## One for the mathematicians

- Subscribe to RSS Feed
- Mark Topic as New
- Mark Topic as Read
- Float this Topic for Current User
- Bookmark
- Subscribe
- Printer Friendly Page

- Plusnet Community
- :
- Forum
- :
- Other forums
- :
- General Chat
- :
- One for the mathematicians

##
One for the mathematicians

23-04-2017 8:54 AM - edited 23-04-2017 8:56 AM

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Report to Moderator

For my bowls club I run a monthly 100 Club draw with three cash prizes. The numbers being randomly drawn from a bag..

Yesterday evening at the annual club opening social a draw was made;

Against probability two members drew their own numbers out of the bag. I wonder what the odds on this happening are?

There are 79 numbers in "play" not all consecutive, the numbers missing from 1 to 100 are, should it make any difference to the calculation:

6,10,15,22,25,26,27,30,33,35,36,38,40,41,44,51,58,66,67,70,71,72,74,75,76,77,80,90,93,and 94.

Nine numbers included over 100 are 104,105,106,107,110,113,115,120 and 121.

Over the years there used to be a lot more players hence the numbers over 100, those remaining preferred to keep their numbers.

When faced with two choices, simply toss a coin. It works not because it settles the question for you. But because in that brief moment while the coin is in the air. You suddenly know what you are hoping for.

##
Re: One for the mathematicians

23-04-2017 9:04 AM

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Report to Moderator

Ah! the bit I forgot to add is: there maybe 79 numbers in play but, some members have more than one number.

So there are 73 actual players.

When faced with two choices, simply toss a coin. It works not because it settles the question for you. But because in that brief moment while the coin is in the air. You suddenly know what you are hoping for.

##
Re: One for the mathematicians

23-04-2017 11:29 AM - edited 23-04-2017 11:41 AM

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Report to Moderator

Do the members with more than one ticket all have two or do some have more? I think that matters. Assuming that the three draws are made by different members... A ticket is drawn by an owner with two tickets. That means that their remaining ticket cannot be drawn by its owner. However, if a ticket is drawn by an owner with five tickets then there are four remaining that cannot be drawn by their owner.

I'm afraid I've forgotten most of the statistics that I learned at school and this problem is too complicated for me to solve.

BBC Radio 4's More Or Less may be of interest to you if you want a better understanding of statistics. It also occasionally answers listeners' questions.

##
Re: One for the mathematicians

23-04-2017 12:03 PM

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Report to Moderator

No alanf, both players have only one number in their own name. It happens that one of them has numbers for other family members in their own names amounting to six numbers in all, but only one in this winners name.

When faced with two choices, simply toss a coin. It works not because it settles the question for you. But because in that brief moment while the coin is in the air. You suddenly know what you are hoping for.

##
Re: One for the mathematicians

23-04-2017 3:37 PM

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Report to Moderator

An interesting article In Excel is the Combin function ( under Maths icon)

##
Re: One for the mathematicians

23-04-2017 4:31 PM

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Report to Moderator

If there are 79 numbers in play the odds that 1 player will pick his own number are 1 in 79.

The odds for a second player picking his own number are 1 in 78 as 1 ball has already been picked.

Multiplying them together gives odds of 1 in 6,162 of this happening together.

Now you're stuck with me because my new ISP doesn't run a forum

##
Re: One for the mathematicians

23-04-2017 4:49 PM

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Report to Moderator

Y'know Browni you make that seem rediculously simple. I would guess it has to be right though. Thank you.

When faced with two choices, simply toss a coin. It works not because it settles the question for you. But because in that brief moment while the coin is in the air. You suddenly know what you are hoping for.

##
Re: One for the mathematicians

23-04-2017 5:17 PM

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Report to Moderator

##
Re: One for the mathematicians

23-04-2017 6:24 PM

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Report to Moderator

@Browni wrote:

If there are 79 numbers in play the odds that 1 player will pick his own number are 1 in 79.

The odds for a second player picking his own number are 1 in 78 as 1 ball has already been picked.

Multiplying them together gives odds of 1 in 6,162 of this happening together.

Very authoritative, but incorrect! You have to divide results by the factorial of 2 i.e 2x1. were there three numbers from 79 you would need to divide the product of 79x78x77 by factorial of 3 i.e. 3x2x1.

Someone didn't read the link I supplied!

##
Re: One for the mathematicians

23-04-2017 6:47 PM - edited 23-04-2017 7:52 PM

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Report to Moderator

Prove it!

(bear in mind that factorials only come into play when combinations/permutations are involved )

Now you're stuck with me because my new ISP doesn't run a forum

##
Re: One for the mathematicians

23-04-2017 7:57 PM

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Report to Moderator

@Luzern Having done the football pools before the Lotto started I'm familiar with the calculation in your post, but that is for drawing any two from 79. Here we are talking about two separate draws of one from 79 followed by one from 78.

Now here's something I've wondered about: On Deal or No Deal 22 boxes are each selected at random by contestants at the beginning of the game. One contestant out of the 22 is then selected to play the game. Is the chance of that player having the top prize in his box one in 22 or one in 22x22?

##
Re: One for the mathematicians

23-04-2017 10:48 PM

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Report to Moderator

Oops! So we're talking permutations, not combinations. Apologies for misreading! BTW Excel formula is called Permut and works same way as Combin.

@Jonpe Ah! The Pools, what they called permutations were actually combinations, I think. I suppose that misnomer was to stop confusion when the Missus said she was putting her combinations down..

##
Re: One for the mathematicians

24-04-2017 1:19 AM - edited 24-04-2017 1:21 AM

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Report to Moderator

@Jonpe wrote:

Now here's something I've wondered about: On Deal or No Deal 22 boxes are each selected at random by contestants at the beginning of the game. One contestant out of the 22 is then selected to play the game. Is the chance of that player having the top prize in his box one in 22 or one in 22x22?

If the contestant has already been chosen then there is 1 in 22 chance that they have the top prize.

If there are 22 potential contestants then there is a 1 in 22 chance that they will be chosen and then a 1 in 22 chance that they have the top prize so then the odds are 1 in 22x22.

##
Re: One for the mathematicians

24-04-2017 9:37 AM

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Report to Moderator

There's one way to settle it, email Carol Vordermen and save yourself a headache.

Growing old is inevitable...But growing up is optional.

##
Re: One for the mathematicians

24-04-2017 10:46 AM

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Report to Moderator

@petlew - Given that you didn't tell us the draw order there are three possible draw orders here with the Y representing a member drawing their own ball (i.e. not a family member ball) from the bag:

Order A Y Y N Order B N Y Y Order C Y N Y

So given the above the probability of two members drawing their own are :

Order A = 0.000160177% Order B = 0.000164393% Order C = 0.000162285%

- Subscribe to RSS Feed
- Mark Topic as New
- Mark Topic as Read
- Float this Topic for Current User
- Bookmark
- Subscribe
- Printer Friendly Page