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Benny
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probability challenging prob.
« on: Jun 13th, 2008, 12:02pm » |
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Problem (1)
Take a stick and cut it twice so that you have three sticks. If you make the cuts randomly, what are the chances you can make a triangle out of your three sticks?
Is the answer 50%?
(2)
Alice tosses 849 fair coins and Bob tosses 850. What is the probability that Alice gets more heads than Bob?
Is the problem worded correctly? I understand it better if the problem would have asked:
What is the probability that Bob gets more heads than Alice?
Then my answer would be: 1/2
Do you agree? Is there something I'm missing here?
(3)
I've worked on the 4 Door Monty Hall Problem
Behind 1 of 4 closed doors is a prize. You pick door 1. Monty opens the 4th door and it is empty. You switch your choice to door 2. Now monty opens door 3 and it is also empty. Given the choice, should you switch back to door 1, and what are your chances of winning if you do?
My answer: chances are 5/8 if you switch
Has anyone worked on an N door generalization of the problem?
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jollytall
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Re: probability challenging prob.
« Reply #1 on: Jun 13th, 2008, 2:11pm » |
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(1)
If I define the two random cuts as such that I generate to numbers (0-1) randomly and then cut at those places then I would go for 1/4.
If both are <0.5 or both >0.5 surely you cannot make a triagle. This is already 1/2. But even when a<0.5 and b>0.5 you can have that b-a>0.5. This is again half of the remaining, i.e. 1/4. So you cannot make 3/4.
(2) The two questions are not the same. They can have the same number of heads. Both questions can be answered though if you calculate the probability of A having a heads and B having b heads. Then you sum for all possible outcome of p(a) the sum of p(b) where b<a.
p(a)= (a|849)/2^849 = a!*(849-a)!/849!/2^849
p(b)= (b|850)/2^850 = a!*(850-a)!/850!/2^850
sum a=1 to 849: p(a)*( sum b=0 to a-1 p(b) )
(3)
I guess you had:
Step 0
1/4 1/4 1/4 1/4.
Step 1
1/4 3/8 3/8 0
Step 2
5/8 3/8 0 0
This sequence is misleading though. It assumes that in step 2 he can show W3. But it is not necessary the case. He can do it 5/8 of the time, while he can show W1 in 6/8 of the time (3/8 he can show either of them). We should know (or assume) something about his strategy when he can show both.
If we assume he shows W1 if he can then showing W3 means we must change back 1/1 (he could not show W1 because it is with the gift).
If we assume he shows W3 if he can then I would not change back 3/5 to stay.
If we assume that if he can show both then he changes random (1/2-1/2) then showing W3 is either because only that he could show (1/4 probability, after changing back 1/1) or both were free 3/8 of the time, he chose it 1/2, i.e. 3/16 and you should not change back. So changing back is winning 4/7 and not 5/8.
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| « Last Edit: Jun 13th, 2008, 2:35pm by jollytall » |
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jollytall
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Re: probability challenging prob.
« Reply #2 on: Jun 13th, 2008, 2:51pm » |
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A bit more explanation on (3).
Let's imagine N is very big and in the first step he shows N-3. We are left with three windows, W1, W2, W3. W1 was our first choice, but now we change to W2 (since almost 50% that it is there). Now he shows W3.
If we assume his strategy is to show W1 if he can then we must change back to W1.
If we assume his strategy is to show W3 if he can then we are either in the case with 1/N probability that it is under W1 or the (N-1)/2N that it is under W2. So staying with W2 is (N-1)/(N+1) chance the winner.
If he choses whichever he can random, then we are either in the case with 1/N when he could only show W3 or we are in the case with probability (N-1) /4N that the gift is under W2 AND he chose W3. So staying is better with a probability (N-1)/(N+3). For N=4 it means changing back is better, for N=5 it is 50-50 and for N>5 it is better to stay with W2.
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Benny
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Re: probability challenging prob.
