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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........? :huh: |
Re: Maths/Reasoning/Logic/Probability Exercise
Originally Posted by The Dean
(Post 9480102)
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........? :huh: 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. |
Re: Maths/Reasoning/Logic/Probability Exercise
Originally Posted by The Dean
(Post 9480102)
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........? :huh: 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 !! |
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. |
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 :D
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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. |
Re: Maths/Reasoning/Logic/Probability Exercise
Originally Posted by typical
(Post 9480349)
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. 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 |
Re: Maths/Reasoning/Logic/Probability Exercise
Originally Posted by Scamp
(Post 9480365)
Is it not an infinite type of problem?
"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 :D ) For what it's worth, I'd rate this of similar difficulty to harder Further Maths A-Level questions or my uni entrance exam. |
Re: Maths/Reasoning/Logic/Probability Exercise
Originally Posted by typical
(Post 9480367)
Yes, hence
"sigma (n=0, inf) (5/6)^n * 15/6" Where "inf" means "infinity" :) |
Re: Maths/Reasoning/Logic/Probability Exercise
Originally Posted by Scamp
(Post 9480378)
so the answer 15 applies to any number of rounds? or based on the game never ending?
Any one instance of the game could yield winnings of anything from 0 to infinity. The 15 is just the weighted average outcome. |
Re: Maths/Reasoning/Logic/Probability Exercise
Originally Posted by typical
(Post 9480387)
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. Intersting and difficult question. I like my offer of a token nugget the best. |
Re: Maths/Reasoning/Logic/Probability Exercise
Do you get free drinks at this casino?
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Re: Maths/Reasoning/Logic/Probability Exercise
Originally Posted by typical
(Post 9480349)
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. |
Re: Maths/Reasoning/Logic/Probability Exercise
Originally Posted by Millhouse
(Post 9480500)
Do you get free drinks at this casino?
Best way to end a night. |
Re: Maths/Reasoning/Logic/Probability Exercise
Originally Posted by typical
(Post 9480349)
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. |
Re: Maths/Reasoning/Logic/Probability Exercise
He should offer nothing, as it is clearly gambling and that just is not allowed here now is it!
Tut tut. :sneaky: |
Re: Maths/Reasoning/Logic/Probability Exercise
4s/3d
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Re: Maths/Reasoning/Logic/Probability Exercise
Originally Posted by typical
(Post 9480349)
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. |
Re: Maths/Reasoning/Logic/Probability Exercise
Originally Posted by Millsyisland
(Post 9480606)
My head hurts just looking at this.
The first go, B wins whatever is on the dice unless it's 6, in which case nada and game over. Hence the expected (average) win for one round is: 1*1/6 + 2*1/6 + 3*1/6 + 4*1/6 + 5*1/6 + 0*1/6 = 15/6. Green being the colour of money. There's then a 5/6 chance to play round two, a 5/6 chance from there to go on to round three, etc, so the winnings expected would be: Code:
5/6*(15/6 +This is 15/6 + 5/6*15/6 + 5/6*5/6*15/6 + 5/6*5/6*5/6*15/6 + ... and on and on (and Ariston). But this is 15/6 * ∑n=0,…,∞ (5/6)^n To calculate the series, note that it looks like this: 1 + 5/6 + 25/36 + ... Call this s = 1 + 5/6 + 25/36 + ... But if we multiply each term by 5/6, we get the same progression but without the first term (of 1): s*5/6 = 5/6 + 25/36 + ... Now through the wonder of infinite series, we can substitute the second equation into the first: s = 1 + s*5/6 s/6 = 1 s = 6 Therefore the expected value of B's winnings is 6*15/6 = 15 B should pay at most 15 to play. |
Re: Maths/Reasoning/Logic/Probability Exercise
Originally Posted by typical
(Post 9481346)
Really? Let me try to explain then, now it's evening!
