Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta
 You are Here: Home >< Maths

# Probabilites AS level standard question watch

1. A fairground game involves taking three throws to get a ring over two poles in the ground at different distances from the throwing position. Throws must be taken alternately at the two pols, but you may start with either one. You win a prize if your ring lands over a pole in two successive throws out of three.

Clearly, it is easier to throw the ring over the nearer pole than the farther one. Is it better to make your attempts in the order 'near,far,near' or 'far,near,far', or doesnt it matter?

Here is the answer given in the textbook, i cant answer the further questions is is asking:

Let us suppose that the probability of hitting the nearer pole is 1/2 and the probability of hitting the farther pole is 1/3. (If the question can be answered, it clearly does not matter what the exact probabilities are or we would have been given them).

If we throw near,far,near, the probabilities of throwing two in a row are as follows:

Hit,hit,miss: 1/2 x 1/3 x 1/2 = 1/12
Miss,hit,hit 1/2 x 1/3 x 1/2 = 1/12
Hit, hit , hit: 1/2 x 1/3 x 1/2 = 1/12

Total probability = 3/12 or 1/4 (25%)
If we throw far,near,fa, the probabilities of throwing two in a row are as follows:

Hit, hit, miss 1/3 x 1/2 x 2/3 = 2/18
Miss, hit, hit 2/3 x 1/2 x 1/3 = 2/18
Hit, hit , hit: 1/3 x 1/2 x 1/3 = 1/18

The total probability of winning is 5/18 or about 28%.The second strategy is better. Some may regard this as counter intuitive as it involves two throws at the harder target. Did you expect this answer? Can you rationalise why the second strategy should work the best? Can you prove that it works for all probabilities?

Also another question from myself - why is hit hit hit in the equation - how would the probability be affected assuming the thrower stops after the second throw as they have already won?
2. (Original post by 2cool)
A fairground game involves taking three throws to get a ring over two poles in the ground at different distances from the throwing position. Throws must be taken alternately at the two pols, but you may start with either one. You win a prize if your ring lands over a pole in two successive throws out of three.

Clearly, it is easier to throw the ring over the nearer pole than the farther one. Is it better to make your attempts in the order 'near,far,near' or 'far,near,far', or doesnt it matter?

Here is the answer given in the textbook, i cant answer the further questions is is asking:

Let us suppose that the probability of hitting the nearer pole is 1/2 and the probability of hitting the farther pole is 1/3. (If the question can be answered, it clearly does not matter what the exact probabilities are or we would have been given them).

If we throw near,far,near, the probabilities of throwing two in a row are as follows:

Hit,hit,miss: 1/2 x 1/3 x 1/2 = 1/12
Miss,hit,hit 1/2 x 1/3 x 1/2 = 1/12
Hit, hit , hit: 1/2 x 1/3 x 1/2 = 1/12

If we throw far,near,fa, the probabilities of throwing two in a row are as follows:

Hit, hit, miss 1/3 x 1/2 x 2/3 = 2/18
Miss, hit, hit 2/3 x 1/2 x 1/3 = 2/18
Hit, hit , hit: 1/3 x 1/2 x 1/3 = 1/18

The total probability of winning is 5/18 or about 28%.The second strategy is better. Some may regard this as counter intuitive as it involves two throws at the harder target. Did you expect this answer? Can you rationalise why the second strategy should work the best? Can you prove that it works for all probabilities?

Also another question from myself - why is hit hit hit in the equation - how would the probability be affected assuming the thrower stops after the second throw as they have already won?
Spoiler:
Show
I'm really not good with understanding probability or these types of math questions so please forgive me if I answer stupidly, it's just these things really really intrigue me and I wish I could talk about them xD
Shouldn't it not matter?
This is what I think. The reason why you've got that 2/18 is because you've multiplied by that 2/3 which is the probability of missing on the far pole.