« Reply #3 on: Jun 13th, 2008, 3:51pm » |
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Quote:
Problem (1)
Take a stick and cut it twice so that you have three sticks. If you make the cuts randomly, what are the chances you can make a triangle out of your three sticks?
Is the answer 50%?
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I think 50% is the wrong answer. I believe now that the answer is 100% !!
Because the basic principle at work here is that no single side of a triangle can be longer than the other two sides combined. If you take a stick and then cut it in 3, you won't violate the basic principle.
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Benny
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Re: probability challenging prob.
« Reply #4 on: Jun 13th, 2008, 4:07pm » |
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Quote:
(2)
Alice tosses 849 fair coins and Bob tosses 850. What is the probability that Alice gets more heads than Bob?
Is the problem worded correctly? I understand it better if the problem would have asked:
What is the probability that Bob gets more heads than Alice?
Then my answer would be: 1/2
Do you agree? Is there something I'm missing here?
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The way I understand this is either Bob has more heads than Alice or Bob has more tails than Alice, but not both.
It is 1/2 because suppose Bob has 1 coin and Alice has 0, then the probability is 1/2. The additional 849 tosses each makes should not affect this, right?
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SMQ
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Re: probability challenging prob.
« Reply #5 on: Jun 13th, 2008, 7:12pm » |
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on Jun 13th, 2008, 3:51pm, BenVitale wrote:| I believe now that the answer is 100% !! |
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I believe you are mistaken. If I have a stick 1m long and I cut two 10cm pieces from it, clearly no triangle can be made with these two pieces and the remaining 80cm piece.
--SMQ
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Benny
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Re: probability challenging prob.
« Reply #6 on: Jun 13th, 2008, 7:38pm » |
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on Jun 13th, 2008, 7:12pm, SMQ wrote:
I believe you are mistaken. If I have a stick 1m long and I cut two 10cm pieces from it, clearly no triangle can be made with these two pieces and the remaining 80cm piece.
--SMQ |
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You're thinking of triangles that connect.
I was thinking of triangles that don't necessarily connect, and used the basic principle.
And, in your case, what is the probability?
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| « Last Edit: Jun 13th, 2008, 7:41pm by Benny » |
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towr
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Re: probability challenging prob.
« Reply #7 on: Jun 14th, 2008, 4:27am » |
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on Jun 13th, 2008, 7:38pm, BenVitale wrote:You're thinking of triangles that connect.
I was thinking of triangles that don't necessarily connect, and used the basic principle. |
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What does that mean? Connecting triangles?! We're trying to make one triangle from three pieces of stick.
If you have sides of length a,b,c and a+b < c, you can't make a triangle with those sides.
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Benny
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Re: probability challenging prob.
« Reply #8 on: Jun 14th, 2008, 10:39am » |
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on Jun 14th, 2008, 4:27am, towr wrote:
What does that mean? Connecting triangles?! We're trying to make one triangle from three pieces of stick.
If you have sides of length a,b,c and a+b < c, you can't make a triangle with those sides.
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How about
/\
_________________________
One could place them such that a triangle appears, but with one side much longer.
I was trying to be tricky.
But, if we are not trying to be tricky then we need to calculate the probability the longest piece is shorter than the two other pieces combined.
Right side --------------------------------- Left side
(a) the first cut is either somewhere close to the left side or close to the right side.
(b) the first cut is at the center.
If (b) you won't be able to make a triangle.
The second cut:
If it is at the center you won't be able to make a triangle.
Since the first cut has to be on one side or the other, you have a 50% chance of getting the same side with the second cut.
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jollytall
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Re: probability challenging prob.
« Reply #9 on: Jun 14th, 2008, 1:02pm » |
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BenVitale,
Read what I said re (1). You did not specify precisely how you make two random cuts. One interpretation was to make two random numbers 0-1 first and then do the cutting at those two positions. Then the probability is 1/4.