The first go, B wins whatever is on the dice unless it's 6, in which case nada and game over. Hence the expected (average) win for one round is: 1*1/6 + 2*1/6 + 3*1/6 + 4*1/6 + 5*1/6 + 0*1/6 = 15/6. Green being the colour of money. There's then a 5/6 chance to play round two, a 5/6 chance from there to go on to round three, etc, so the winnings expected would be: Code:
5/6*(15/6 +This is 15/6 + 5/6*15/6 + 5/6*5/6*15/6 + 5/6*5/6*5/6*15/6 + ... and on and on (and Ariston). But this is 15/6 * ∑n=0,…,∞ (5/6)^n To calculate the series, note that it looks like this: 1 + 5/6 + 25/36 + ... Call this s = 1 + 5/6 + 25/36 + ... But if we multiply each term by 5/6, we get the same progression but without the first term (of 1): s*5/6 = 5/6 + 25/36 + ... Now through the wonder of infinite series, we can substitute the second equation into the first: s = 1 + s*5/6 s/6 = 1 s = 6 Therefore the expected value of B's winnings is 6*15/6 = 15 B should pay at most 15 to play. |
Re: Maths/Reasoning/Logic/Probability Exercise
Originally Posted by al dente
(Post 9481407)
Now you're talking dirty....
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Re: Maths/Reasoning/Logic/Probability Exercise
Originally Posted by Morecambe Mindy
(Post 9481652)
He's a maths slut!
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Re: Maths/Reasoning/Logic/Probability Exercise
Originally Posted by typical
(Post 9481668)
Well, a degree in mathematics has to be good for something - might as well be that...
Typical, the wording in the problem ("....willing to pay....") is significant. If I understand your answer correctly, 15 is the fee at which B would be 50% likely to lose (i.e. he wins 14 or less) and 50% likely to win (i.e. he wins 16 or more). The answer then should surely be 14, not 15? He should only be willing to pay a fee that gives him a probability, albeit small, of winnning, surely? |
Re: Maths/Reasoning/Logic/Probability Exercise
Originally Posted by The Dean
(Post 9482009)
Typical, the wording in the problem ("....willing to pay....") is significant. If I understand your answer correctly, 15 is the fee at which B would be 50% likely to lose (i.e. he wins 14 or less) and 50% likely to win (i.e. he wins 16 or more). The answer then should surely be 14, not 15? He should only be willing to pay a fee that gives him a probability, albeit small, of winnning, surely?
(E.g. imagine a very simple game where you throw two coins: HH wins 200 but any other outcome wins 100. What should you pay to play? The expected outcome is: HH: 200 * 1/4 + HT: 100 * 1/4 + TH: 100 * 1/4 + TT: 100 * 1/4 = 125 So you should pay at most 125. But note that you're not equally likely to win more than this and less than this in any single game!) However, if you played the game an infinite number of times, and calculated your average winnings from all those games, you would find the average to be 15. Now, as a mathematician I say you should pay 15 at the most, since at 15 you would be (in probability terms) ambivalent about playing. Your expected net win is 0, but that's still ok - you won't care between playing or not playing so there's no reason not to play. As an economist, it comes down to attitudes towards risk. Some people will pay much more than the expected outcome if the probability density function means there is a chance of a massive upside and limited downside (as there is in this game). See: any lottery that doesn't pay out all its revenue as prize money. |
Re: Maths/Reasoning/Logic/Probability Exercise
Originally Posted by typical
(Post 9482316)
It's not a good idea to think of expected value in that way. For a start, it ignores the probability of actually getting 15 - but also it can be incorrect.
