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 Don't lose too fast (Posted on 2011-01-25)
A game is played starting with 6 fair coins laid out with heads face up.

Each round consists of flipping all of the coins showing heads.

If fewer than half of the flipped coins come up heads the player loses.

Rounds continue until the player either loses or has one heads remaining.

The player wins by getting to one heads without losing.

What is the probability of winning this game?

Examples: 6→4→1 would be a loss. (1 is less than half of 4.) 6→5→3→2→2→2→1 would be a win.

 No Solution Yet Submitted by Jer Rating: 4.0000 (1 votes)

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 Solution (using Excel) Comment 5 of 5 |

18.638%

The object is to figure out the probability of having 2 coins remaining.  Once at that place, there is a two-thirds probability of success.

After the first flip, the probabilities of having coins remaining are:

3 - 20/63
4 - 15/63
5 -  6/63

After the second flip, the probabilities are

2 - 0.2313
3 - 0.0942
4 - 0.0154

After the third flip, the probabilities are

2 - 0.0465
3 - 0.0041

After the fourth flip

2 - 0.0018

Adding up the total probability of having two coins left:

0.2796

Multiply this by two-thirds yields 0.18638

 Posted by hoodat on 2011-01-26 00:41:31

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