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Coin tossing (Posted on 2004-06-11) Difficulty: 3 of 5
I threw a coin n times, and never got three tails in a row. I calculated the odds of this event, and found out they were just about even; 50%-50%. How many times did I throw the coin?

A second question: what were the chances of having not gotten three heads in a row either?

See The Solution Submitted by Federico Kereki    
Rating: 3.6667 (6 votes)

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Solution Markov chain solution - 2nd part | Comment 10 of 26 |
For the second part, we need extra states HH (we got two heads in a row) and HHH (we got three heads in a row). Now the equations are P(i+1,HHH)=P(i,HHH)+0.5xP(i,HH), P(i+1,HH)=0.5xP(i,H), P(i+1,H)=0.5x(P(i,T)+P(i,TT)), P(i+1,T)=0.5x(P(i,HH)+P(i,H)), P(i+1,TT)=0.5xP(i,T), P(i+1,TTT)=P(i,TTT)+0.5xP(i,TT). As there were i=10 tosses, calculating we get P(10,HHH)=P(10,TTT)=0.413 so the probability of not getting three heads or tails in a row are 0.174. As the problem asked for the probability of not getting three heads given that we didnt get three tails, the answer is 0.174/(1-0.413)=29.6%.

Note: I edited this to fix a typing mistake that was pointed out by Charlie -- thanks!


Edited on June 13, 2004, 10:20 am
  Posted by Oskar on 2004-06-12 23:09:11

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