Tashyboy
Please don’t ask to see my tatts 👍
So the more one putts you have on the trot lessens the chance of you one putting on the next hole. Eh.
Maths / Probability Puzzle
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Blue in Munich
Crocked Professional Yeti Impersonator
You have 3 doors and 1 prize; a 1 in 3 chance. Or a 10 in 30 chance if you prefer to look at it that way. Now the statistical proof from the article
The switchers are normally roughly twice as successful. Last time we had 60 pairs in 30 of which the contestants were always stickers and in the other 30 pairs always switchers:
- Among the 30 switcher contestants, the Cadillac was won 18 times out of 30 - a strike rate of 60%
- Among the 30 sticker contestants, there were 11 successes out of 30, a strike rate of about 36%
So this demonstrates that 11 out of 30 chose the prize right the first time, or roughly 1 in 3, and that 18 out of 30 won it by swapping; but surely this means that in that group only 12 out of 30 chose the prize first time, or roughly 1 in 3……...
It's all well & good saying that over 30 attempts you improve your chances of winning by switching if you've got 30 attempts, but you haven't; you've got one chance.
Anyone starting to see where the phrase "lies, damned lies & statistics" came from?
SwingsitlikeHogan
Major Champion
Yes I did...error well spotted.
Slime
Tour Winner
Phew, for a moment there, I thought I was missing something :thup:.
Ethan
Money List Winner
Assuming the prizes were placed in position before you made your choice, then changing your choice now that you know one of the other doors has one of the two booby prizes makes no difference. The error is to think that switching your choice improves your odds. It doesn't. Your odds have changed with the information that you have dodged one of the booby prizes. The prize is equally likely to be behind either the door you chose or the other unopened one.
In the original version of this question, you do not know that another door is opened to reveal a booby prize.
In the original version of this question, you do not know that another door is opened to reveal a booby prize.
Ethan
Money List Winner
There is a variation on this, using different statistical rules. You meet a person with 2 children. They tell you one is a boy. How likely is it the other is also a boy? Assume there is no biological factor in either parent favouring having a child of a given gender.
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CheltenhamHacker
Journeyman Pro
Must be 1/3?
Logic being there is a 1/4 chance of each combo (b/b, b/g, g/g, g/b)
We've got the first part, which leaves the remaining 3. Only one of these has the other sibling being a boy.
The only issue I spot with my working is if you can use b/g and g/b as two different options
Ethan
Money List Winner
Exactly right.
I see the same logic applying to the car/goat question (known as The Monty Hall problem)
car/goat/goat
goat/car/goat
goat/goat/car
We know it isn't the last one, so it is a toss up between the others and assuming no skullduggery, no benefit in switching.
I see the same logic applying to the car/goat question (known as The Monty Hall problem)
car/goat/goat
goat/car/goat
goat/goat/car
We know it isn't the last one, so it is a toss up between the others and assuming no skullduggery, no benefit in switching.
Hacker Khan
Yurt Dwelling, Yoghurt Knitter
You are looking at the odds on a one off guess. The statistics look at the increased probability of success if you change over many attempts once you are told that the goat is not in one of the ones you have not chosen.
Also not sure what you mean when you say 'we know it is not the last one'?
Foxholer
Blackballed
It's much easier to work out the probabilities and the reason for the uplift in probability by using the (highlighted) possibilities Ethan has quoted!
Initially, the chance of selecting the car is 1 in 3 (33.3%). I don't believe anyone has an issue with that - and the probability doesn't change if a goat is revealed and the player 'sticks' to the original choice!
However, lets look at the 3 options and what the host must do after the player's choice - and how switching affects the result (therefore probability of winning).
Lets assume (s)he chose the first door/screen. It doesn't actually matter which door the player chooses as the options are the same for each player choice - just in a different order.
In line 1 the host can select either of the others to show the goat. By switching, the player loses.
In line 2 or 3, the host cannot show the car, so must select the other door/screen. So by switching, the player wins.
So overall, switching gives a probability of 2/3, while sticking gives a probability of 1/3!
