UNSW competition senior division (1 Viewer)

[seanieg89]

Ohh I see, is this similar to what you had in mind:











Add them all side by side, take limit n to infinity

a+b+c+d+e = 1

1 = \frac{7}{2} L

????

------

Also for problem 4, can you explain reasoning for part (a)?

From what I can see, the fact that the coin is tossed with 3 heads, is not dependent on the probability that he picked the biased coin?

Thanks for the help

Yep. Your notation is a little different to mine but the answer is correct unless I have messed up.

4a) Either we have {biased coin & 3 consec heads} or {unbiased coin & 3 consec heads}.

These have probability (1/2)*(1) and (1/2)*(1/8) respectively.

So the odds that the first happens given that one of them happens is: (1/2)/(1/2+1/16).

 

[seanieg89]

I just thought of something, since there's a 50/50 chance of him going back to town D, then wouldn't it just cycle (long term) as so:

A > B > C > D > E > A > B > C > D > E > D > E

meaning the answer would be

Nah this isn't valid. Interesting that it is so close numerically though.

 

[RealiseNothing]

Nah this isn't valid. Interesting that it is so close numerically though.

Yep I thought so, I remember getting something like 2/7 so Sy is correct. But I'm curious as to where the logic goes wrong for this method. Over the long term, I would have thought that the man's trip follows the above cycle (not over short term, but surely over long term?) since there's a 50/50 chance of E > A and 50/50 chance of E > D.

Maybe it gets skewed a little as the cycle might not be "complete" when you visit the town he is in (i.e. you might visit when he is in town B and so has not finished the cycle). And this skewing is what makes the 2 answers so close in numerical value?

 

[Sy123]

4)a) If you chose the 2 headed coin, the probability of flipping 3 heads is 1. If you chose the fair coin, the probability of flipping 3 heads is 1/8.

So since you did flip 3 heads, it is 8 times more likely that it is the 2 headed coin (1 vs 1/8). Hence the probabilities must be 8/9 and 1/9.

Yep. Your notation is a little different to mine but the answer is correct unless I have messed up.

4a) Either we have {biased coin & 3 consec heads} or {unbiased coin & 3 consec heads}.

These have probability (1/2)*(1) and (1/2)*(1/8) respectively.

So the odds that the first happens given that one of them happens is: (1/2)/(1/2+1/16).

Thank you both, I think I get it

 

[seanieg89]

Yep I thought so, I remember getting something like 2/7 so Sy is correct. But I'm curious as to where the logic goes wrong for this method. Over the long term, I would have thought that the man's trip follows the above cycle (not over short term, but surely over long term?) since there's a 50/50 chance of E > A and 50/50 chance of E > D.

Maybe it gets skewed a little as the cycle might not be "complete" when you visit the town he is in (i.e. you might visit when he is in town B and so has not finished the cycle). And this skewing is what makes the 2 answers so close in numerical value?

Don't know. It's just a very vague and non-rigorous thing to talk about "long-term cycles" as opposed to the limit of a concrete sequence of probabilities. If you tried to write up a rigorous solution based on this heuristic it would probably be a lot more clear as to what's going wrong.