21 Apr 08
Originally posted by kbaumenI see the other point that has been made. If you only have two cards: g/s and s/s, and you pull out one and show the silver side of it, what are the chances that the opposing side is also silver?
If it's poorly worded, then how come so many people understand it clearly?
It's 50/50.
Because showing the silver side is not a random event, but choosing one card or the other is.
21 Apr 08
Originally posted by forkedknightBy probability problem convention, anytime a random draw is made the results are considered equiprobable unless additional information is supplied to the contrary. How the phrase "which happens to be silver" can be interpreted to mean "chosen to be silver on purpose" is beyond me.
I see the other point that has been made. If you only have two cards: g/s and s/s, and you pull out one and show the silver side of it, what are the chances that the opposing side is also silver?
It's 50/50.
Because showing the silver side is not a random event, but choosing one card or the other is.
23 Apr 08
Originally posted by geepamoogle1) 50%
Pretty close to 100%, if I applied Bayes formula right..
Anyways, (moving on,) here's another pair of probability problems.
Suppose a family has two children. The children are not twins, and each child is either a boy or a girl, and equally likely to be either.
What are the chances the family has a girl if...
[b]1) the oldest is a boy?
2) they have at least one boy?[/b]
2) 50%
23 Apr 08
1 edit
Originally posted by LemonJelloIf you don't spin, then the chance of the bullet is 25%. If you do spin, it's 33.333...%. Don't spin.
Nah, don't pretend. We all know you got swindled.
Anyway, moving on, here's another problem for you to keep the thread going.
I have a six-chambered revolver in which I have loaded two bullets into successive chambers. The other four chambers are empty. I spin the chamber, then I point the gun at your head and pull the trigger. No shot fires ...[text shortened]... her or not the chamber is spun before I pull the trigger a third time. What is your decision?
In the second case it's 33.333...% either way I think, so it doesn't matter.
23 Apr 08
1 edit
Originally posted by LemonJello1) Position can be XX000, 0XX00, 00XX0, or 000XX. 25% chance of getting shot, versus 33%.
I have a six-chambered revolver in which I have loaded two bullets into successive chambers. The other four chambers are empty. I spin the chamber, then I point the gun at your head and pull the trigger. No shot fires. I tell you that I am going to pull the trigger again, but I give you the option of whether the chamber is spun or not beforehand. Wha ...[text shortened]... ether or not the chamber is spun before I pull the trigger a third time. What is your decision?
Don't spin.
2) Position can be XX00, 0XX0, or 00XX. 33% chance of getting shot. Spin or don't spin. It's
all the same.
It seems odd, though. that the initial odds of getting shot are 33%, and after two shots,
the odds are the same. One would intuitively think they would get worse.
I don't know how to show this mathematically, just graphically.
Nemesio
23 Apr 08
Originally posted by Nemesioyeah it's a cool problem - the interesting fact is that the event "not already being dead" makes your next trigger pull pretty safe (because of the consecutive placement of the bullets). even though there are two bullets in the gun, only one of them comes directly after an empty chamber, so 3/4 of the original empty chambers are safe, whereas only one of those 4 empty chambers is friends with dick cheney.
1) Position can be XX000, 0XX00, 00XX0, or 000XX. 25% chance of getting shot, versus 33%.
Don't spin.
2) Position can be XX00, 0XX0, or 00XX. 33% chance of getting shot. Spin or don't spin. It's
all the same.
It seems odd, though. that the initial odds of getting shot are 33%, and after two shots,
the odds are the same. One would intuitively t ...[text shortened]... ld get worse.
I don't know how to show this mathematically, just graphically.
Nemesio
that's another way of looking at the 25%. you discount that you spun and chose a bullet, because clearly your face is still connected to your head, otherwise we wouldn't be thinking about another pull of the trigger. and so you must have chosen one of the four empty chambers, and like i said, only one of those four has a bullet coming after it because of your consecutive bullet placement
25 Apr 08
New gambling problem! Simple, straightforward probability calculations.
Two players, A and B, are playing a game in which they take turns throwing a pair of dice. Player A wins if he rolls a 6, and player B wins if she rolls a 7. If neither player hits their winning point, the game continues until there's a winner.
What is the probability that A wins this game if:
(a) B goes first?
(b) A goes first?
25 Apr 08
Originally posted by PBE6I arrive at the following figures.
New gambling problem! Simple, straightforward probability calculations.
Two players, A and B, are playing a game in which they take turns throwing a pair of dice. Player A wins if he rolls a 6, and player B wins if she rolls a 7. If neither player hits their winning point, the game continues until there's a winner.
What is the probability that A wins this game if:
(a) B goes first?
(b) A goes first?
1) 25/61 if B goes first.
2) 30/61 if A goes first.
25 Apr 08
Originally posted by PBE6The tricky part is eliminating the part of the equation where the game keeps going.
Correct! Told ya it was simple 😉
However, it's easy if you can find a point down the line where you essentially loop back to an identical situation. Since the odds for this second point are the same as at the beginning, then you can ignore the part you have left (the odds the entire thing loops around) and look at the established final results part from the first action to the last one before the whole game repeats itself.
In this case, you have one player cast dice. If they fail, the second player cast dice. If they fail also, then you're back where you started two rolls later.
25 Apr 08
Originally posted by geepamooglecan you show me the calculations.
The tricky part is eliminating the part of the equation where the game keeps going.
However, it's easy if you can find a point down the line where you essentially loop back to an identical situation. Since the odds for this second point are the same as at the beginning, then you can ignore the part you have left (the odds the entire thing loops aroun ...[text shortened]... econd player cast dice. If they fail also, then you're back where you started two rolls later.