You mean pr(each at least once) or pr(x at least once), for all x?

But after n games, you have a (1/3) ^ (n-1) chance of having always played the same race, and a (2/3) ^ (n-1) chance of having only played two different races. Sum them, and subtract from 1. So it takes 7 games. Right?

Well, for the n = 3 case it's easy enough to enumerate all the ways we can get one of each
TPZ
TZP
PZT
PTZ
ZTP
ZPT

out of 3^3 = 27 total possible outcomes. 6/27 = 2/9
But if I understand your formula, you're suggesting
1 - ( (2/3)^2 + (1/3)^2) = 1 - (4/9 + 1/9) = 4/9
So something is off

F(n) = ( 3^(n-1) - 2^n + 1 ) / 3^(n-1)

It requires 9 tries.

I haven't read and considered all the replies, but some of them seem wrong to me. Here's my take:

If 1 of the 3 options is never selected in n attempts, there are clearly 2^n ways this can happen. That's for 1 specific, but arbitrarily chosen, option. For all 3 of them, we need to multiply by 3, but then subtract the cases we've counted twice. Clearly, the only cases we've counted twice are those where we've continually selected the same option, so there are only 3 of these.

Thus, the probability we want, after n attempts, is 1-[(3*2^n-3)/3^n], or (3^n-3*2^n+3)/3^n. I now notice this is the same as Joaz's formula (divide top and bottom by 3), so I can stop here and refer to his post

0

1/8 (Some sources say 1/24 but I assume they mean probability per matchup)

OK those were trick questions.

http://www.gamethreat.net/forums/starcr ... post546873

Probability of 'at least once' is 1 minus probability of 'never'.

After n games, probability of (for example) never being assigned Z is (2/3)^n
Note that this includes both the probabilities of always being assigned T as well as always being assigned P.

As there are three possibilities of leaving out one race, I initially thought the overall probability was three times the former value, that is 3*(2/3)^n or (2/3)^(n-1).

But after cross-checking with Joaz' results and seeing his were right, I found that simply adding the tree probabilities ends up counting those of being assigned only one race twice:
The probability of never being assigned T contains those for 'only Z' and 'only P' and the one of never being assigned P contains those for 'only Z' and 'only T'.
So one has to subtract three times the probability of being assigned a single race, which is 3*(1/3)^n or (1/3)^(n-1).

Putting it all together you get 1 - ( (2/3)^(n-1) - (1/3)^(n-1) ) which is incidentally identical to Joaz' result.

By the way, even the wrong formula without the last term yields the result n=9 for getting over 90%, apparently because the probability of being assigned only one race falls off rather quickly for large n.

After 3 flips, the two are obviously equal.

The trick is that starting with the fourth flip, some of the patterns that give you HTH are "blocked" by the HHT patterns.

Consider the fourth flip:

(1) HHH 1/2 chance of finishing HHT
(4) HTT
(5) THH 1/2 chance of finishing HHT
(6) THT 1/2 chance of finishing HTH
(7) TTH
(8) TTT

Missing from the list are
(2) HHT
(3) HTH
because you stop after three flips for these two.

I'm not sure how deeply related they are, but this reminds of an even more counterintuitive problem: http://mathoverflow.net/questions/17960 ... -want-boys with answers at http://mathoverflow.net/questions/17960 ... -want-boys. I don't pretend to understand even a fraction of that--the only part that made sense was to consider what happens if the country is just one family.

Edit: You actually have to show that the advantage HHT has after the fourth flip will persist. It seems obvious, but I don't have a good argument for it.

P.S. Oh, the question was about Starcraft. It's PvT, PvT, PvZ is more likely?

Isn't it fairly obvious? If on average a coin comes up heads 50% of the time, then after it comes up heads on one flip, it needs to come up heads less than fifty percent of time on subsequent flips in order for the asserted probability to hold true. If you doubt me, flip a coin five or six times to test this.