Re: Maths/Reasoning/Logic/Probability Exercise
 

This is a key part. While this is very true, it does not change the fact that it would take an expected 6 rolls to get a 6 (not four). Ain't probability great?

Why is this?

Well, the first roll you have probability 1/6 of getting a six.
The first roll doesn't change the dice though, so it doesn't change what'll happen next roll.

Therefore, if we call the number of rolls x, you have a 1/6 chance of needing one roll:

1 * 1/6

and a 5/6 chance of needing to start again - but having 'wasted' a roll.

(x+1) * 5/6

x = 1*1/6 + (x+1)*5/6
x = 1/6 + 5x/6 + 5/6
x = 1 + 5x/6
x/6 = 1
x = 6

Fair play to your daughter though, breaking down the individual probabilities is a good way to approach it...

Re: Maths/Reasoning/Logic/Probability Exercise
 

It reminded me of the old horse racing "trick" question........

You go to the races, and notice two particular horses (in separate races), each of which you believe has a 70 pct chance of winning its race.

It is tempting to assume the chance of BOTH winning must be 70 pct, but of course, it's (0.70 x 0.70) 49%, and therefore "unlikely".

My daughter followed that principle............

Re: Maths/Reasoning/Logic/Probability Exercise
 

But you'd take odds of 2-1 to bet on them both winning, I assume :)

Probabilities play strange tricks on people.

You only need 23 people in a room to have a 50% chance of a pair of them sharing a birthday. You only need 57 for that to be a 99% chance...

Ignoring people with birthdays on Feb 29 that is :o And assuming uniform distribution of birthdays which is not quite true but doesn't change things tooo much.

Re: Maths/Reasoning/Logic/Probability Exercise
 

I showed your calculations to a mate who wants to be A if you are B......

His explanation:

Expected return on any given throw £2.50 { £(1+2+3+4+5+0) / 6 }

Expected number of throws before you hit a 6 = 3.5 { (1+2+3+4+5+6) / 6 }

£2.5 * 3.5 = £8.75

You can't both be right.............. :huh:

Re: Maths/Reasoning/Logic/Probability Exercise
 

3.5 is the expected value of a single throw of the dice, not the expected number of throws before getting a 6.

(3.5 may also be the median number of throws - can't be bothered to work out the probability density function and check. Nevertheless this isn't that useful. Yes, around half the time you will throw three times or fewer (assuming that is the median). But half the time you will throw four or more - and by "or more" I mean "up to infinity".)

Re: Maths/Reasoning/Logic/Probability Exercise
 

OK - this is how I see it (as a non-mathematician)....... the 'four rolls' was a rough estimate............

B should consider how likely it is that the game will still continue after one roll, then after two rolls, etc.......

After 3.80176 rolls, the probability of the game continuing falls to 50%. (0.8333333, or 5/6, to the power of 3.80176 = 0.500002). After 3.80176 rolls of non-6 results, the maximum that could have been achieved is 3.80176 x 5. The minimum is 3.80176 x 1. The probable outcome is the mean of all combinations, which is 11.4053.

Therefore B should be willing to pay no more than 11.40, and preferably less.

Works for me..............

Re: Maths/Reasoning/Logic/Probability Exercise
 

The thread keeps going! :)

Again, the problem with the median is one side you have the set {1 throw, 2 throws, 3 throws} and the other side you have {4 throws, 5 throws, ...} - all the way to infinity.

That's why the mean is higher than the median. The range of winnings values on one side of the median is 0-15, and on the other side 3-infinity.

You also can't throw dice 3.8 times :)

But if you want to miss the chance to win INFINITE MONEY for the sake of three pounds sixty... :D

(There are any number of ways to get to the expected number of rolls to get a six, but they all equal 6.)

Re: Maths/Reasoning/Logic/Probability Exercise
 

OK......... but if I was C, and wanted to bet on A-v-B, and I knew B had paid 15 pounds to A, I'd bet on A every time.

And surely I still have the chance to win an infinite amount even if I "only" pay 11.40?

OK - let's stop now, or else we'll either (a) get banned by the Mods, or (b) get accused of being boring by fellow Bored (sic) members...........

Re: Maths/Reasoning/Logic/Probability Exercise
 

:rofl:

You'll only have the chance to win if the other party will let you play :)

Final question... who came up with this? :sneaky:

Re: Maths/Reasoning/Logic/Probability Exercise
 

A law firm my little angel is going to be internshipping with this summer. It was one of a few questions they put in a question paper she had to complete.

Re: Maths/Reasoning/Logic/Probability Exercise
 

Let us know if they got it right...

Re: Maths/Reasoning/Logic/Probability Exercise
 

B should pay the maximum amount that he would lose in one throw, which is 6...Because as soon as there's a 6, it's game over...

Can't be arsed dry with all the maths and calculus broohahaaa....

Re: Maths/Reasoning/Logic/Probability Exercise
 

Correct - and your mate is wrong...

In fairness the answer isn't really 15 - it should be less than 15 if he is rational, or greater if he gets free beer and ham sandwiches as part of the deal.

Re: Maths/Reasoning/Logic/Probability Exercise
 

I'm still waiting to read a convincing rebuttal to my approach (post 36).

If the question is about B's willingness to pay to play, he would surely want to measure the probabilities attached to how long the game is likely to last?

I certainly wouldn't pay 15 pounds to play, but 11.40 feels about right......

Re: Maths/Reasoning/Logic/Probability Exercise
 

Hell, if that's pork ham and beer in the UAE, it's got to be worth more than 15.

He will, yes. Again, the problem is taking the median - because of the "long tail" the 50% he'd be giving up on your strategy has the potential to be very, very, very lucrative. A would want compensated for that, even if B doesn't think it to be likely that he'd get there.