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1. How unlucky was that!?
The UK Deal or No Deal game 29/7/2012 was called the worst game of DoND ever played. The player eliminated the boxes in this order:

first round removed { 250K, 75k, 50k, 3k, 5} bankers offer £99
second round removed { 20k, 15k, 1k} bankers offer £13
third round removed {100k, 35k, 10k} bankers offer 13p

This left an 11 box game {5k, 750, 500, 250, 100, 50, 10, 1, 50p, 10p, 1p} i.e 10 blue boxes (low values) and 1 red box (high value).

The next choice removed {5k, 750, 250} leaving an all blue 8 box: {500, 100,50,10,1, 50p, 10p, 1p} for which the offer was a packet of peanuts.

There was a lot of comment about how unusual this sequence of events was and how especially unlucky. But was it really less probable than a sequence of events leading to a particular 8 box { 250k, 10k, 3k, 1k, 500, 100, 10, 10p} in which there are 4 red and 4 blue boxes?
2. Re: How unlucky was that!?
Technically any combination of boxes is possible and each is of the same probably but as you remove each box the probably of each combination changes accordingly.
3. Re: How unlucky was that!?
Any order is as probable as the next. Correct me if I'm wrong, but isn't there the same probability that the same lottery numbers could be drawn week after week?
4. Re: How unlucky was that!?
(Original post by Pitt1988)
Any order is as probable as the next. Correct me if I'm wrong, but isn't there the same probability that the same lottery numbers could be drawn week after week?
Total number of possible combinations of box colours for 8 boxes is 2^8 = 256, therefore the probability of him getting those exact boxes is 1/256. This is the same as any other specific combination however although any order is as probable as the next, only one possible combination of 8 boxes that gives all 8 being blue, whereas there are a large number of possibilities giving 4 of each.

You are correct that the probability of any given set of lottery numbers being drawn is the same each week. Probability for Lotto (6 main balls, 1 bonus) will be
6/49 x 5/48 x 4/47 x 3/46 x 2/45 x 1/43 x 1/42 = 1 in 25,254,771,696

The probability of the next week's set producing the same numbers is the same, as long as you don't mind what the first set of numbers were. The probability will be exactly the same as if you picked any other random set of numbers

However if you were to pick a set of numbers that you wanted to appear on more than one week, rather than just wanting the first set repeated regardless of what it was, the combined probability is going to be the product of the 2 separate probabilities, i.e. 25,254,771,696 x 25,254,771,696.

So the probability depends on whether you are choosing the balls beforehand, or just wanting a repeat of the randomly given previous week's numbers.

Sorry if that's a little confusing!
5. Re: How unlucky was that!?
6. Re: How unlucky was that!?
So the probability of getting the specific all blue 8 box obtained in the game is the same as the probability of getting the specific 4 red and 4 blue box used as an example above. (and the same as the probability of getting any specific 8 box)

But the probability of getting an all blue 8 box irrespective of the actual values of the boxes (other than they are low) is much smaller than the probability of getting a (4 red, 4blue) 8 box irrespective of the particular values. i.e. there are many more ways of making a 4 red 4 blue 8 box from 11 red and 11 blue values than there are of making an all blue 8 box from 11 blues.

We shouldnt confuse these two arguments.

So the 'unlucky game' is not so surprising after all!
7. Re: How unlucky was that!?
(Original post by nixbits)
So the probability of getting the specific all blue 8 box obtained in the game is the same as the probability of getting the specific 4 red and 4 blue box used as an example above. (and the same as the probability of getting any specific 8 box)

But the probability of getting an all blue 8 box irrespective of the actual values of the boxes (other than they are low) is much smaller than the probability of getting a (4 red, 4blue) 8 box irrespective of the particular values. i.e. there are many more ways of making a 4 red 4 blue 8 box from 11 red and 11 blue values than there are of making an all blue 8 box from 11 blues.
Exactly right, worse if you consider that it wasn't just any 8 boxes of the 11, but the lowest 8.

(Original post by nixbits)
So the 'unlucky game' is not so surprising after all!
Overall, very unlucky due to the overall probability, but no more unlucky than any other specific combination of boxes.

Realised my calculation was more than a little off. That specific combination (or any specific combination) of 8 boxes out of 22 would have a probability of 1/319770 (8/22 x 7/21 x 6/20 x 5/19 x 4/18 x 3/17 x 2/16 x 1/15). The probability of an all-blue (or red) combination would be 1/1938. I can't be assed to work out the probability of a 4/4 split combination as it's a little more complicated!
8. Re: How unlucky was that!?
Number of possible 8 box combinations = ways of choosing 8 from 22
= 22!/(8!14!)= 319770

so any particular (specific) 8 box combination occurs with prob = 1/319770

Number of possible blue 8 boxes = ways of choosing 8 from 11
= 11!/(8!3!)= 165

Prob of blue 8 box = 165/319770 = 1/1938 = 1/6 . 1/17 . 1/19

number of possible 4 red 4 blue boxes= (ways of choosing 4 from 11)^2

= 11!/(4!7!) . 11!/(4!7!) = 108900

Prob of 4 red and 4 blue = 108900/319770 = 110/(17.19)

so prob 4 red and 4 blue/ prob all blue 8 box = 660
9. Re: How unlucky was that!?
(Original post by nixbits)
Number of possible 8 box combinations = ways of choosing 8 from 22
= 22!/(8!14!)= 319770

so any particular (specific) 8 box combination occurs with prob = 1/319770

Number of possible blue 8 boxes = ways of choosing 8 from 11
= 11!/(8!3!)= 165

Prob of blue 8 box = 165/319770 = 1/1938 = 1/6 . 1/17 . 1/19

number of possible 4 red 4 blue boxes= (ways of choosing 4 from 11)^2

= 11!/(4!7!) . 11!/(4!7!) = 108900

Prob of 4 red and 4 blue = 108900/319770 = 110/(17.19)

so prob 4 red and 4 blue/ prob all blue 8 box = 660
I figured there'd be an easy way of calculating it. I'd have ended up having to manually work out the ways of possible choosing the boxes.