I guess you also want to say the same logic of cutting, but your following reasoning is incorrect. It is true that if they are on the same half then you cannot make a triangle and that has 1/2 probability. But even if they are in the opposite sides it does not guarantee that you can make a triangle. It can happen that the middle piece is too long. This is 1/2 of the cases when the two cuts are at the opposite sides, or 1/4 of all the cases. This leaves 1/4.
Actually you can also define two random cuttings a bit different. First you make one cut at a random place and keep aside the left part. Then from the remaining (right) side again with a linear distribution you cut down the second piece at what is left is the third (just like in the Sausage duel). In this case if the first cut is >0.5 you do not have a chance. If it is less then 0.5 (x) then the probability that the second one result a triabgle is x/(1-x). You simply have to integral it 0-0.5 to get the total probability of a triangle made.
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Benny
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Re: probability challenging prob.
« Reply #10 on: Jun 14th, 2008, 3:03pm » |
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Thanks for the feedback.
The problem was presented to me as such
Quote:
Take a stick and cut it twice so that you have three sticks. If you make the cuts randomly, what are the chances you can make a triangle out of your three sticks?
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It did not specify precisely how one can make 2 random cuts.
on Jun 13th, 2008, 2:11pm, jollytall wrote:(1)
If I define the two random cuts as such that I generate to numbers (0-1) randomly and then cut at those places then I would go for 1/4.
If both are <0.5 or both >0.5 surely you cannot make a triagle. This is already 1/2. But even when a<0.5 and b>0.5 you can have that b-a>0.5. This is again half of the remaining, i.e. 1/4. So you cannot make 3/4. ............
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I need time to process your solution. I was so convinced of my solution, i need time, change my mind-set.
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jollytall
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A two parameter random situation is best to imagine graphically and then compare the areas. Earlier I was lazy to draw, but here is a very simple one.
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Benny
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Re: probability challenging prob.
« Reply #12 on: Jun 15th, 2008, 12:45pm » |
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Jollytall, I now understand your solution. Thanks.
If we increase the number of doors from 3 to 100. If there are 100 doors, and Monty Hall shows that 98 of them are valueless.
the chance the prize behind the remaining door is
Using the formula in the following linked document
http://en.wikipedia.org/wiki/Monty_Hall_problem
N door generalization of the original problem in which the host opens p losing doors and then offers the player the opportunity to switch; in this variant switching wins with probability
(N-1)/N(N-p-1)
N=100, p=98 so the probability is 99/100
How to establish (N-1)/N(N-p-1) ?
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towr
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Re: probability challenging prob.
« Reply #13 on: Jun 15th, 2008, 1:41pm » |
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on Jun 15th, 2008, 12:45pm, BenVitale wrote:| How to establish (N-1)/N(N-p-1) ? |
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The not-selected doors have a sum probability of (N-1)/N of containing the price. There are (N-p-1) identical doors left. So divide the total probability by this to get (N-1)/N(N-p-1) per door for switching.
But note that this is unlike your earlier generalization where doors where opened one at a time and you could switch at every step.
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| « Last Edit: Jun 15th, 2008, 1:42pm by towr » |
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Benny
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Re: probability challenging prob.
« Reply #14 on: Jun 16th, 2008, 3:19pm » |
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Quote:
(3)
I've worked on the 4 Door Monty Hall Problem
Behind 1 of 4 closed doors is a prize. You pick door 1. Monty opens the 4th door and it is empty. You switch your choice to door 2. Now monty opens door 3 and it is also empty. Given the choice, should you switch back to door 1, and what are your chances of winning if you do?
My answer: chances are 5/8 if you switch |
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Sorry, i forgot about this one. This is how I reasoned:
The chances are 5/8 if you switch. The initial chances of your first choice were 1/4, and opening another door without the prize doesn't change that, so the remaining 2 doors now have a 3/8 chance because the chances of all possibilities must sum to 1. But then when Monty opens door 3, using the same reasoning, door 2 stays at 3/8 chance and so the remaining door changes to 5/8.
