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Expected value in tossing coins
By Gale Greenlee on Friday, December 27, 2002 - 09:37 pm:

I read an article in a puzzle book which mentioned that the expected value for the number of times you would have to toss a fair coin to get the result head-head was six, whereas the number required for head-tail is only four. Could anyone explain this? Gale

By Demetres Christofides on Saturday, December 28, 2002 - 09:42 am:

Hope this helps. (Case HH). If you still can't show that the expected number of tosses is 6 then tell me.

prob

Demetres

By Gale Greenlee on Monday, December 30, 2002 - 03:27 pm:

Thanks Demetres.

Looking at Demetres flowchart I see, we have: M= trying for HH; N= trying for HT; h=H occurred in the first toss; t=T occurred at the first. We can also have th for TH occured first, htt for HTT occured first etc.

For M -- the expected number of tosses to reach HH -- we have

M = (Mh + Mt)/2.

Now Mh = (Mhh + Mht)/2 = (2 + (1+Mt))/2 = (3+Mt)/2,
and Mt = (Mth + Mtt)/2 = ((1+Mh) +(1+Mt))/2 = (2 + Mh + Mt)/2.

Solving the two above equations gives Mh = 5 and Mt = 7, so M = 6.

Similarly, for N we have N = ((Nh + Nt)/2 with

Nh = (Nhh + Nht)/2 = ((1+Nh) + 2)/2, and Nt = (Nth + Ntt)/2 = ((1+Nh) + (1+Nt))/2.

Solving gives Nh = 3 and Nt = 5, so N = 4.

Thanks again, GALE