to infinity and beyond...
...well from 10.0 to 9.4 this year would be nice.
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.
2016 handicap 15.16.15.14.15.14.132017 handicap 13.12.13.12 https://www.justgiving.com/fundraising/martyn-vaughan3
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.
Tinkety tonk old fruit, and down with the Nazis.
2016 handicap 15.16.15.14.15.14.132017 handicap 13.12.13.12 https://www.justgiving.com/fundraising/martyn-vaughan3
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.
Last edited by Beezerk; 31-Aug-2017 at 11:40.
2016 handicap 15.16.15.14.15.14.132017 handicap 13.12.13.12 https://www.justgiving.com/fundraising/martyn-vaughan3
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.
I discuss forum threads, therefore I am
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.
No, he's right. http://puzzles.nigelcoldwell.co.uk/fortyfive.htm
Tinkety tonk old fruit, and down with the Nazis.
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.
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