But that shouldn't be done because you're only focussing on the probability of getting two consecutive throws. Since in any case it's going to be near-far (or far-near, doesn't matter) the probability will be 1/2 x 1/3 = 1/6
And that's in both cases, near-far, or far-near, your chances of getting two consecutive throws is 1/6.

The higher probability relates to the fact that you have a higher chance of missing on the last throw.
For instance for [hit, hit, miss], near-far-near is 1/12 and for far-near-far it is 2/18. But it's higher only because you're more likely to miss on the third throw for the far pole. In both cases you still got the two hits required and the probability of both of them is the same.

Maybe I've just completely disregarded something or misunderstood something (remember the spoiler!) but that's how I saw it...
Haven't had a thought about your other question
3. Well I'd look at it like this: in order to win, you need to get the middle one. So having the near pole - which you're more likely to hit - in the middle, gives you an advantage.

Looking at it more mathematically, let the probability of hitting the near pole be p and the far pole q. An important point here is that p is greater than q.

The chance of winning with FNF (far-near-far) is:
// add up all the options and factorise
// simplify

The chance of winning with NFN is:
// add up all the options and factorise again.

Since p>q, (2-p) < (2-q), so FNF is more likely to give you a win.
4. (Original post by TLDM)
Well I'd look at it like this: in order to win, you need to get the middle one. So having the near pole - which you're more likely to hit - in the middle, gives you an advantage.

Looking at it more mathematically, let the probability of hitting the near pole be p and the far pole q. An important point here is that p is greater than q.

The chance of winning with FNF (far-near-far) is:
// add up all the options and factorise
// simplify

The chance of winning with NFN is:
// add up all the options and factorise again.

Since p>q, (2-p) < (2-q), so FNF is more likely to give you a win.

That equation is a little difficult for me to understand I think I have a better method that I can visualize better - is the following correct?:

Having two shots at a million to one chance reduces the probability to 500,000-1 , so by having two goes at it you make a huge difference by shortening your chances by 500,000

whereas the easy shot you have a 95/100 of making, so two attempts only reduces it to 90/100, so two shots at a hard target makes a bigger difference to the overall probability.

Is this correct?
5. (Original post by 2cool)
That equation is a little difficult for me to understand I think I have a better method that I can visualize better - is the following correct?:

Having two shots at a million to one chance reduces the probability to 500,000-1 , so by having two goes at it you make a huge difference by shortening your chances by 500,000

whereas the easy shot you have a 95/100 of making, so two attempts only reduces it to 90/100, so two shots at a hard target makes a bigger difference to the overall probability.

Is this correct?
It doesn't quite reduce the probability to 1/500000, but close enough (you're only one in a trillion out!). Yes, that's another way you can look at it.

The equations were a mathematical proof that it's always the better choice, no matter what the probabilities are (as long as the near one is easier than the far one). I'll try to break the first equation down to make it more understandable; even if you don't care hopefully someone else might find it helpful.
Spoiler:
Show
The probability of hitting the near pole is , and for the far pole it's . These just represent the probabilities.

There are three possible options that let you win:
- Hitting all three posts,
- Missing the first shot and hitting the other two,
- Hitting the first two and missing the third.

The chance of each happening (in the same order as the list above), aiming for the far post first, is:
-
-
-

So to find the total probability of winning, you just add them all together.

All of these have a term in them, so we can factorise this out.

And this simplifies to:
.

A similar thing can be done to show that if you aim for the near post first, the probability of winning is .

If you want to find which is the better way to win, we need to find which of these two options is more likely. Remember that p is bigger than q, since p is the probability of hitting the near pole. (2-p) is therefore smaller than (2-q). Even after multiplying them both by pq, the first of those is still smaller, so the probability of winning is smaller if you aim for the near post first.

Reply
Submit reply
Turn on thread page Beta
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: February 27, 2016
Today on TSR

### Loughborough better than Cambridge

Loughborough at number one

Poll
Useful resources

## Make your revision easier

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

Can you help? Study help unanswered threads

## Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.