Probability?

Tetsuo_Shima

Tetsuo_Shima

In Cryo Sleep
Not technically a puzzle post, just trying to clear something up that's been preying on my mind for days :)

One of my mates sprung this on me the other night -

If you flip a coin three times in a row, what are the odds that heads will come up at least once? (assuming a complete 50/50 chance each flip, none of your '2p coin heads weigh more than tails' or 'act of god' rubbish :p)

At first I answered '150%!' before realising the error of my ways. Finally, I came up with the solution of 87.5%. The first flip is 50%, would the second and third flips cut the difference (compared to 100) in half each time? i.e. The second flip would be 50+[(100-50)/2] = 75% chance that heads will come up at least once in 2 flips. This would make sense in that each consecutive flip brings the chances closer to absolute certainty without ever reaching 100.

It kinda sounds right, but I think there might be some broke logic in there somewhere.

First person to correctly answer will pay a visit to the reputation station, if my original solution is indeed correct then you all have to give me rep :D Jokes, lads, jokes.
 
thatbloke

thatbloke

Junior Administrator
You could think of this another way:

Think of it as the probability that 3 Tails in a row will NOT happen, i.e. 1 - the probability of 3 tails in a row. That probability is 0.5*0.5*0.5, which is 0.125, and thereby gives a result of 0.875 (87.5%) when subtracted from 1, which is indeed the result that you obtained :D
 
Dragon

Dragon

Well-Known Member
Both solutions should be correct according to mathematic rules ... but however I do doubt that you have such a coin :P ... and furthermore ... what about the chances the coin has for landing on its side? ... So actually (if you really have that "ideal coin") you would have to lower the chance for heads in one throw down to 33,3% ;)
 
Tetsuo_Shima

Tetsuo_Shima

In Cryo Sleep
Dragon said:
Both solutions should be correct according to mathematic rules ... but however I do doubt that you have such a coin :P ... and furthermore ... what about the chances the coin has for landing on its side? ... So actually (if you really have that "ideal coin") you would have to lower the chance for heads in one throw down to 33,3% ;)
Click to expand...

Tetsuo_Shima said:
(assuming a complete 50/50 chance each flip, none of your '2p coin heads weigh more than tails' or 'act of god' rubbish :P)
Click to expand...

Now now, young Dragon. What did I mention earlier? :p If we took the coin landing on its side into consideration, I should hardly think that the chance for heads in one throw will be 33.3%, probably more along the lines of 49.99997% or something. I guess even then it depends on the method of flipping, where if you tend to flip the coin and catch it face-down on the back of your hand the chance would indeed be 50%. If you flip and let the coin land on the ground then it would depend how soft the ground is, I suppose. Let's not overcomplicate matters, shall we? ;)

EDIT: Although, some really really weird things happened to me recently, like I attempted to throw a teabag into my (recently deceased :() teacup and the teabag actually landed on its side (ie. on the edge, not flat) on the rim of the teacup. It was wonderous.
 
BiG D

BiG D

Administrator
Staff member
Since we got that figured out, here's one of my favourites. :D
---------------------------------
You're a guest on the set of Monty Hall's game show Let's Make a Deal. You are presented with three doors. Behind one of these doors is glorious riches, the other two contain nothing.

You pick door #3. Instead of opening that door, Monty opens door #1 and shows you that it is empty.

"Let's make a deal!" he says. "Now that you know that door one is empty, I'll give you the option of sticking with door three or switching to door two!"

So. What do you do? What is the probability of winning if you switch? Think it through, glorious riches are at stake!
 
Dragon

Dragon

Well-Known Member
I'd stick ... I don't exactly remember how we solved that problem, but we did it in one of our math lessons ^^
 
DocBot

DocBot

Administrator
Staff member
CBA to do the actual math, but switching gives you a 50% chance, as opposed to the 33% you have if you stay where you are. Really weird but proven true iirc
 
thatbloke

thatbloke

Junior Administrator
Yea I remember this one... Can't quite remember how to prove it but my stats teacher in college did this one day:

Basically like Docbot, it can be proved (though I cannot quite remember how) that there is more chance of you winning if you switch as opposed to sticking.
 
BiG D

BiG D

Administrator
Staff member
Yes well, it's easy enough to assume that, but this is math! I want numbers!
 
waterproofbob

waterproofbob

Junior Administrator
switching is more likely to win. I'll try to explain best i can. I may also add in the probability equations as well if i can be arsed.
switching gives you a 2/3 chance of winning.

It relies on the fact you have 3 starting choices: I'll assume the win is in #1.

1. you choose #1 :#3 is opened. for this if you stick you win, switch you lose.

2. choose #2 :#3 is opened. sticking loses, switching wins.

3. choose #3 : #2 is opened. sticking loses, switching wins.

Thus you have a 2/3 chance of winning if you switch.

Anyone wants the probabilty equations can pm me and i'll give them to you.
 
BiG D

BiG D

Administrator
Staff member
That's it. :D

For some reason I found it easier to understand this way:
When you choose the door, there is a 2/3 chance the prize is behind one of the other two doors. If one of those doors is opened, you can see it has a 0 chance, so that 2/3 chance must apply to the third door now.

If it's still not making sense, imagine it with a bigger number of doors. 100 works well.

When you choose a door, there was a 99% chance you guessed wrong. If Monty then opens 98 doors, there's still a 99% chance you were wrong, and this means a 99% chance the other door is the winner. :)
 
Tetsuo_Shima

Tetsuo_Shima

In Cryo Sleep
My mate tried this one on me as well, at the same time as he said the coin toss one :D Except that he explained using three cups and a ball rather than 3 doors and glorious riches. It's a little hard to get your mind round (we were arguing in the car for half an hour), but now I feel enlightened :)

Easy one:

Imagine you are somehow forced into playing Russian Roulette with a six-shooter revolver, one chamber is loaded. The chambers are spun and the weapon is handed to you. You grit your teeth and fire - *click* - empty chamber. What is the best thing to do to increase your chance of survival? Fire again, or respin the chambers and then fire again?

Easy indeed. To give yourself the best chance, the idea is to spin the chambers again before firing. The reasoning is that once you have fired and found an empty chamber, to get that empty chamber back into circulation you must spin the chambers again. If not, you decrease your chances to one-in-five.
 
E

elDiablo

Guest
That was indeedy easy Tets, but thanks for the nice maths problems :)

And the Monty Hall problem was given to us in our AI lecture (I think), hence how tb knew the answer :)
 
E

elDiablo

Guest
We got it in some lecture, yes. Though, then again, you didn't go to many did you? :D
 
You must log in or register to reply here.
Top