talkmagic.co.uk

Having a nice debate on the JREF forum (Iknow I shouldn't, but they're so easy to wind up). A poster on the forum claims that doing a test with a one in a thousand chance followed by another test with a chance of one in a thousand means that the overall test has a one in a million chance. I have suggested that this is wrong and that it is in fact similar to the coin toss thing whereby no matter how often one tosses a coin and gets heads, the odds on the next toss being heads again is still 50/50.

Am I wrong? probably, but thought I'd ask the learned folk in here for their opinion.

As far as I'm aware, the odds under those conditions would reset before each attempt and the previous result is irrelevant - just like flipping the coin.

The odds of getting both right is one in a million, in the same way that getting two consecutive coin tosses right is a one in four chance.

The odds of getting the second right having already got the first right is one in a thousand.

themagicwand wrote:Having a nice debate on the JREF forum (Iknow I shouldn't, but they're so easy to wind up). A poster on the forum claims that doing a test with a one in a thousand chance followed by another test with a chance of one in a thousand means that the overall test has a one in a million chance. I have suggested that this is wrong and that it is in fact similar to the coin toss thing whereby no matter how often one tosses a coin and gets heads, the odds on the next toss being heads again is still 50/50.

Am I wrong? probably, but thought I'd ask the learned folk in here for their opinion.

The other bloke's right when the overall outcome is predicated on the outcomes of a sequence of events.

Hello!

I'll go back to the coins here, test odds = 1/2 chance of a head right?

so, one test then another, 1/2*1/2 = 1/4. The odds of the second test are 1/2 (yes) but the chance of both occuring is 1/4.

[repaly the 2's with 1000's in the above (and the 4 by 1000000 of course) as per your test]


...Now, you're on a game show and there's these 3 doors...

Lawrence wrote:...Now, you're on a game show and there's these 3 doors...

Uncle Monty, is it really you?

Thanks for the replies! I now understand. I think...

Ah, but no! Wait!! If I say I can toss a coin twice and gets heads both times, then I toss the coin and get a head, what are the odds of me getting heads on the second toss?

My brain hurts. It's my own fault. I experimented too much in my twenties.

themagicwand wrote:Ah, but no! Wait!! If I say I can toss a coin twice and gets heads both times, then I toss the coin and get a head, what are the odds of me getting heads on the second toss?

Because it's the second toss, it still means that it's a sequence, therefore it's 1/2 * 1/2 = 1/4.

the odds of you getting heads on the second toss is 1/2.
It's 1/4 if you're only going to make the second toss if the first one was a head (i.e, if it's dependant upon the sequence)

Mr Thompson, I challenge you to a probability off!

Right that's it! I'm doing a basic probability for magicians guide in the next few weeks!!! All this misunderstanding of probability drives me mad!!! The worst was the LHC detractor that the Daily Show parodied brilliantly!!!

Harry Guinness wrote:Right that's it! I'm doing a basic probability for magicians guide in the next few weeks!!! All this misunderstanding of probability drives me mad!!!

Any chance that you could work out the probability that two cards of different chosen values are next to each other in a (properly) shuffled deck of cards, disregarding suits, such as a Jack and a 4 next to each other?

I'm wrecked and I haven't done this out properly but I believe it's 2 in 13. The first one is always successful and there is a 4/52 chance the card below it is the second named card and a 4/52 chance the card above is the second chosen one. That is the chance that at some point in the deck there is any Jack next to any 4. This is rough as the top and bottom cases are not considered but that'll only knock it by a very small bit (tops 2%) so it's not worth the extra hassle calculating!

Can someone confirm?

2/13 is right, yeah.

Anybody who struggles with statistics, I'd recommend the book Struck By Lightning or The Tiger That Isn't. It ironed it out for me.

Fun fact for those who do Out Of This World: assuming we use the "Schrodinger's Card" model where every card's odds of being either red or black is 50/50, the chances of a spectator dealing a shuffled (hmm...) deck into two piles at random and accurately separating them into their respective colours with no mistakes is 4.5035996274 x10^15. That's 4,503,599,627,400,000 (4.5 quadrillion) to 1 . The population reference bureau estimates there have been 106 billion people born since the dawn of Homo sapiens as a species, which means that even if - presuming the first people on earth had access to playing cards - every single person who had ever lived had each taken 40 million shots at doing it, statistically speaking nobody in the history of the known universe would have ever done it.

If using probability I can predict the future, does that make me closed-eye or open-eye?

I'm thinking partly of Asimov's Psychohistory (which is I understand a work of fiction ...) and partly of a superstition quiz in today's Guardian (http://www.guardian.co.uk/science/2009/ ... erstitions) where the first question was

"It is probably true that certain people can predict the future quite accurately."

Now I can predict the future using probability (ish) and I can certainly predict the future when the worlds of chocolate and my children collide. But I couldn't predict random events.

So, is it true that I can predict the future? And if I can, is that closed-eye or open-eye?

I think I need to lie down now doctor, but I'm not going to reveal what my score was.