Markov chains and coin tosses
Eτ A = Eτ + E ( τ A − τ )
= Eτ + p A E ( τ A − τ | τ = τ A ) + p B E ( τ A − τ | τ = τ B ) = Eτ + p B E ( τ A − τ | τ = τ B )
The last expression is the average time one has to wait for A after B has shown. Hence we may write
A : A = E τ + p B ( A : A − B : A )
Also, p B = 1 − p A , and, after rearranging:
p A · A : A + p B · B : A = Eτ
Eτ does not change if we interchange A and B and thus:
p A · A : A + p B · B : A = E τ = p B · B : B + p A · A : B
Factoring out p A and p B , we find that the odds in favour of A are
Since p A + p B = 1, it follows that
For A = HTH and B = HHT, this matches with the computed value of .
It is an interesting thing to note that if A is better than B and B is better than C , then A is not necessarily better than C , as demonstrated by the example HHTH, HTHH, and THHH [2]. In fact, one might prove that if the opponent chooses the pattern a 1 a 2 a 3 ...a n , we will gain advantage over him by picking a 2 a n a 1 ...a n − 1 ( a 2 is the symbol other than a 2 ). 4 Using this fact, one can easily win bets.
References
[1] Li S.- Y. R. (1980) ’A Martingale Approach to the Study of Occurrence of Sequence Patterns in Repeated Experiments’, Annals of Probability 8: 11711176
[2] Graham, R. L., Knuth, D. E., Patashnik, O. (1994), Concrete Mathematics. Reading, Ma. [3] Jakubowski, J., Sztencel, R. (2010) Wstep do teorii prawdopodobien´stwa. Warsaw
4 A proof can be found in [2].
225
Made with FlippingBook - Online catalogs