- #36
KLscilevothma
- 322
- 0
Do you mind if you answer oxy's question and I post the other question ?
In summary: Lat: 90Long: 0as John's starting position.So that's one place.There would be infinitely many such points (all close to the north pole).Thank you for the elegant solution, Hurkyl! In summary, in the conversation, the participants discussed a math quiz game where the first person to answer correctly gets to ask the next question. They also talked about a picture puzzle and a carpeting problem. Eventually, they started discussing a riddle involving a man walking one mile in different directions and seeing a bear. The solution to the riddle is that the bear is white, but the question was then posed of how many different ways the man could have walked to end up back where he started
Do you mind if you answer oxy's question and I post the other question ?
- #37
marcus
Science Advisor
Gold Member
Dearly Missed
- 24,775
- 792
Guys! we are doing double threading, it is BOTH Oxy and Kam turn to ask question
and anyone can answer one or the other or both
with no preferred order
Kam please go ahead and pose a Q: we are eagerly looking forward to it.
Oxy's question is already before us and we may suppose that this
is how they play the stock market in Australia, and I would guess tho it remains to be proven, that the large investor
has a high probability of wiping out the small investor. How hard can it be? Somebody take a whack at it!
and anyone can answer one or the other or both
with no preferred order
Kam please go ahead and pose a Q: we are eagerly looking forward to it.
Oxy's question is already before us and we may suppose that this
is how they play the stock market in Australia, and I would guess tho it remains to be proven, that the large investor
has a high probability of wiping out the small investor. How hard can it be? Somebody take a whack at it!
Last edited:
- #38
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
- 14,983
- 28
There is a finite state space and from any intermediate state, the endstate is reachable in n turns with some fixed nonzero probability for some values of n, so by the law of large numbers, the end state is reached with probability 1.
IOW you're guaranteed the game ends, no matter what x and y are.
Now, an interesting question is what is the probability the player with x pennies beats the player with y pennies...
IOW you're guaranteed the game ends, no matter what x and y are.
Now, an interesting question is what is the probability the player with x pennies beats the player with y pennies...
- #39
KLscilevothma
- 322
- 0
Geometry
In the figure, construct a line through p cutting triangle ABC into 2 parts with equal areas.
In the figure, construct a line through p cutting triangle ABC into 2 parts with equal areas.
Last edited:
- #40
Oxymoron
- 870
- 0
Correct thus far. However...
...is the real problem!
This is an example of a one-dimensional random walk (actually quite an easy one!).
If anyone needs some help just ask!
Cheers
- #41
Oxymoron
- 870
- 0
Suppose you have $20 and need $50 for a bus ride home. Your only chance to get more money is by gambling in a casino. There is only one possible bet you can make at the casino--you must bet $10, and then a fair coin is flipped. If "heads" results, you win $10 (in other words, you get your original $10 back plus an additional $10), and if it's "tails", you lose the $10 bet. The coin has exactly the same 50% probability of coming up heads or tails. What is the chance that you will get the $50 you need?
This is very similar to my original question, and I have changed it here for simplicity. In this case the casino is player A and the person is player B. Note that casino (Player A) has MUCH more coins than player B. I will now prove the question...
This problem is known as the ``Gambler's Ruin Problem''--a gambler keeps betting until he goes broke, or until he reaches a certain goal, and there are no other possibilities. Here is a nice matrix-based approach.
At any stage in the process, there are six possible states, depending on your ``fortune''. You have either $0 (and you've lost), or you have $10, $20, $30, $40 (and you're still playing), or you have $50 (and you've won). You know that when you begin playing, you are in a certain state (having $20 in the case of the latest problem). After you've played a while, lots of different things could have happened, so depending on how long you've been going, you have various probabilities of being in the various states. Call the state with $0 ``state 0'', and so on, up to ``state 5'' that represents having $50.
At any particular time, let's let P0 represent the probability of having $0, P1 the probability of having $10, and so on, up to P5 of having won with $50.
We can give an entire probabilistic description of your state as a column vector like this:
P0
P1
P2
P3
P4
P5
Now look at the individual situations. If you're in state 0, or in state 5, you will remain there for sure, because that is where the game ends. If you're in any other state, there's a 50% chance of moving up a state and a 50% chance of moving down a state. Look at the following matrix multiplication:
1 0.5 0 0 0 0 x P0 = P0 + 0.5P1
0 0 0.5 0 0 0 x P1 = 0.5P2
0 0.5 0 0.5 0 0 x P2 = 0.5P1 + 0.5P3
0 0 0.5 0 0.5 0 x P3 = 0.5P2 + 0.5P4
0 0 0 0.5 0 0 x P4 = 0.5P3
0 0 0 0 0.5 1 x P5 = 0.5P4 + P5
Clearly, multiplication by the matrix above represents the change in probabilities of being in the various states, given an initial probabalistic distribution. The chance of being in state 0 after a coin flip is the chance you were there before, plus half of the chance that you were in state 1. Check that the others make sense as well.