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jollytall
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Re: probability challenging prob.
« Reply #15 on: Jun 17th, 2008, 12:08am » |
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BenVitale,
It was clear how you calculated. I also mentioned it in my first reply. But you also see there and then more in details in the second one, why that is not correct.
You also have to have some initial assumption about Monthy's actions. In case of four windows it is very important.
It is also important in the original problem. We always assume that Monthy MUST show an empty window after your first choice. In a real game it would not necessarily be the story. The producer might have other instructions. If you assume that they want to save cost then he will only show an empty window and offer to swap if you found the gift. So with this assumption you should never swap. But the opposite might be true as well. Their sponsor wants to gift given. So if you found it, he will simply give it and will only offer the change if you missed. So you must change. This is so obvious that we never mention.
But in case of four windows it is not that obvious, what the instructions are.
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Hippo
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Re: probability challenging prob.
« Reply #16 on: Jun 17th, 2008, 2:15am » |
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The modified monty hall is formulated wrongly. By these rules the moderator have no choice in fourth step with probability 3/8. ... so as mentioned by jollytall ... without the knowledge of moderators strategy you cannot compute the win probabilities (in range from 2/5 to 1 for switching).
If by the rules he will open 2nd doors even in the case you don't swith, you should prefere not to swith before the last door opening.
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towr
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Re: probability challenging prob.
« Reply #17 on: Jun 17th, 2008, 2:52am » |
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How about these four versions for the modified monty hall with N doors
1) The host must uniformly randomly open a (no-prize) door amongst the closed doors, excluding your current pick. (And of course the game ends when the host has no door left to open.)
2) The host must uniformly randomly open a (no-prize) door amongst the doors that have yet never been selected (either in the current or previous rounds)
3) Like 2, but you also cannot switch back to a door you chose previously.
4) The host has to open a closed (no-prize) door until you commit to a choice (or there are no other doors left), but he tries to minimize your chance of winning.
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| « Last Edit: Jun 17th, 2008, 2:53am by towr » |
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jollytall
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Re: probability challenging prob.
« Reply #18 on: Jun 17th, 2008, 5:29am » |
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2) Which one do you get, the last chosen, or do you have still a chance to change when he announces that he cannot legally open a window? (you choose 1, he opens 4, you choose 2, he announces - It is behind 3).
1) As Hippo says, wait until only 2 windows left. Change then. Although if it is allowed to change once he announces that although there is a closed door, but he is not allowed to open it, you can improve further. Choose and stick to it until either he opens all, or he announces that he cannot legally open the other.
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towr
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Re: probability challenging prob.
« Reply #19 on: Jun 17th, 2008, 5:58am » |
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on Jun 17th, 2008, 5:29am, jollytall wrote:| 2) Which one do you get, the last chosen, or do you have still a chance to change when he announces that he cannot legally open a window? (you choose 1, he opens 4, you choose 2, he announces - It is behind 3). |
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Well, typically you get to switch after the host opened another door. So if he can't open another one for whatever reason, you don't get to switch either.
Actually, to make any of the 4 cases interesting, we really need to encourage switching in some way; for instance include the rule that you either switch to a another/new door, or the game ends and you get what's behind your door.
Because otherwise you just wait till (N-2) doors have been opened and switch then.
I really should have put a bit more thought into it.
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Benny
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Re: probability challenging prob.
« Reply #20 on: Jul 3rd, 2008, 8:05pm » |
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Taking a break from theoretical physics, i propose the following fun probability problem:
Suppose you are traveling alone (let's say on a business trip) in coach, or any other flight class that deals in rows of three seats per aisle side.
When buying your ticket, you are presented with a choice of any seat you want. We will assume that you are the first person to get to select your seat - that is, you can have your choice of any seat in coach.
