Can someone please explain the statistics of 5 box.
Today i was confused when The Banker said there was a 62.5% chance of removing the highest amount, i've always thought its 60% as you open 3/5 boxes. But if the Banker says its 62.5% then it must be, but why?

It's because the Banker is an idiot who can't even turn a fairly simple fraction (3/5) into a percentage (60%) correctly. If I'd have been in the audience for Suzi's game, I wouldn't have been able to keep quiet -- and I'd have probably been thrown out of the studio because of it...

I'm going to appear very thick to those who know about maths now. But when people say that it's a 60% chance of taking out the highest value, am I right in saying that it's a 60% chance of taking out ANY value? (Or 62.5% chance if you're the banker.)

Yes. So long as you're referring to the final round from 5-box to 2-box. (You're opening 3 boxes out of the 5 left, so each remaining amount has a 3/5 chance of being revealed. Are you reading, Banker?! It's that simple!)

I see... therefore I assume that the chances of taking out any particular amount in the next box (from 5-box) is 20%...?

You're getting the hang of it!

Well, probability was never my strong point in maths at school. I remember a Year 7 lesson where we were told that the chances of the Sun rising in the morning were 1 in 1. Someone challenged the teacher (the DREADFUL RIGMAROLE one) by saying that the sun wouldn't rise if the universe imploded etc etc.

I believe that led on to a fascinating discussion about the chances of that happening, followed by a discussion of how the chance was so small that it could be rounded down to 0% and therefore the sun would always rise.

That's about all I remember about probability from school, I'm afraid.

Everybody makes mistakes... ...:/

Hmm, I'll try and link this to what I'm currently doing. Think it's called the Binomial Distribution.

So you have 10 possible rounds you can have, and you need to choose 3 boxes from a set of 5. You're looking for the largest value in the first box OR second box OR third box.

[5C1 x (0.2) x (0.8)^4] + [4C1 x (0.25) x (0.75)^3] + [3C1 x (0.33) x (0.66)^2] = 1.27591944444444

Erm, I've done something wrong... =/

Are you forgetting the conditions for a distribution to be Binomial there? One condition is that the trials are identical. This means that the probability of a success doesn't change from trial to trial. Here, the first trial is picking from five boxes and the second is picking from four, so clearly this condition doesn't hold and the distribution cannot be Binomial.

Another condition for a distribution to be Binomial is that the trials are independent. This means that the probability of a success isn't affected by what's happened in prior trials. I'll leave it to you to work out why that condition doesn't hold here either.

So, here's a question for you. One of the following random variables can be modelled with a Binomial distribution, and the other two cannot. Which one's the Binomial one?

1. The number of red amounts still on the board at 5-box in tomorrow's game.
2. The number of Power 5 amounts on the west wing tomorrow.
3. The number of times the £250,000 is on the table between now and the end of this month.

You lost me with the first sentence there.

I think he's talking about bisexual gnomes.

(I've already got my coat and left the building...)

Waves hand up in the air frantically.

The Binomial only works on independent trials, but in Deal what is opened in one box affects the chances on the next turn, like taking a card our of a deck of cards



Okay, so 1. It's a progressive chance and every box affects the next possibility.

1. The number of red amounts still on the board at 5-box in tomorrow's game.
2. The number of Power 5 amounts on the west wing tomorrow.
3. The number of times the £250,000 is on the table between now and the end of this month.

The number of Power 5 amounts on the west wing in tomorrows game. The bag-of-balls system changes the chances of the next box on the west wing being selected having a Power 5.

3 sounds plausible though, because the boxes have already been distributed before the box is selected to come to the table. It's an equal chance each day.

Good work, Aaron! If I've understood you correctly, you're going for option 3, which is indeed the only Binomial one. It's the only option where the trials are identical and independent, since the number of boxes and the amounts in them are the same each day (identical) and what happens in one game has no bearing at all on what happens in other games (independent).

And your explanation as to why the original situation wasn't independent was rather nice too.

(Bad luck, Simon. Option 2 isn't Binomial since each trial there -- a trial being the picking of a box to go on the west wing -- depends on what's already been chosen, so they're not independent of each other.)

Yeah, I find it quite interesting really. Saw this thread and jumped right into it. But I get it now This is all my AS Stuff, maybe I could come to you if I had any problems?

Dear god...this is painful reading material...our resident mathematician and our resident gamer both putting their heads together and coming up with winning formulae!

Of course! PM me whenever you like. There's nothing more enjoyable than doing A-level maths.

Whoops. Not good for someone with a masters degree in Mathematics to get a basic probability question wrong.