Puzzles with numbers and things

[elvaago]

I completely impartially submit the notion I should get a point for this. :-)

[davis_greatest]

Here's another (very similar) way! We look instead at the chance that Charlie wins

P(Charlie wins)

= P(Charlie pots) + P(Charlie misses).P(Oliver misses).P(Charlie wins)

[because if they both miss, we are back to where we started]

= 2/3 + 1/3 x 2/3 x P(Charlie wins)

= 2/3 + 2/9 x P(Charlie wins)

So 7/9 x P(Charlie wins) = 2/3

So P(Charlie wins) = 2/3 x 9/7 = 6/7

If Charlie has a 6/7 chance of winning, then Oliver must have a 1/7 chance of winning (because someone must win eventually, so the chances must add up to 1).

[davis_greatest]

At last!

1/7 ????????

Yes! There are many ways to solve this. If you follow the method from the hint I gave, you would get:

Prob(Oliver wins if Charlie starts)

= Prob(Oliver wins on his first shot if Charlie starts) + Prob(Oliver wins on a shot later than his first if Charlie starts)

= Prob(Charlie misses and then Oliver pots) + Prob(Charlie and Oliver both miss and then Oliver wins if Charlie starts)

= Prob(Charlie misses) . Prob(Oliver pots) + Prob(Charlie misses) .Prob(Oliver misses) . Prob(Oliver wins if Charlie starts)

= 1/3 x 1/3 + 1/3 x 2/3 x Prob(Oliver wins if Charlie starts)

= 1/9 + 2/9 x Prob(Oliver wins if Charlie starts)


So 7/9 x Prob(Oliver wins if Charlie starts) = 1/9

So Prob(Oliver wins if Charlie starts) = 1/7


There were so many guesses that I think I'll give half a point to snookersfun and half a point to elvaago and a point to davis_greatest!

[davis_greatest]

you are adding them all up, so:
1/9+1/3*2/3*1/9 (incidentally =11/81, of my random fraction generator)
+1/3*2/3*1/3*2/3*1/9+ ...

now these start to be really neglible: (n-1)(1/3 * 2/3) * 1/9

so, when can one stop adding??

You can never stop adding! If you do it that way, it is a sum of an infinite number of terms - but it has a finite answer!

... which you have given in your subsequent post!

[elvaago]

OK, I'm ending up at 2/7th. That's my final final answer!

[snookersfun]

1/7 ????????

[snookersfun]

you are adding them all up, so:
1/9+1/3*2/3*1/9 (incidentally =11/81, of my random fraction generator)
+1/3*2/3*1/3*2/3*1/9+ ...

now these start to be really neglible: (n-1)(1/3 * 2/3) * 1/9

so, when can one stop adding??

[davis_greatest]

The bigger you make n, the smaller the chance gets that Oliver wins.

Is it 0?

Hehe... No. If it were 0, that would mean that Oliver had no chance whatsoever. He must have SOME chance, because Charlie sometimes misses and Oliver sometimes pots.

[elvaago]

The bigger you make n, the smaller the chance gets that Oliver wins.

Is it 0?

[davis_greatest]

Ha! Take another one then 11/81

No, but I admire your random fraction generator.

[davis_greatest]

Yes, it should be (1/3 * 2/3) to the (n-1)th power * 1/9.
Or 2/9 to the (n-1)th power * 1/9.

I don't know how to give you just one number without using n where n represents number of attempts! Doesn't the chance change the more often they play?

Well, they only play once... they just keep going until someone wins! So there must be a chance that someone wins eventually (without using "n")

[elvaago]

I would say 'Great minds think alike' but I wouldn't claim my mind to be as great as yours, sir!

[davis_greatest]

Sorry, elvaago, I may have misunderstood what you meant by

"(n-1)(1/3 * 2/3) * 1/9"

Do you mean

(1/3 * 2/3)^(n-1) * 1/9 ?

Because if so, that's better!


EDIT - oh, you just answered this in the post above, at the same time as I was writing that!

[elvaago]

Yes, it should be (1/3 * 2/3) to the (n-1)th power * 1/9.
Or 2/9 to the (n-1)th power * 1/9.

I don't know how to give you just one number without using n where n represents number of attempts! Doesn't the chance change the more often they play?

[snookersfun]

No! But you're close... in that it is a fraction.

Ha! Take another one then 11/81