So if the matrix on the left in equation (3) is called P, each time you multiply the vector corresponding to your initial situation by P, you'll find the probabilities of being in the various states. So after 1024 coin-flips, your state will be represented by P1024.
Here is a probability matrix of probabilities of various states after 1024 flips:
1 0.8 0.6 0.4 0.2 0
0 0.0 0 0.0 0 0
0 0 0.0 0 0.0 0
0 0.0 0 0.0 0 0
0 0 0.0 0 0.0 0
0 0.2 0.4 0.6 0.8 1
I write 0.0 because in actual fact that coordinate is 1.549x10^-93 which is an extremely small number (EXTREMELY SMALL!)
So for practicality I will write it like this
1 0.8 0.6 0.4 0.2 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0.2 0.4 0.6 0.8 1
So what does this probability matrix tell us? Well it tells us that we have a 100% chance of winning! Look at the 1 in the bottom right hand corner. This coordinate represents where the game finishes by winning. It says 1 which represents 100% chance. But there is also a 100% of losing?!
Can anyone tell me why this is?
Also know think of why Casino's always end up on top!
This is very similar to my original question, and I have changed it here for simplicity. In this case the casino is player A and the person is player B. Note that casino (Player A) has MUCH more coins than player B. I will now prove the question...
This problem is known as the ``Gambler's Ruin Problem''--a gambler keeps betting until he goes broke, or until he reaches a certain goal, and there are no other possibilities. Here is a nice matrix-based approach.
At any stage in the process, there are six possible states, depending on your ``fortune''. You have either $0 (and you've lost), or you have $10, $20, $30, $40 (and you're still playing), or you have $50 (and you've won). You know that when you begin playing, you are in a certain state (having $20 in the case of the latest problem). After you've played a while, lots of different things could have happened, so depending on how long you've been going, you have various probabilities of being in the various states. Call the state with $0 ``state 0'', and so on, up to ``state 5'' that represents having $50.
At any particular time, let's let P0 represent the probability of having $0, P1 the probability of having $10, and so on, up to P5 of having won with $50.
We can give an entire probabilistic description of your state as a column vector like this:
P0
P1
P2
P3
P4
P5
Now look at the individual situations. If you're in state 0, or in state 5, you will remain there for sure, because that is where the game ends. If you're in any other state, there's a 50% chance of moving up a state and a 50% chance of moving down a state. Look at the following matrix multiplication:
1 0.5 0 0 0 0 x P0 = P0 + 0.5P1
0 0 0.5 0 0 0 x P1 = 0.5P2
0 0.5 0 0.5 0 0 x P2 = 0.5P1 + 0.5P3
0 0 0.5 0 0.5 0 x P3 = 0.5P2 + 0.5P4
0 0 0 0.5 0 0 x P4 = 0.5P3
0 0 0 0 0.5 1 x P5 = 0.5P4 + P5
Clearly, multiplication by the matrix above represents the change in probabilities of being in the various states, given an initial probabalistic distribution. The chance of being in state 0 after a coin flip is the chance you were there before, plus half of the chance that you were in state 1. Check that the others make sense as well.
So if the matrix on the left in equation (3) is called P, each time you multiply the vector corresponding to your initial situation by P, you'll find the probabilities of being in the various states. So after 1024 coin-flips, your state will be represented by P1024.
Here is a probability matrix of probabilities of various states after 1024 flips:
1 0.8 0.6 0.4 0.2 0
0 0.0 0 0.0 0 0
0 0 0.0 0 0.0 0
0 0.0 0 0.0 0 0
0 0 0.0 0 0.0 0
0 0.2 0.4 0.6 0.8 1
I write 0.0 because in actual fact that coordinate is 1.549x10^-93 which is an extremely small number (EXTREMELY SMALL!)
So for practicality I will write it like this
1 0.8 0.6 0.4 0.2 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0.2 0.4 0.6 0.8 1
So what does this probability matrix tell us? Well it tells us that we have a 100% chance of winning! Look at the 1 in the bottom right hand corner. This coordinate represents where the game finishes by winning. It says 1 which represents 100% chance. But there is also a 100% of losing?!
Can anyone tell me why this is?
Also know think of why Casino's always end up on top!
- #42
KLscilevothma
- 322
- 0
What should we do if nobody would like to answer our questions ?
- #43
Oxymoron
- 870
- 0
I guess we win! and get to ask another question.
Take a helium balloon (the kind that floats in air ) and attach the string to floor of you car. You should have a floating balloon tied to a string attached to the floor in your car. Got it? Good let's continue...
Now start to accelerate, which way will the balloon go?
a) BACKWARD
b) FORWARD
c) WILL NOT MOVE
d) UPWARD
e) WHAT BALLOON?
(It is not exactly a maths question, more of a physics question but let's see who can get it first!).
PS. KD I did honestly have a go at your question but was quite unsure of how to answer it?
