seanie

I'm an occassional gambler.

And I usually lose.

Principia

I understand, seanie. It is a serious problem as one of the real world and rather insidious applications of mathematics -- to confuse the intuition most people have on the concept of uncertainty and probability.

seanie

Scientiae[retired]

As I understand Parkladian (bear with me here) everything hinges on the role of my friend. What he sees or is allowed to impart to me.

There is no intrinsic difference between him handing me a dice or telling me about it.

That's not the crux.

It's what he's seeing or rather what information he is allowed to use.

A) He can't tell me anything.

I've got a 1 in 36 chance of winning if I bet on a double 6.

B) He gets to see both dice.

Anytime he sees a 6 in the combination he must tell me there is one or he hands one to me. Either way it doesn't matter. If there is a double 6 he must still only tell me about or hand me one 6. Now we've improved my odds of a succesfull bet. He's narrowed it down from 1 in 36 to 1 in 11. Because he's looking at both dice for a 6, 11 out of 36 times a 6 will appear. This will trigger him which will then trigger me to bet.

Of those bets I'd expect to win 1 in 11 times.

C) He only get's to see the result of one of the dice.

Whether it's the same dice over and over or just randomly selected I don't think it matters. When a six occurs he tells or hands a 6 to me. This reduces the number of times I'll bet to 1 in 6. Because he's not looking at both dice for a 6 (11 possibilites out of 36) but only one (and with 1 dice you'd expect a 6 1 in 6 times).
This means I'll bet less. Betting less is good.

When I do bet I've got a 1 in 6 chance of getting a double 6.

seanie

Just to clarify.

He will always see a 6-6 and trigger me to bet.

But because he only gets to see one dice he won't spot half of the other combinations where a 6 occurs.

If he's only looking at dice A he'll spot

6-1
6-2
6-3
6-4
6-5

But miss

1-6
2-6
3-6
4-6
5-6

So he'll trigger me to bet 6 times out of 36 And when I do I'll have a 1 in 6 chance of being right.

seanie

parkdalian

Sorry I spelt your name wrong earlier.

Principia

I think you got it in the next to last post!

Congratulations.

Now, try some of the other problems, like the boy-girl one on the top of the first page. And check your reasoning again.

Good luck.

seanie

I have to go lie down now.

Ratatosk

Although not necessary, I decided to simulate this for a few trials (which may or may not be related to the procrastination of one abstract algebra assignment) and got the following data:

Sum of hits: 195196292 Total trials: 2147483647 Hit rate : 0.09089535665274381 Inverse : 11.001662096122194

possibly errors coming from the fact the numbers generated by a computer are not truly random

It is also likey the last time I will undergo 2147483647 (2^31 - 1) trials (bleh). My poor old computer took far too long to finish that.

Synaesthesia

I just realized that Monty <img src="graemlins/banghead.gif" border="0" alt="[Bang Head]" /> Hall <img src="graemlins/banghead.gif" border="0" alt="[Bang Head]" /> is <img src="graemlins/banghead.gif" border="0" alt="[Bang Head]" /> Satan.

Of course, when they open the other doors (n-2) they take the sample that you did not select and narrow the range to only those that could contain it.

In the initial selection, there was no restriction(1/n) and so the probability is different ((n-1)/n). The math clears it up quite a bit.

Cool cool.

Beausoleil, thanks for the insanity.

Seanie, Have you seen the Monty hall link yet?

Where's my asprin?