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Flipping Coins (Posted on 2004-10-13) Difficulty: 4 of 5
You play a coin flipping game with 5 coins. On round 1 you flip all of them. On round 2, you pick up all the ones that came up tails (leaving all the heads alone) and flip them again. You continue to do this until all the coins are heads. For example:
Round 1:  H T T H T
Round 2:  - H T - H
Round 3:  - - T - -
Round 4:  - - T - -
Round 5:  - - H - -
Done in 5 Rounds.

What is the expected number of rounds you'll need to finish the game?
What is the probability you will finish the game in 3 rounds or less?

See The Solution Submitted by Brian Smith    
Rating: 3.7143 (7 votes)

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Solution solution (raw form) | Comment 7 of 26 |

The second part is easier than the first.

The probability that a given column (individual coin) will have reached its heads by round n is (1 - (1/2)^n) as (1/2)^n is the probability of having no heads.  The columns (coins) are independent, so the probability that all will have reached their goal by a given round is (1 - (1/2)^n)^5.  Substituting n=3 gives .512908935546875 as the probabilityof having finished in 3 rounds or less.

To get the expected length of the game we need to multiply each possible length by the probability of completion at exactly that number of tosses.  As we have the cumulative probability of having completed the set by round n, we have to subtract the previous probability of completion (i.e., by the previous round). We get:

Sigma{n=1 to infinity} n*((1-1/2^n)^5 - (1-1/2^(n-1))^5)

Summing thes just n=1 to 33 yields a total of 3.79416280604821, with the latest term being .0000000192085280889.

However, trying to figure the summation to infinity looks like each (1-1/2^i)^5 is subtracted out once more than it's added in, except for (1-1/2^0)^5, which is equal to zero anyway.

That would yield a negative result, but that's impossible.  It certainly looks as if it's something like 3.794....


  Posted by Charlie on 2004-10-13 18:37:50
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