|
|
|
On February 19 2009 14:50 paper wrote:P14. Return 4 hats randomly back to 4 people (n=4). Let X=# matches. Find: a. P(X=4) b. P(X=3) c. P(X>=1) d. P(X=-) + Show Spoiler [My Solution] +
a. 1/4P4 = 1/24
b. if you have 3 matching, you must also have 4 matching, so same as (a)? is there overlap?
c. 1-(6/24)
d. 6/24
will add more as i continue doing practice problems...^^
+ Show Spoiler [part b] +
Actually, you're on the right track for this one but wrong. As you said if you have 3 matching the 4th must match too therefore P(X=3) = 0. You can never have exactly 3 pairs matching.
|
Wait, what was that picture for?
Funny how I had a blog today with help on probability/statistics. 
I don't think you posted in there to help me though haha x_X
|
nice pic ^^
I thought it was going to be along the lines of "what's the probability of me banging her at the back of the club next time i see her"
|
On February 19 2009 15:13 OmgIRok wrote:Wait, what was that picture for? Funny how I had a blog today with help on probability/statistics. I don't think you posted in there to help me though haha x_X
hahah i actually did your problems on my scratch paper but 93819023 people answered before i had even started on them anyway :s
btw thanks abydos
|
On February 19 2009 15:16 Cambium wrote:
nice pic ^^
I thought it was going to be along the lines of "what's the probability of me banging her at the back of the club next time i see her"
Lol, the pic didn't even load for me the first time I viewed this.
|
Here's a better probability question:
You have a regular fair die and you are allowed three rolls, and you can stop whenever you want. I will give you X dollars at the end where X is the face value of the die. Find the EV.
|
Hey paper! I can't answer ur Questionz but hi!
|
On February 19 2009 15:38 Cambium wrote:
Here's a better probability question:
You have a regular fair die and you are allowed three rolls, and you can stop whenever you want. I will give you X dollars at the end where X is the face value of the die. Find the EV.
doesn't the value a person would generally stop at change the EV for them? oh well
its probably around seventy
|
On February 19 2009 15:17 paper wrote:Show nested quote +On February 19 2009 15:13 OmgIRok wrote:Wait, what was that picture for? Funny how I had a blog today with help on probability/statistics. I don't think you posted in there to help me though haha x_X hahah i actually did your problems on my scratch paper but 93819023 people answered before i had even started on them anyway :s btw thanks abydos
ohoh haha nice i'll refer to you next time i need help with this probability/stats you look like the expert :O
So what was that picture for?
|
On February 19 2009 16:27 OmgIRok wrote:Show nested quote +On February 19 2009 15:17 paper wrote:On February 19 2009 15:13 OmgIRok wrote:Wait, what was that picture for? Funny how I had a blog today with help on probability/statistics. I don't think you posted in there to help me though haha x_X hahah i actually did your problems on my scratch paper but 93819023 people answered before i had even started on them anyway :s btw thanks abydos ohoh haha nice i'll refer to you next time i need help with this probability/stats you look like the expert :O So what was that picture for?
to unexpectedly brighten your day!
|
United States47024 Posts
On February 19 2009 16:24 SpiritoftheTunA wrote:Show nested quote +On February 19 2009 15:38 Cambium wrote:
Here's a better probability question:
You have a regular fair die and you are allowed three rolls, and you can stop whenever you want. I will give you X dollars at the end where X is the face value of the die. Find the EV. doesn't the value a person would generally stop at change the EV for them? oh well its probably around seventy
I assume the problem implies that a person will always make the statistically favorable choice, e.g. stop when the expected value of the remaining rolls is lower than or equal to their current roll.
+ Show Spoiler +
Working backwards:
EV of 1 roll = 7/2
For 2 rolls, keep roll on 4, 5, or 6, re-roll on 1, 2 or 3. EV is therefore:
1/2 * 5 + 1/2 * 7/2 = 17/4
For 3 rolls, keep roll on 5 or 6, re-roll on 1, 2, 3, or 4. EV is therefore:
2/3 * 4.25 + 1/3 * 5.5 = 14/3
|
On February 19 2009 16:40 TheYango wrote:Show nested quote +On February 19 2009 16:24 SpiritoftheTunA wrote:On February 19 2009 15:38 Cambium wrote:
Here's a better probability question:
You have a regular fair die and you are allowed three rolls, and you can stop whenever you want. I will give you X dollars at the end where X is the face value of the die. Find the EV. doesn't the value a person would generally stop at change the EV for them? oh well its probably around seventy I assume the problem implies that a person will always make the statistically favorable choice, e.g. stop when the expected value of the remaining rolls is lower than or equal to their current roll. + Show Spoiler +
Working backwards:
EV of 1 roll = 7/2
For 2 rolls, keep roll on 4, 5, or 6, re-roll on 1, 2 or 3. EV is therefore:
1/2 * 5 + 1/2 * 7/2 = 17/4
For 3 rolls, keep roll on 5 or 6, re-roll on 1, 2, 3, or 4. EV is therefore:
2/3 * 4.25 + 1/3 * 5.5 = 14/3
For those that are interested, this was given to me during a Google interview. I gave the "correct" answer, which, later, after discussing with friends, I realized was actually probably incorrect (I haven't gotten around to work it out yet).
I said exactly what you said...
+ Show Spoiler +
I think that's incorrect because you should stop only when you roll is lower than your future potential EV.
EV of 1 roll = 7/2
So, technically, you should re-roll if you get 1, 2 or 3.
However, if you look at it from the other end, you still have two rolls left, which will yield an EV of 17/4, so I think you should actually stop rolling only if you get 5 or 6 on your first roll.
Okay, I just tried to work it out with what I said in the spoiler, and I just cannot put it to work as I got really confused with forward and backward EV. (but in my defense, it's 3:30 AM), let me know if you figure it out, or if it doesn't make sense at all.
|
United States47024 Posts
On February 19 2009 17:16 Cambium wrote:+ Show Spoiler +
I think that's incorrect because you should stop only when you roll is lower than your future potential EV.
EV of 1 roll = 7/2
So, technically, you should re-roll if you get 1, 2 or 3.
However, if you look at it from the other end, you still have two rolls left, which will yield an EV of 17/4, so I think you should actually stop rolling only if you get 5 or 6 on your first roll.
+ Show Spoiler +
Um, you did the exact same thing I did.
Your first roll keeps 5s or 6s, meaning that the EV of all 1st rolls that are kept is 5.5. If you choose to re-roll, your second roll is kept on 4, 5, or 6, meaning the EV of all 2nd rolls that are kept is 5. If you roll a 3rd time, you have an EV of 3.5.
So the total EV is:
(1/3)*5.5 + (2/3)*(1/2)*5 + (2/3)*(1/2)*3.5 = 14/3
|
|
|
|
|
|
EPT![[image loading]](http://img18.imageshack.us/img18/7448/taonyefg4.jpg)
