Play the Probability Game: Test Your Luck!

[Shackleford]

I'm not sure how to proceed in the last line.

http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled.png?t=1302307209

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110408_185752.jpg?t=1302307222

 

[praharmitra]

Well, firstly

P(win| R_0 = 1) = P(win| R_0 = 2) = P(win| R_0 = 3) = P(win| R_0 = 4).

So you just need to find anyone of them and that's your probability.

Now do it for special cases -

1. What is the probability that he wins in 1 try?
2. 2 tries?
3. 3 tries?
4. 4 ... ?

Do you see a pattern?

The answer is then the infinite sum

P(win) = \sum_{n=1}^{\infty}P(win in n tries)

 

[Shackleford]

Well, I suspected some sort of pattern last night. If you break up one of those tries into "equals 1" and "not equals 1," the "not equals 1" will translate into R₂=R₁, R₂=R₀, and again R₂ not = R₁.

 

[praharmitra]

Well, so did you work it out? Kindly post what you have been able to do.

 

[Shackleford]

Well, so did you work it out? Kindly post what you have been able to do.

I just got home.

Oh, I'm supposed to work out when you win in a certain number of tries. Let me do that in a few minutes.

 

[Dick]

I just got home.

Oh, I'm supposed to work out when you win in a certain number of tries. Let me do that in a few minutes.

This problem is actually fairly tricky unless you are used to Markov chain sorts of problems. It branches depending on whether you win or continue at stages. Took me a few false tries to get it right. But praharmitra is correct. Doing some special cases of n will really help to see the general pattern.

 

[Shackleford]

I have heard of Markov chains, but I have no idea what they are nor have I encountered them in coursework.

This is probably not correct.

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110409_224337.jpg?t=1302407152

 

[Dick]

I have heard of Markov chains, but I have no idea what they are nor have I encountered them in coursework.

This is probably not correct.

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110409_224337.jpg?t=1302407152

Ok, so what is that blurry thing you posted supposed to tell me? What is the probability of winning at say n=1, n=2, n=3 and n=4? Solving the whole question may be a little subtle, but those special cases are easy. Do them. Do something.

 

[Dick]

I have heard of Markov chains, but I have no idea what they are nor have I encountered them in coursework.

This is probably not correct.

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110409_224337.jpg?t=1302407152

It's not only not correct. It has no content whatsoever. Give us some numbers. A Markov chain is a sequence of events where you have a potentially infinitely recurring state. This is one. There may be a more elementary way to solve this but I haven't thought of it. Maybe it you just write out enough special cases of n you can see the pattern.

 

[Shackleford]

It's not only not correct. It has no content whatsoever. Give us some numbers. A Markov chain is a sequence of events where you have a potentially infinitely recurring state. This is one. There may be a more elementary way to solve this but I haven't thought of it.

Well, I looked at a previous problem in the homework. This is the probability of winning in exactly n tries in this game:

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110409_231023.jpg?t=1302408812

 

[Dick]

Stop posting these blurry photobuckets and give me some straight answers. What is the probability of winning at n=1? That shouldn't even require much thought, much less calculation. Now n=2. You might have to think about that for a few seconds. Not much more. If you get that then do n=3. If I see another vague, blurry photobucket shot I'm out of here.

 

[Shackleford]

stop posting these blurry photobuckets and give me some straight answers. What is the probability of winning at n=1? That shouldn't even require much thought, much less calculation. Now n=2. You might have to think about that for a few seconds. Not much more. If you get that then do n=3. If i see another vague, blurry photobucket shot I'm out of here.

Of course, you can look at each roll separately, and it's always 1/4. Right?

 

[Dick]

Of course, you can look at each roll separately, and it's always 1/4. Right?

Wow, you evaded any meaningful answer again. Yes, your probability of getting any given number is 1/4. Now tell me something about how that would translate into a probability of winning the game at any given n. Any n, you pick.

 

[praharmitra]

Let me give you an example for n = 4.

Given that in the zeroth try we got R0. In the first try, we can have only 3 allowed throws with probability 3/4. In the second try, you're not allowed to get either R0 or R1, so you have two allowed throws with probability 2/4. Again third try, you are not allowed to get R0 and R2, so again probability is 2/4. Finally in the 4th try, the only allowed throw is R0, with probability 1/4. So probability for winning in 4 throws is

(3/4)(2/4)(2/4)(1/4).

Now, use the same logic and find it for more numbers (n=1 is a slightly special case, so make sure you do that too)

 

[Shackleford]

Shouldn't it always be 2/4 for every k not equal to 1 or n?

3*(1/4)ⁿ*2^n-2

 

[Shackleford]

I'm off just a bit.