Take a helium balloon (the kind that floats in air
) and attach the string to floor of you car. You should have a floating balloon tied to a string attached to the floor in your car. Got it? Good let's continue...Now start to accelerate, which way will the balloon go?
a) BACKWARD
b) FORWARD
c) WILL NOT MOVE
d) UPWARD
e) WHAT BALLOON?
(It is not exactly a maths question, more of a physics question but let's see who can get it first!).
PS. KD I did honestly have a go at your question but was quite unsure of how to answer it?
- #44
bogdan
- 191
- 0
Is it a convertible ? (the car)
If it's not then I guess that the baloon should not move...
...because the air moves with the car...and if the acceleratin is not too great than the air should not compress...I guess...
If it's not then I guess that the baloon should not move...
...because the air moves with the car...and if the acceleratin is not too great than the air should not compress...I guess...
Last edited:
- #45
bogdan
- 191
- 0
For that triangle problem...KL_KAM...
I suppose that 2*AP=PB...
http://www.angelfire.com/pro/fbi/images/trio.bmp
Sorry for the delay...
Because the link above may not work...
Here is the solution:
Take on AC a segment CE=2*AC on the opposite side to A...then let F be the intersection between PE and BC...now S(ACFP)=S(PBE)...good enough ?
I suppose that 2*AP=PB...
http://www.angelfire.com/pro/fbi/images/trio.bmp
Sorry for the delay...
Because the link above may not work...
Here is the solution:
Take on AC a segment CE=2*AC on the opposite side to A...then let F be the intersection between PE and BC...now S(ACFP)=S(PBE)...good enough ?
Last edited by a moderator:
- #46
Oxymoron
- 870
- 0
bogdan, the type of car doesn't really matter but for simplicity let's just say it is a normal car with all its windows up. Thanks for having a go anyway!
(c) is incorrect.
If anyone else would like to have a try please remember to put in your reasoning like bogdan did.
(c) is incorrect.
If anyone else would like to have a try please remember to put in your reasoning like bogdan did.
- #47
bogdan
- 191
- 0
Are you sure that c is incorrect ?
Remember:the baloon is attached to the floor...not to a weight that floats in the air and is attached to the floor by a string...soooo...why should c be incorrect ?
If it was attached to a weight than it should move forward...
And why do you think that the type of the car doesn't matter ?
If it is a convertible (opened) than the baloon moves backward...
Soo...what's the problem with my answer ?
Remember:the baloon is attached to the floor...not to a weight that floats in the air and is attached to the floor by a string...soooo...why should c be incorrect ?
If it was attached to a weight than it should move forward...
And why do you think that the type of the car doesn't matter ?
If it is a convertible (opened) than the baloon moves backward...
Soo...what's the problem with my answer ?
Last edited:
- #48
bogdan
- 191
- 0
Aaaa...and for that probability problem...the first on the 3rd page...with x and y pennies...
P(x)=x/(x+y)...P(x)->the probability that the player with x pennies wins...
P(y)=y/(x+y)...P(y)->...
Proof ? Somesort of recursivity...too long to be written here...
P(x)=x/(x+y)...P(x)->the probability that the player with x pennies wins...
P(y)=y/(x+y)...P(y)->...
Proof ? Somesort of recursivity...too long to be written here...
- #49
KLscilevothma
- 322
- 0
I didn't say 2*AP=AC.
P can be any point on AB.
Hint: Think of the special case when P=A.
Last edited by a moderator:
- #50
bogdan
- 191
- 0
Okay...I'll think about it...
- #51
MathNerd
I have infact found a method for proving the probability question using matrices and vectors, no recursivity. So I will be able to prove whether your answer is correct or not
.- #52
brum
- 81
- 0
b) FORWARD
- #53
Oxymoron
- 870
- 0
FINALLY! Brum is correct. Could you give me a reason why you chose B?
- #54
hypnagogue
Staff Emeritus
Science Advisor
Gold Member
- 2,285
- 3
Darn, my guess was E.
Oxymoron, care to explain the solution to your probability matrix problem since no one seems to be answering?
Oxymoron, care to explain the solution to your probability matrix problem since no one seems to be answering?
- #55
KLscilevothma
- 322
- 0
Solution to the geometry problem
Since I'll be busy, so I think I better post the solution now.
1. Construct a line AD such that CD = BD. Therefore [ADC] = [ADB]
([ADC] means area of triangle ADC)
2. Draw a line to join PD
3. Construct a line AF such that AF // PD
4. Join PF
Now, [APG] = [GFD] (note that AP does not necessary equal to FD)
and [APCF] = [PBF]
The line PF is all I want.
http://www.angelfire.com/alt2/antiwork/sol.bmp
Since I'll be busy, so I think I better post the solution now.
1. Construct a line AD such that CD = BD. Therefore [ADC] = [ADB]
([ADC] means area of triangle ADC)
2. Draw a line to join PD
3. Construct a line AF such that AF // PD
4. Join PF
Now, [APG] = [GFD] (note that AP does not necessary equal to FD)
and [APCF] = [PBF]
The line PF is all I want.
http://www.angelfire.com/alt2/antiwork/sol.bmp
Last edited by a moderator: