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Maths/Reasoning/Logic/Probability Exercise

Maths/Reasoning/Logic/Probability Exercise

Old Jul 7th 2011, 2:05 am
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Default 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........?

Last edited by The Dean; Jul 7th 2011 at 2:07 am. Reason: .
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Old Jul 7th 2011, 4:41 am
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Default Re: Maths/Reasoning/Logic/Probability Exercise

Originally Posted by The Dean View Post
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........?
Toughie.

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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Old Jul 7th 2011, 5:01 am
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Default Re: Maths/Reasoning/Logic/Probability Exercise

Originally Posted by The Dean View Post
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........?
Depends on who is setting the price

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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Old Jul 7th 2011, 5:02 am
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Default 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.
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Old Jul 7th 2011, 5:33 am
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Default 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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Old Jul 7th 2011, 5:48 am
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Default 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.
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Old Jul 7th 2011, 6:00 am
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Default Re: Maths/Reasoning/Logic/Probability Exercise

Originally Posted by typical View Post
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.
Is it not an infinite type of problem?

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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Old Jul 7th 2011, 6:01 am
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Default Re: Maths/Reasoning/Logic/Probability Exercise

Originally Posted by Scamp View Post
Is it not an infinite type of problem?
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.

Last edited by typical; Jul 7th 2011 at 6:07 am.
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Old Jul 7th 2011, 6:08 am
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Default Re: Maths/Reasoning/Logic/Probability Exercise

Originally Posted by typical View Post
Yes, hence

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

Where "inf" means "infinity"

so the answer 15 applies to any number of rounds? or based on the game never ending?
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Old Jul 7th 2011, 6:17 am
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Default Re: Maths/Reasoning/Logic/Probability Exercise

Originally Posted by Scamp View Post
so the answer 15 applies to any number of rounds? or based on the game never ending?
The number 15 (again assuming my maths is right) is the expected value of winnings for player B: that is the (weighted) average of all possible outcomes. Those outcomes include the game being over in one throw (winning nothing), and the game going on forever, and everything inbetween.

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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Old Jul 7th 2011, 6:40 am
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Default Re: Maths/Reasoning/Logic/Probability Exercise

Originally Posted by typical View Post
The number 15 (again assuming my maths is right) is the expected value of winnings for player B: that is the (weighted) average of all possible outcomes. Those outcomes include the game being over in one throw (winning nothing), and the game going on forever, and everything inbetween.

Any one instance of the game could yield winnings of anything from 0 to infinity. The 15 is just the weighted average outcome.
Roger that. I'm with you now.

Intersting and difficult question.

I like my offer of a token nugget the best.
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Old Jul 7th 2011, 7:49 am
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Default Re: Maths/Reasoning/Logic/Probability Exercise

Do you get free drinks at this casino?
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Old Jul 7th 2011, 7:52 am
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Default Re: Maths/Reasoning/Logic/Probability Exercise

Originally Posted by typical View Post
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.
Looks sensible to me...
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Old Jul 7th 2011, 8:04 am
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Default Re: Maths/Reasoning/Logic/Probability Exercise

Originally Posted by Millhouse View Post
Do you get free drinks at this casino?
Never got free drinks at my old favourite casino, but they used to let me have as many ham and cheese sandwiches as I could eat for nothing.

Best way to end a night.
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Old Jul 7th 2011, 9:12 am
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Default Re: Maths/Reasoning/Logic/Probability Exercise

Originally Posted by typical View Post
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.
My head hurts just looking at this.
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