# 2015_II_12_ProbStats

i) We note that as soon as T is tossed A cannot win anymore as A needs 2 consecutive H, which would make B win due to the previous T.

Therefore the first 2 throws have to be H, after which A must win, for tossing a T in any subsequent throw will cause the sequence HHT to emerge. This has a probability of \frac{1}{4} of occurring.

ii) As before, once T is tossed A cannot win. So the first 2 throws have to be H for A to win. P(A wins) = \frac{1}{4}

The case for C is analogical:

Once H is tossed C cannot win as C will need two consecutive T, which will make D win, due to the previous H.

Therefore for C to win the first 2 throws have to be T, after which C must win, for tossing a T in any subsequent throw will cause the sequence HHT to emerge.
Therefore P(C wins) =\frac{1}{4}.

If the first 2 throws are HT or TH, which has probability \frac{1}{2} of happening, then only B or D can win. By symmetry, B and D must have the same probability of winning, namely \frac{1}{4}.

Therefore all players have a probability of \frac{1}{4} of winning.

iii) If the first two throws are C, then the probability of C winning is 1 because as soon as H is tossed C wins before D can complete the sequence.

Consider HT is tossed:

• If the next toss is T then C must win as TT is formed and only the last two throws are relevant.
• If the next toss is H then we have the same case as when TH is tossed, again because only the last two throws matter in deciding the winner.

Each of these two events happen with probability \frac{1}{2}
Hence p = \frac{1}{2}+\frac{1}{2} q

Consider TH is tossed:

• If the next toss is T If the next toss is H then we have the same case as when HT is tossed.
• If the next toss is H then B wins

Therefore, q= \frac{1}{2} p + 0

Consider HH is tossed:

• If the next toss is T we have the same case as when HT is tossed.
• If the next toss is H it is as if nothing happened as the last two tosses are still HH.

Therefore: r =\frac{1}{2} p + \frac{1}{2} r
This simplifies to \frac{1}{2} p = \frac{1}{2} r, therefore, p = r.

Now we solve for p using the first equation:
p = \frac{1}{2}+\frac{1}{2} q
p = \frac{1}{2}+\frac{1}{4} p
\frac{3}{4} p = \frac{1}{2}

p = r = \frac{2}{3}, q =\frac{1}{3}

Each of the 4 initial cases has a probability of 1/4 of happening. Therefore the overall probability of C winning is:
\frac{1}{4}(1+\frac{2}{3}+\frac{2}{3}+\frac{1}{3})=\frac{2}{3}

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