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Turn a coin and if it falls heads move three places to the right otherwise move 2 places left.

After 20 times you turn the coin, in what positions might you be and what is the probability to be in each of them?

Also calculate the expected position at the end of the process. I really need some help with this problem...

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If you get $N$ heads, you'll get $20-N$ tails, so the position after $20$ tosses with $N$ heads is $X(N) = 3N - 2(20-N) = 5N - 40.$

There are $20 \choose N$ ways to get $N$ heads with $20$ (fair) coin tosses, and $2^{20}$ possible outcomes (including order). So the probability of getting exactly $N$ heads is

$$P(N) = \frac{20 \choose N}{2^{20}}.$$

The most probable position is $X(10) = 5(10) - 40 = 10.$

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  • $\begingroup$ So 10 is the most probable position.But what about in what positions might you be and what is the probability to be in each of them $\endgroup$ – Ioannis Mar 4 '14 at 21:19
  • $\begingroup$ The answers to both of those questions are already in my answer. $\endgroup$ – John Mar 4 '14 at 23:07
  • $\begingroup$ You re right.Thank you a lot $\endgroup$ – Ioannis Mar 4 '14 at 23:31
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Would it help to think about this exercise:

Move $\frac12$ place to the right, flip the coin, and then move $2\frac12$ places right or left, depending on the result of the flip. Repeat 20 times.

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Hint: what is the furthest right you can be? What is the furthest left? Note that your final location only depends on how many heads/tails you get, not the sequence. How much does it change if you change a head to a tail? Can you now describe the possible ending locations? What is the expected number of heads out of $20$ flips?

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