# Maths/Reasoning/Logic/Probability Exercise

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**1**Lost in BE Cyberspace

Thread Starter

Joined: Jul 2007

Posts: 13,553

**Maths/Reasoning/Logic/Probability Exercise**

Interesting one they threw at my daughter last week............. your views please?

A and B are playing a dice game. The dice is rolled, and if there is a number from 1 to 5, A pays that amount to B. (i.e. if a 3 is rolled, A pays 3 pounds to B, etc).

If a 6 is rolled at any time, including on the first roll, the game ends.

So A cannot win, and B cannot lose...........

Just one catch: B has to pay an 'entrance fee' to A before he is allowed to play the game.

Question: how much should B be willing to pay........?

A and B are playing a dice game. The dice is rolled, and if there is a number from 1 to 5, A pays that amount to B. (i.e. if a 3 is rolled, A pays 3 pounds to B, etc).

If a 6 is rolled at any time, including on the first roll, the game ends.

So A cannot win, and B cannot lose...........

Just one catch: B has to pay an 'entrance fee' to A before he is allowed to play the game.

Question: how much should B be willing to pay........?

*Last edited by The Dean; Jul 7th 2011 at 2:07 am. Reason: .*

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**2****Re: Maths/Reasoning/Logic/Probability Exercise**

A and B are playing a dice game. The dice is rolled, and if there is a number from 1 to 5, A pays that amount to B. (i.e. if a 3 is rolled, A pays 3 pounds to B, etc).

If a 6 is rolled at any time, including on the first roll, the game ends.

So A cannot win, and B cannot lose...........

Just one catch: B has to pay an 'entrance fee' to A before he is allowed to play the game.

Question: how much should B be willing to pay........?

Probability is 1in6 getting a 6 first go, so perhaps pay 5/6ths of the potential revenue....but then do you do that as a total 1+2+3+4+5 or as a max 5?

You could argue that you should pay the mid range - 2.50 as there is equal chance of getting anything up to 5 pounds or nothing.

You could say that based on the equal probability of getting any number then you should pay no money.

^^^ All of the above is my typing as it came into head ^^^

There are loads of potential ways to look at it, bit of a killer question for kids at school. If it was me, i'd offer a pound, because I'm nice.

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**3****Re: Maths/Reasoning/Logic/Probability Exercise**

A and B are playing a dice game. The dice is rolled, and if there is a number from 1 to 5, A pays that amount to B. (i.e. if a 3 is rolled, A pays 3 pounds to B, etc).

If a 6 is rolled at any time, including on the first roll, the game ends.

So A cannot win, and B cannot lose...........

Just one catch: B has to pay an 'entrance fee' to A before he is allowed to play the game.

Question: how much should B be willing to pay........?

If its an independent price - then scamps theory holds true (plus a 5% mark up for the House)

If B - then the lowest winning prize which is $1

If its A - then im buggered as to know why A creates the game in the first place !!

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**4**Lost in BE Cyberspace

Joined: Jan 2008

Posts: 41,518

**Re: Maths/Reasoning/Logic/Probability Exercise**

Do A and B have to agree on the entrance fee?

edit: that's weird, UKCityGent was thinking along the same lines.

edit: that's weird, UKCityGent was thinking along the same lines.

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**5****Re: Maths/Reasoning/Logic/Probability Exercise**

C. comes along and offers to sell B. a loaded dice that never allows a 6 to roll for 50% of B's winnings. B then pays A 5.00 and wipes the floor with him

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**6****Re: Maths/Reasoning/Logic/Probability Exercise**

The "fair" price would be B's expected winnings.

Quick calculation, these are:

First round: (1+2+3+4+5)/6

Second round: 5/6 * (1+2+3+4+5)/6

Third round 5/6 * 5/6 * (1+2+3+4+5)/6

...

or sigma (n=0, inf) (5/6)^n * 15/6

or 15/6 * sigma (n=0,inf) (5/6)^n

How can we calculate the series?

Note this is

1 + 5/6 + 25/36 + ... = s

But: 5/6 + 25/36 + ... = s * 5/6 (from multiplying through)

So s = 1 + 5s/6

s/6 = 1

s = 6

Then substituting in above, 15/6 * 6 = 15.

The answer is 15.

Probably an error somewhere but this is the gist of how to solve.

