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 18^{th}, 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 1^{st} 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 17^{th}, 18^{th} or 198th position (or the 1^{st}, 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 19^{th} 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 1^{st} 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 1^{st} 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 €)

- 20 fixed in first position
- 19 fixed in first position
- 20 fixed in 19
^{th}position - 19 fixed in 19
^{th}position - 20 fixed in 18
^{th}position - 19 fixed in 18
^{th}position - 20 fixed in 17
^{th}position - 19 fixed in 17
^{th}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 20^{th} position, and vice versa, between the sequences that have the 19 in 20^{th} 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 19^{th} position are **18!**.

Sequences with 20 in first position AND 19 in 18^{th} position are still **18!**.

And more, sequences with 20 in first position AND 19 in 17^{th} 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**":

- 20 fixed in first position, but not those that have also the 19 in 17
^{th}, 18^{th}or 19^{th}position, because those will be counted later - 19 fixed in first position
- 20 fixed in 19
^{th}position, but not those that have also the 19 in 1^{st}, 17^{th}or 18^{th}position, because those has been counted before, or will be counted later - 19 fixed in 19
^{th}position - 20 fixed in 18
^{th}position, but not those that have also the 19 in 1^{st}, 17^{th}or 19^{th}position, because those has been counted before, or will be counted later - 19 fixed in 18
^{th}position - 20 fixed in 17
^{th}position, but not those that have also the 19 in 1^{st}, 18^{th}or 19^{th}position, because those has been counted before - 19 fixed in 17
^{th}position

Let's count them:

- are
**19!****minus**(3 multiplied by**18!**) [=19!-(3*18!)] - are 19!
- are
**19!****minus**(3 multiplied by**18!**) [=19!-(3*18!)] - are 19!
- are
**19!****minus**(3 multiplied by**18!**) [=19!-(3*18!)] - are 19!
- are
**19!****minus**(3 multiplied by**18!**) [=19!-(3*18!)] - 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...)