 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Flipping Coins (Posted on 2004-10-13) 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) Comments: ( Back to comment list | You must be logged in to post comments.) My approach - matches penny's | Comment 12 of 26 | Here�s the way I approached this problem

For each round I made two tables. Table A is simple: Depending on if you are starting with 5,4,3,2 or 1 coins in that turn, what are the various chances of leaving 0,1,2,3,4 or 5 tails. So if you leave 0 tails that means you won in that round. (If you want to know the exact construction of this table: I put 5,4,3,2,1 across the top for the #of coins at the start of the round, call this S. Then I put 0,1,2,3,4 and 5 down the left for the number of tails left at the end of the round, call this E. Then inside the table, the formula for the chances of having left E tails starting with S coins is S!/[E!(S-E)!]/[2^S].)

Table B relies on information from the previous round. It takes into account the total chances of even starting with S coins. So the entries in Table B is (the corresponding value in Table A) * (chances of starting with S coins, found in Table B of the previous round). Then you just sum all the chances for having E coins at the end, and that is your total chance of leaving E coins in that particular round.

I hope this makes sense. By the way, round 1 can only start with S=5 coins, and no need to relate to a previous round =)

Anyways, the point is that when I did this, I got a 3.125% chance of winning round 1, a 20.605% chance in round 2, a 27.56% chance in round 3, a 21.129% chance in round 4, and a 12.902% in round 5. That means that there is a 14.678% chance of winning in round 6 or later.

So to answer the original questions, the expected number of rounds needed to finish the game is 3, since that has the highest chance (27.56%). And the probability of finishing the game in 3 rounds or less is again 27.56%.

I�m sorry but what is this 3.7�. number that everyone else is talking about? I couldn�t catch what was even trying to be calculated in those. The only answer that resembles mine is Penny�s simulation, I think.

 Posted by nikki on 2004-10-14 20:22:04 Please log in:

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