(E.g. imagine a very simple game where you throw two coins: HH wins 200 but any other outcome wins 100. What should you pay to play? The expected outcome is: HH: 200 * 1/4 + HT: 100 * 1/4 + TH: 100 * 1/4 + TT: 100 * 1/4 = 125 So you should pay at most 125. But note that you're not equally likely to win more than this and less than this in any single game!) However, if you played the game an infinite number of times, and calculated your average winnings from all those games, you would find the average to be 15. Now, as a mathematician I say you should pay 15 at the most, since at 15 you would be (in probability terms) ambivalent about playing. Your expected net win is 0, but that's still ok - you won't care between playing or not playing so there's no reason not to play. As an economist, it comes down to attitudes towards risk. Some people will pay much more than the expected outcome if the probability density function means there is a chance of a massive upside and limited downside (as there is in this game). See: any lottery that doesn't pay out all its revenue as prize money. You've given me some food for thought though........ my (non-mathematical) gut feeling told me to pay a lot less than 15........... |
Re: Maths/Reasoning/Logic/Probability Exercise
Originally Posted by The Dean
(Post 9482399)
But in your game, the two coins are tossed, and the game is then definitely over - not the same thing. Your 125 reflects the total of ALL possible outcomes divided by the sample size - not the same thing.
Instead of looking at it as you write above, look at it as "the sum of all possible outcomes multiplied by the chance of each outcome occuring". This is the same thing as you suggest, but writing it this way helps understand why the massive outcomes with tiny probabilities in the infinite game case don't change the overall answer that much. In my second game above, the sum of all possible outcomes multiplied by the chance of each outcome occuring is 125. In the original game, the sum of all possible outcomes multipled by the chance of each outcome occuring is 15. :) |
Re: Maths/Reasoning/Logic/Probability Exercise
The expected value of the number of rounds played is 6. That might help you think about why the number may be higher than expected.
(I had something in here about how to use this to calculate expected winnings but haven't proved it so decided best not to risk saying it :) ) Also, if you really want to rake your noodle: There is a higher probability of rolling a six and the game ending on the first go, than on the second go. :) |
Re: Maths/Reasoning/Logic/Probability Exercise
Originally Posted by typical
(Post 9482409)
Ah but it IS (effectively) the same thing. That's the beauty.
Instead of looking at it as you write above, look at it as "the sum of all possible outcomes multiplied by the chance of each outcome occuring". This is the same thing as you suggest, but writing it this way helps understand why the massive outcomes with tiny probabilities in the infinite game case don't change the overall answer that much. In my second game above, the sum of all possible outcomes multiplied by the chance of each outcome occuring is 125. In the original game, the sum of all possible outcomes multipled by the chance of each outcome occuring is 15. :) Probability of the game reaching a second roll = 0.8333 Ditto..... third roll = 0.6944 Ditto..... fourth roll = 0.5787 Ditto..... fifth roll = 0.4822 B should therefore assume that is unlikely (i.e. less than 50:50) that there will be more than four rolls of the dice..... Four rolls of 5 = max 20 pounds winnings Four rolls of 1 = min 4 pounds winnings ... etc...... Therefore, B would pay a MAXIMUM of 12 pounds. |
Re: Maths/Reasoning/Logic/Probability Exercise
Originally Posted by The Dean
(Post 9482425)
A simplistic way of looking at it.................
Put it this way, as A I would not charge LESS than 15 quid to play... |
Re: Maths/Reasoning/Logic/Probability Exercise
Originally Posted by typical
(Post 9482442)
:)
Put it this way, as A I would not charge LESS than 15 quid to play... Right, I'm off to the pub............. (the above was my daughter's answer by the way - she's a Law student, not a mathematician - not bad I thought) |
Re: Maths/Reasoning/Logic/Probability Exercise
Originally Posted by The Dean
(Post 9482425)
B should therefore assume that is unlikely (i.e. less than 50:50) that there will be more than four rolls of the dice.....
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
Originally Posted by typical
(Post 9482459)
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... 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
Originally Posted by The Dean
(Post 9482460)
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".
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
Originally Posted by The Dean
(Post 9482647)
Expected number of throws before you hit a 6 = 3.5 { (1+2+3+4+5+6) / 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
Originally Posted by typical
(Post 9483358)
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.) 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
Originally Posted by typical
(Post 9483416)
: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: |
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