SwingsitlikeHogan
Major Champion
My point was simply that while you have a 50% chance of same birthdays with 23 folks - two lots of 23 folks (46) don't mean that there will be two with same birthday in one or other of the two group. The probabililty remains the same,
Beezerk
Money List Winner
That depends though, I imagine in real life any regular person would take the first door out of the equation and say "right I now have a 50-50 chance".
I don't think many would look at it as having a 2-3 chance although it can be portrayed that way depending on how you do your sums.
It's like finding 3 golf balls in the rough, you pick one up an it isn't yours, you stick it on your pocket (unless it's a Top Flite
) and forget about it and move onto the other 2, it's then 50-50 which is yours.
Hacker Khan
Yurt Dwelling, Yoghurt Knitter
That depends though, I imagine in real life any regular person would take the first door out of the equation and say "right I now have a 50-50 chance".
I don't think many would look at it as having a 2-3 chance although it can be portrayed that way depending on how you do your sums.
It's like finding 3 golf balls in the rough, you pick one up an it isn't yours, you stick it on your pocket (unless it's a Top Flite
Statistically no it does not, which is the whole point. It is not how you do your sums, it is statistical proof.
It is a bit like casinos, all the games you play in there 'against the casino' are statistically weighted so the casino over time has a slight edge in that they will have better slightly odds in winning than punters do. Yes some punters will win big but over time the Casino will make money as the stats are loaded in their favour.
Beezerk
Money List Winner
Over 30 or 60 people or whatever it was, I'm simplifying to a real life situation of a single person playing it on the tv.
Does one person have a better chance of winning by changing on that given day?
ColchesterFC
Journeyman Pro
Statistically, yes. If you accept that when you initially pick you have a 1/3 chance of choosing the correct one and there is a 2/3 chance of it being one of the other two. The host revealing one wrong answer doesn't alter the fact that it's still a 1/3 chance of it being the one you chose and a 2/3 chance of it being the one the host didn't reveal.
Beezerk
Money List Winner
I need a lie down
*edit*
I watched that video in post #14, makes sense now.
Switching is a winner unless you choose the car first time (1/3 chance of choosing car, 2/3 of choosing goat).
Sticking is always a 1/3 chance of choosing car.
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londonlewis
Tour Rookie
Swap. Every time.
Watch 21 (Kevin Spacey film - based on the book 'bringing down the house').
It explains it based on variable change and probabilities. It increases the odds of winning.
Watch 21 (Kevin Spacey film - based on the book 'bringing down the house').
It explains it based on variable change and probabilities. It increases the odds of winning.
Pro Zach
Assistant Pro
This is wrong. This is the same ‘gamblers fallacy’ that I tried to explain in post 34. If having a girl or a boy is a 50/50 chance then that is what it is. The probability of having a boy is 50% and it ALWAYS is. Knowing the person has one boy does not change this. The probability of the other child being a boy is and ALWAYS is 50%.
You can work it out from the known possibilities, as you tried to, but it is unnecessary and more difficult. As you seemed to suspect you have to dismiss either b/g or g/b.
As we don’t know whether the boy was born 1st or 2nd we don’t know which one can’t be true. But one of them can’t be true because the boy must have been born either 1st OR 2nd. So we will be left with b/b, g/b or b/b, b/g. Both of which give a 50% probability of the other child being a boy.
Which is exactly what we already knew because there is ALWAYS a 50% chance of any INDIVIDUAL child being a boy.
Hacker Khan
Yurt Dwelling, Yoghurt Knitter
Pro Zach
Assistant Pro
An interesting link, but is he correct?
He asks the question ‘how does this change if you are told the oldest child is a boy?’
He points out if we know the eldest child is a boy then we are left with two possibilities – b/b or b/g – giving a 50% probability of the other child being a boy.
If he had then asked the question ‘how does this change if you are told the youngest child is a boy?’
He could then point out if we know the youngest child is a boy then we are left with two possibilities – b/b or g/b – giving a 50% probability of the other child being a boy.
IF he then asked himself the question ‘what is the probability of the boy being either the eldest or the youngest child? He may realise that it is a 100% certainty that he is one or the other.
He could then conclude as both possibilities produce a 50% probability of the other child being a boy and one of the possibilities is a certainty then it is in fact a 50% chance of the other child being a boy.