You want to maximize the probability of ending up with an empty seat next to you. We'll assume the flight is near, but not at, capacity.
Which seat do you choose?
Of course, there are other "quality-of-flight" factors we can ignore that mess up a perfectly good plan, such as:
- the crying baby factor
- the little-kid-kicking-the-back-of-your-seat factor
- the incessant talker factor
- the fat-guy-that-ought-to-take-two-seats factor
- whatever-else-you-can-think-of factor
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Re: probability challenging prob.
« Reply #21 on: Jul 4th, 2008, 12:18am » |
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on Jul 3rd, 2008, 8:05pm, BenVitale wrote:| Taking a break from theoretical physics, i propose the following fun probability problem: |
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It's not really a probability problem if there isn't any data about the probabilities.
You may as well start a rumour that you're a serial killer and let people know what seat you're in. Then they'll 'probably' avoid you.
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Benny
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Re: probability challenging prob.
« Reply #22 on: Jul 4th, 2008, 1:08am » |
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on Jul 4th, 2008, 12:18am, towr wrote:
It's not really a probability problem if there isn't any data about the probabilities.
You may as well start a rumour that you're a serial killer and let people know what seat you're in. Then they'll 'probably' avoid you.
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Couldn't we determine the seat to choose in order to maximize the probability of ending up with an empty seat next to you?
How about choosing the least desirable seat on the plane -- a seat that no one would choose if other options are available? So, in my opinion, the worst seat on a commercial flight is the last row at the rear of the aircraft. Those seats will have the highest probability of being empty.
Or how about outside aisle seat ?
no dude likes to sit in middle so next guy that comes will sit in the window (many prefer views and sights of journey)
Now to the middle seat when a person buys ticket and is previously aware of situation then based on human pyschology they have tendencey not to sit beside 2 people for lack of freedom and peace; it is simple: most people don't like sitting crowd with 2 person beside them in a big place like a plane or even coach; yet if you choose window seat then couples of 2 are most likely to sit in the next 2 seats beside you the difference here and choosing aisle seat is that if you sit on outside seat couples would not likely be pleased to be "trap" between you and vehicle walls, sitting on outside also shows hints of domination and power
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Re: probability challenging prob.
« Reply #23 on: Jul 4th, 2008, 1:24am » |
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on Jul 4th, 2008, 1:08am, BenVitale wrote:| Couldn't we determine the seat to choose in order to maximize the probability of ending up with an empty seat next to you? |
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If we had any relevant data on the subject; but we don't. Or I don't, in any case.
Quote:| How about choosing the least desirable seat on the plane -- a seat that no one would choose if other options are available? |
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And which would that be?
(and of course you should pick the seat next to it, not the seat itself)
And what of the motivations of the other people? Do they prefer to sit next to someone, or does everyone want an empty seat next to them.
Quote:| So, in my opinion, the worst seat on a commercial flight is the last row at the rear of the aircraft. Those seats will have the highest probability of being empty. |
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Well, at least no I know where to sit in a plane if I want to avoid you  But what about other people?
Quote:Or how about outside aisle seat ?
no dude likes to sit in middle so next guy that comes will sit in the window (many prefer views and sights of journey) |
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But if you sit in the middle, you have two chances of having an empty seat next to you. What's the chance of someone willing to shuffle past the guy in the middle seat to take a window seat? If there's other less obstructed seats available they may fill up first.
Of course there's a disturbing lack of any statistical data to actual base any probabilistic judgment on.
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ThudnBlunder
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Re: probability challenging prob.
« Reply #24 on: Jul 4th, 2008, 4:51am » |
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on Jul 4th, 2008, 1:24am, towr wrote:
Of course there's a disturbing lack of any statistical data to actual base any probabilistic judgment on. |
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Yeah, but for Ben 'no data' just means it's quite Hard. 
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| « Last Edit: Jul 4th, 2008, 7:42am by ThudnBlunder » |
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