Is Aaron still here? If so, I have a question for you - how would you work out the odds of any two sums of money, for example, 1p and 10p, being left at the end of a game?

Thanks!

I think it's 1 in 231.

Would you accept an explanation from somebody other than Aaron?

There is more than one way of calculating this. The first is to use the theory of probability.

Consider your chosen two amounts (for example, your 1p and 10p). There is a 2 in 22 chance that one of those amounts is in the player's box. As a fraction, that's 2/22, which cancels down to 1/11 -- one eleventh. Within that one eleventh chance, there is a 1 in 21 chance that the other amount is in the other unopened box (as a fraction, that's 1/21). So you have to find one eleventh of 1/21.

1/11 of 1/21 = 1/231

So there is a 1 in 231 chance.

The second way of getting the answer is to consider how many combinations of box pairs there are. In other words, if I were to choose two boxes out of the 22, how many different pairs am I choosing from?

Here are some of the possibilities...

1p/10p
1p/50p
1p/£1
1p/£5
(and so on)

It turns out there are 231 different pairs, each of which is equally likely to end up as the final two. Hence the answer is 1 in 231.

I wonder the what probability is of having the 3 top numbers left on the board with no other sum there?

And what are the chances of taking every blue out without touching a single red?

Well, using the first of the above methods, the calculation you'd have to do would be...

3/22 x 2/21 x 1/20


Yep, you could work that out using the same method, too.

i'm not sure on the 'top three' one - but as for the other one.

When the game starts, you have an 11/22 chance of removing a blue, then 10/21, 9/20, 8/19...

You get:

11/22 * 10/21 * 9/20 * 8/19 * 7/18 * 6/17 * 5/16 * 4/15 * 3/14 * 2/13 * 1/12 which is [be prepared] 1/32,249,7420,020,639,834,881,320,949,432,400,000

That's exactly the right method, but I'm sorry to tell you that the final answer is incorrect.

what should it be?

I think it's 1 in 705,432.

See, it's a game of probability and luck not holding hands;)

Probability of keeping the top 3 is about 1/1500 roughly.....

Just 19/22 x 18/21 etc down to 1/4 not sure if that answer above is correct atall it mite not be....

Listen to MisterAl, he is wise.

Essentially most of these questions deal with combinations i.e. picking a certain number of boxes m from a larger set of boxes n. Generally we don't care about the order in DOND: we are interested in combinations not permutations.

It doesn't matter whether you are picking boxes to keep or boxes to lose i.e. the chances of ending up with the power 3 are exactly the same as the chances of starting the game with them.

MisterAl's multiplications e.g. 3/22 x 2/21 x 1/20 are absolutely correct; they can be simplified to:

nCm the ways of choosing m from n = m! m factorial i.e. 1 x 2 x 3 x ... x m * (n-m)!/n! 1 x 2 x 3 x ... x 19 divided by 1 x 2 x 3 x ... x 22 in this example, leaving 20 x 21 x 22

For example, the chance of having a perfect final round (removing the lowest three still standing) is:

5C3 = 3! * (5-3)! / 5! = 3*2*1 * 2*1 / 5*4*3*2*1 = 12 / 120 = 10%

Note that this is the same result as choosing 2 from 5.

gareth_barry is asking for 22C3 (or 22C19, if you prefer, which is what he has done) which works out as 1 in 1540.

It gets more complicated when you start asking things like "What is the chance of keeping at least 2 of the top 3 in this round?" but these sort of problems can be solved by essentially the same methods. "Should I deal or not?" cannot

Here endeth today's lesson.

Thank you for the compliment. But I'm also an annoying pedant...

Aaron has misused the mCn notation there. The actual formula for mCn is as follows...

mCn = m! / [n! (m-n)!]

This gives the number of combinations. To find the probability of choosing one of those combinations, we must take the reciprocal of mCn. This reciprocal is essentially what Aaron has calculated directly, although I feel duty bound to point out his, er, 'non-standard' notation.

If you've enjoyed reading these posts, and would like to learn more about Combinations and how to calculate them, then please enrol onto your nearest A-level Mathematics course.

I had a feeling something was up there, couldn't quite recall the notation; it's been a while since my A-levels, at which point formal Mathematics and I finally parted ways

Do you think we've scared everyone away from this thread now?

pheeeew you've put me off maths for life, that's for sure...

and theres me thinking it was just opening boxes.
(furrows brow and pouts a little)

Wow...well, thanks for responding, guys. I'll try to understand it now!

For someone to win £250,000 there needs to be the right set of circumstances because if there was a small amount, say £5, with £250,000 left at the end, and the banker offered £100,000 to deal , few would be brave or foolish enough to gamble in that scenario with the risk of so great a loss. The odds of the last two boxes containing the two highest amounts, £250,000 and £100,000 are 1 in 231, the same odds as finding 250K and 75K at the end (as Kirsty had when she went to her final box which revealed 75K). The odds of having the 250K with either the 50K the 75K or the 100K at the end are 1 in 77. The odds of having the 250K at the end with anything between 10K and 100K are 1 in 33, so just over once a month that can be expected to happen. Perhaps not so many would risk the gamble with just £10,000 as the safety net, but I reckon a lot more would if the £250,000 was accompanied by any of the Power 5 (35K up) which represents a 1 in 46 chance approximately (a fraction over 2%) For more on strategies to deal at the right time based upon the odds see my site. Statistically then we are due for someone to win

First I'd like to thank Aaron and MisterAl, as I've been able to calculate various things easily now. It seemed daunting at first, but it's alright once you get used to it.

I was also thinking about what the recent Paul managed to do - take out 1p, £1,000, 10p, £3,000 and 50p in the first round, in that order. This was quite incredible. The odds of taking out those five are 1 in 26334 for a start, but when you take the added complication of orders...I might try and work this out later.

Learning maths is like learning any skill really. The more you do it, the easier it becomes.


As Aaron hinted at above, if you're interested in the order too, then you're dealing with permutations (not combinations).

You can use exactly the same method though. Just remember that for each box pick, this time there is always only 1 'successful' box. So you get the following calculation...

1/22 x 1/21 x 1/20 x 1/19 x 1/18

...which will give you the required probability.

(Compare that with the calculation you've done where the order was not important...

5/22 x 4/21 x 3/20 x 2/19 x 1/18 = 1/26334

...and you should be able to see that the method to use is just the same.)