## Welcome to BrainDen.com - Brain Teasers Forum

 Welcome to BrainDen.com - Brain Teasers Forum. Like most online communities you must register to post in our community, but don't worry this is a simple free process. To be a part of BrainDen Forums you may create a new account or sign in if you already have an account. As a member you could start new topics, reply to others, subscribe to topics/forums to get automatic updates, get your own profile and make new friends. Of course, you can also enjoy our collection of amazing optical illusions and cool math games. If you like our site, you may support us by simply clicking Google "+1" or Facebook "Like" buttons at the top. If you have a website, we would appreciate a little link to BrainDen. Thanks and enjoy the Den :-)
Guest Message by DevFuse

# Coin tossing patterns revisited

Best Answer plasmid, 04 January 2014 - 05:08 AM

I can't find one where I can prove that the wait time for A is longer than B while A is more likely to happen first, but I can prove a case where the wait time for A is longer than B and they are equally likely to happen first.

Spoiler for getting things started
Go to the full post

2 replies to this topic

### #1 bonanova

bonanova

bonanova

• Moderator
• 6160 posts
• Gender:Male
• Location:New York

Posted 28 December 2013 - 07:49 PM

With a nod to harey and Penney and SP
http://brainden.com/...e-you-sure-win/
http://brainden.com/...e-win/?p=337091

We can select a pattern of four coin tosses, say it's HTTH, and flip a fair coin until we observe it.
We can determine the number of tosses on average that will precede that pattern.
We call that the Wait Time. Each pattern has a wait time that may not be unique.
Clearly THHT has the same wait time as HTTH.

We also can select two patterns of four tosses, say TTTT and HTTT.
By extension of the previous puzzle we know that HTTT is likely to happen first.

Is it possible that for two 4-patterns a and b,
a has the longer Wait Time but is more likely to happen first?

- Puzzle due to Martin Gardner.
• 0

Vidi vici veni.

### #2 plasmid

plasmid

Senior Lolcat

• VIP
• 1484 posts
• Gender:Male

Posted 04 January 2014 - 05:08 AM   Best Answer

I can't find one where I can prove that the wait time for A is longer than B while A is more likely to happen first, but I can prove a case where the wait time for A is longer than B and they are equally likely to happen first.
Spoiler for getting things started

• 0

### #3 bonanova

bonanova

bonanova

• Moderator
• 6160 posts
• Gender:Male
• Location:New York

Posted 25 January 2014 - 06:45 AM

I can't find one where I can prove that the wait time for A is longer than B while A is more likely to happen first, but I can prove a case where the wait time for A is longer than B and they are equally likely to happen first.

Spoiler for getting things started

I'll mark this solved if no one else posts a reply.

Plasmid's well-described example shows that "waiting time" is a quantity that is less intuitive than first imagined. A binary number comprising random bits that takes a "long time" to happen (on average) doesn't give equal likelihood for other binary numbers to occur first. It also points out that waiting times and precedence (between two numbers) are not simple things to calculate.

I'll leave the thread open in case anyone wants to find a quadruplet that is likely to precede a quadruplet having a shorter wait time. Then if anyone is interested I can share a quick method (one that I read) of getting wait times and ordering probabilities for arbitrary sequences.

• 0

Vidi vici veni.

#### 0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users