Quick calculation, these are:

First round: (1+2+3+4+5)/6

Second round: 5/6 * (1+2+3+4+5)/6

Third round 5/6 * 5/6 * (1+2+3+4+5)/6

...

or sigma (n=0, inf) (5/6)^n * 15/6

or 15/6 * sigma (n=0,inf) (5/6)^n

How can we calculate the series?

Note this is

1 + 5/6 + 25/36 + ... = s

But: 5/6 + 25/36 + ... = s * 5/6 (from multiplying through)

So s = 1 + 5s/6

s/6 = 1

s = 6

Then substituting in above, 15/6 * 6 = 15.

The answer is 15.

Probably an error somewhere but this is the gist of how to solve.

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**7****Re: Maths/Reasoning/Logic/Probability Exercise**

Quick calculation, these are:

First round: (1+2+3+4+5)/6

Second round: 5/6 * (1+2+3+4+5)/6

Third round 5/6 * 5/6 * (1+2+3+4+5)/6

...

or sigma (n=0, inf) (5/6)^n * 15/6

or 15/6 * sigma (n=0,inf) (5/6)^n

How can we calculate the series?

Note this is

1 + 5/6 + 25/36 + ... = s

But: 5/6 + 25/36 + ... = s * 5/6 (from multiplying through)

So s = 1 + 5s/6

s/6 = 1

s = 6

Then substituting in above, 15/6 * 6 = 15.

The answer is 15.

Probably an error somewhere but this is the gist of how to solve.

If there was a round limit you could do it on the probability and pay the 15 quoted...

It's a head scratcher, but I wouldn't offer 15

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**8****Re: Maths/Reasoning/Logic/Probability Exercise**

Yes, hence

"sigma (n=0, inf) (5/6)^n * 15/6"

Where "inf" means "infinity"

Oh, and the most you'd want to offer is 15 (assuming my maths is right, someone else can check )

For what it's worth, I'd rate this of similar difficulty to harder Further Maths A-Level questions or my uni entrance exam.

"sigma (n=0, inf) (5/6)^n * 15/6"

Where "inf" means "infinity"

Oh, and the most you'd want to offer is 15 (assuming my maths is right, someone else can check )

For what it's worth, I'd rate this of similar difficulty to harder Further Maths A-Level questions or my uni entrance exam.

*Last edited by typical; Jul 7th 2011 at 6:07 am.*

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**10****Re: Maths/Reasoning/Logic/Probability Exercise**

Any one instance of the game could yield winnings of anything from 0 to infinity. The 15 is just the weighted average outcome.

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**11****Re: Maths/Reasoning/Logic/Probability Exercise**

Any one instance of the game could yield winnings of anything from 0 to infinity. The 15 is just the weighted average outcome.

Intersting and difficult question.

I like my offer of a token nugget the best.

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**13****Re: Maths/Reasoning/Logic/Probability Exercise**

Quick calculation, these are:

First round: (1+2+3+4+5)/6

Second round: 5/6 * (1+2+3+4+5)/6

Third round 5/6 * 5/6 * (1+2+3+4+5)/6

...

or sigma (n=0, inf) (5/6)^n * 15/6

or 15/6 * sigma (n=0,inf) (5/6)^n

How can we calculate the series?

Note this is

1 + 5/6 + 25/36 + ... = s

But: 5/6 + 25/36 + ... = s * 5/6 (from multiplying through)

So s = 1 + 5s/6

s/6 = 1

s = 6

Then substituting in above, 15/6 * 6 = 15.

The answer is 15.

Probably an error somewhere but this is the gist of how to solve.

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**15****Re: Maths/Reasoning/Logic/Probability Exercise**

Quick calculation, these are:

First round: (1+2+3+4+5)/6

Second round: 5/6 * (1+2+3+4+5)/6

Third round 5/6 * 5/6 * (1+2+3+4+5)/6

...

or sigma (n=0, inf) (5/6)^n * 15/6

or 15/6 * sigma (n=0,inf) (5/6)^n

How can we calculate the series?

Note this is

1 + 5/6 + 25/36 + ... = s

But: 5/6 + 25/36 + ... = s * 5/6 (from multiplying through)

So s = 1 + 5s/6

s/6 = 1

s = 6

Then substituting in above, 15/6 * 6 = 15.

The answer is 15.

Probably an error somewhere but this is the gist of how to solve.