SAB 28.10.2006: 100.000 € in the last two boxes
DOM 29.10.2006: 500.000 € in the last box
MAR 31.10.2006: win: 250.000 €; 500.000 € in the last four boxes
GIO 02.11.2006: 250.000 € in the last three boxes
SAB 04.11.2006: 250.000 € in the last three boxes
LUN 06.11.2006: 500.000 € in the last two boxes
MAR 07.11.2006: 250.000 € in the last two boxes
MER 08.11.2006: 500.000 € in the last four boxes
GIO 09.11.2006: 500.000 € in the last five boxes
VEN 10.11.2006: Special "Affari Tuoi" prime time (26 boxes), 1.000.000 € in the last two boxes, 250.000 € in the last six boxes, 500.000 € in the last seven boxes
SAB 11.11.2006: 250.000 € in the last three boxes
LUN 13.11.2006: 250.000 € in the last five boxes
MER 15.11.2006: "Affari Tuoi" didn't broadcast
GIO 16.11.2006: 250.000 € in the last two boxes
VEN 17.11.2006: Special "Affari Tuoi" prime time (26 boxes), 2.000.000 € in the last two boxes, 500.000 € in the last six boxes
DOM 19.11.2006: "Affari Tuoi" didn't broadcast
MER 22.11.2006: "Affari Tuoi" didn't broadcast
GIO 23.11.2006: 500.000 € in the last two boxes, 250.000 € in the last four boxes.
VEN 24.11.2006: Special "Affari Tuoi" prime time (26 boxes), 2.000.000 € in the last four boxes
SAB 25.11.2006: 500.000 € in the last three boxes
DOM 26.11.2006: 500.000 € in the last five boxes
LUN 27.11.2006: 500.000 € in the last two boxes
VEN 01.12.2006: Special "Affari Tuoi" prime time (26 boxes), 250.000 € in the last four boxes.
LUN 04.12.2006: 500.000 € in the last three boxes
MAR 05.12.2006: 500.000 € in the last three boxes
MER 06.12.2006: "Affari Tuoi" didn't broadcast
GIO 07.12.2006: Rai and Endemol made a press conference to defend the regularity of the show: the same evening no big prizes in the last boxes.
VEN 08.12.2006: 500.000 € in the last three boxes
SAB 09.12.2006: 500.000 € in the last box
DOM 10.12.2006: 250.000 € in the last box
LUN 11.12.2006: New prime times Rai-Mediaset: no big prozes at the end
MAR 12.12.2006: 250.000 € in the last box
VEN 15.12.2006: 500.000 € e 250.000 € in the last two boxes (Telethon episode)
LUN 18.12.2006: 250.000 € in the last three boxes

We see, the sequence describes how sometimes the show didn't broadcast, and how sometimes the "big-prize-boxes" arrived till the end, but it tells nothing about missing days. So, I must assume that in missing days the show DID broadcast and NO BIG-PRIZE-BOXES arrived till the last four boxes.
Let's count: from October 3rd to December 18th, it's 77 days.
From the sequence we can understand:
- 12 days must be removed, because show didn't brooadcast, or it was a "special episode" (26 boxes, I've done the trial on 20 boxes...).
- episodes with "at least 250.000 €, survived till last four boxes" are 31.
So we have 31 events defined "at least 250.000 € till last 4 boxes, out of a total of 65. It's worth the 31/65=47,69%. Too high? Too low? Let's decide...

Probability that at least 250.000 € stay till the last 4 boxes is 36,84%!
Let's demonstrate it:
For the probability, we can assume that the game is equal to a simple extraction of the numbers from 1 to 20, in random order. So, every number corresponds to a prize.
To check the event "at least 250.000 € in the last 4 boxes", let's say that the prize of 250.000 € should be the 19, and the prize of 500.000 € should be the 20. And we say that the 1st number drawn should be the "box assigned to the player", so - of course - it arrives to the end of the game.
Then, we draw the 19 numbers remaining, and these are the boxes (=prizes) in the order of call, and then removed. For this reason, if 19 or 20 are called in the 17th, 18th or 198th position (or the 1st, that means it's the player's box), it means that an high prize has arrived to the last 4 boxes.
This is an example of a sequence, without the "winning" event that we care:

15-12-5-6-13-20-19-1-4-7-9-10-18-16-17-2-3-11-8-14

This is an example of a sequence WITH the "winning" event that we care (the box 20, called at 19th position) [the box 15 is assigned to the player]):

15-12-5-6-13-14-19-1-4-7-9-10-18-16-17-2-3-11-8-20

How many sequences do exist?

Trust me: total number of permutations without repetitions of 20 numbers is the FACTORIAL of 20 (notation: 20!). And it's given by the product 1x2x3x4….x19x20.

If I'd hold a number in a certain position, let's say the 20 in 1st position, how many sequences can I make?

If I've fixed ONE number in ONE position, it means that I have a permutation of 19 numbers in 19 positions, without repetitions and it's Factorial of 19 (19!). Then it's 1x2x3x4x…x18x19 .

Since I've chosen to fix the 20 in the 1st position, and this corresponds to one of the events I'm interested to track (it's the 500.000 € assigned to the player), it means that there are 19! sequences, out of a total of 20!, that give us back an event of "500.000 € in the player's box = 500.000 € will arrive till the last 2 boxes".

Now, we see that 19! divided by 20! is worth exactly 1/20 = 0,05 = 5%

So this is the probability that, in a sequence of 20 random numbers, there is a GIVEN number fixed in a GIVEN position.
What are, and how many are the sequences that give us back "At least 250.000 € in the last 4 boxes"? (remember: the 20 is the 500.000 € and the 19 is the 250.000 €)

  1. 20 fixed in first position
  2. 19 fixed in first position
  3. 20 fixed in 19th position
  4. 19 fixed in 19th position
  5. 20 fixed in 18th position
  6. 19 fixed in 18th position
  7. 20 fixed in 17th position
  8. 19 fixed in 17th position


Each of thhese phrases identifies exactly 19! sequences out of 20!.

Let's see how to calculate the probability that at least ONE phrase would be TRUE.
We can't simply sum the probabilities, because this way we'd sum TWICE some events, i.e. : between the sequences with 20 in the first position, we find also some sequences with the 19 in 20th position, and vice versa, between the sequences that have the 19 in 20th position, we find also some sequences with 20 in the first position.
We must exclude these repetitions, let's see how:

Sequences with TWO numbers fixed in TWO positions are - obviously - all the permutations of remaining 18 numbers in 18 free positions, then they are 18! out of a grand-total that's always 20!.

So: sequences with 20 in first position AND 19 in 19th position are 18!.
Sequences with 20 in first position AND 19 in 18th position are still 18!.
And more, sequences with 20 in first position AND 19 in 17th posiztion are again 18!.

This is the exact list of events that give us back a result of "At least 250.000 € in the last 4 boxes":

  1. 20 fixed in first position, but not those that have also the 19 in 17th, 18th or 19th position, because those will be counted later
  2. 19 fixed in first position
  3. 20 fixed in 19th position, but not those that have also the 19 in 1st, 17th or 18th position, because those has been counted before, or will be counted later
  4. 19 fixed in 19th position
  5. 20 fixed in 18th position, but not those that have also the 19 in 1st, 17th or 19th position, because those has been counted before, or will be counted later
  6. 19 fixed in 18th position
  7. 20 fixed in 17th position, but not those that have also the 19 in 1st, 18th or 19th position, because those has been counted before
  8. 19 fixed in 17th position

Let's count them:

  1. are 19! minus (3 multiplied by 18! )    [=19!-(3*18!)]
  2. are 19!
  3. are 19! minus (3 multiplied by 18! )    [=19!-(3*18!)]
  4. are 19!
  5. are 19! minus (3 multiplied by 18! )    [=19!-(3*18!)]
  6. are 19!
  7. are 19! minus (3 multiplied by 18! )    [=19!-(3*18!)]
  8. are 19!

Then, the sequences that give us the result of "At least 250.000 € in the last 4 boxes", are : out of a total of 20!

For all these reasons, the chances of "At least 250.000 € in the last 4 boxes" happening is: that's equal to 36,84%, QED.



Now, let's focus on the real result of the show, revealed by Striscia la Notizia: 31 episodes out of 65 with at least 250.000€, survived till the last 4 boxes.
We said that the probaility is 36,84%, so, the most likely outcome should be (65 x 36,84%) = 24 episodes.
Instead, we've got it 31 times, may we decide that it is tricky? not at all...
No doubt that we have an outcome "more or less" far from the mean, we can say it's unlikely... BUT NOT IMPOSSIBLE!

Let's show that:
We suppose to repeat a lot of times a "block" of 65 episodes, and to check out how many times we get the event "At least 250.000 € in the last 4 boxes".
The mean is 24, so we expect a value near 24, like 22,23... 24 itself, 25 or 26.
By repeating a lot of times our experiment (a sequence of 65 episodes= "extraction of 65 random sequences of the numbers from 1 to 20"), the random variable "number of times that event happens" (out of 65) will arrange itself according to a Gaussian curve, around the mean.
Here's the proof: this is a graph of repetition for 22.000 time of the experiment.


As you can see the fact "31 times the event under examination, out of 65" has happened 450 times out of 22.000 trials.
It means that what is happened in the TV Show had 2% chances to happen.
Too little or too much, you decide, certainly one and only one sequence as a stat is not enough...

But let's check the last data... See next graph:


We could ask ourselves:"What was the chance that, out of 65 episodes, we get MORE than 30 times the event At least 250.000 € in the last 4 boxes?"
Answer is the sum of all numebers above the 30, then it was 5%.

Here we are, everyone can decide by himself...
If you want, you can download the program in Visual Basic, that simulate the extractions of the boxes at "Deal or not Deal": Deal or not deal simulator (Sorry' it's in Italian language...)