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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

To anyone who's read Jenilia's topic on the Maths Challenges, this is the kind of thing that you will get if you do well enough in the senior one to move on to the next test. I had a whole book of these, lent to me by my teacher, and I remember this one because it was one of the very few that I could do. Anyway...

Bill and Ben decide to play a game that involves tossing a coin. They keep tossing it again and again until one of them wins.

Bill wins when the coin gives a consecutive sequence of head, head, head. Ben wins when the coin gives a consecutive sequence of head, tail, head.

Ben is surprised that Bill has accepted these rules, because he thinks that he has more chance of winning, but he doesn't know that they are playing with Bill's special coin. With this coin, they both have an equal chance of winning.

So, what is the probability of throwing a head with this coin?

Why did the vector cross the road?

It wanted to be normal.

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 23,774

1/2

Because there's nothing special about Bill's coin!

Because the probability of getting (head, head and head),

(head, tail, and head) are the same, that is 1/8 !

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

That would be right if they tossed the coin three times and if no one won, they started again as if nothing happened.

But they play it so that possible chains link with each other.

e.g. (I'm tossing a coin to get these) T, T, H, T, T, H, H, T, T, T, T, H, T, T, T, H, T, H and then Ben would have won because the coin gave a sequence of head, tail, head.

T, H, T, H - Ben wins.

T, T, H, H, H - Bill wins.

You just keep going until the last three coins are either H, H, H or H, T, H in that order.

Hopefully, that's explained it a bit better.

Why did the vector cross the road?

It wanted to be normal.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Theoretically the game could go on forever. Let the probability that this happens (and hence nobody wins) be *P*.

Then the probability that somebody wins is 1−*P*.

Since the bias of the coin has been fixed so that either person is an equal chance of winning, the probability that Bill wins is (1−*P*)⁄2 and the probability that Ben wins is also (1−*P*)⁄2.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Well, in trying to compute *P* (the probability that the game never ends), I notice this pattern:

(i) If the sequence HT appears, the next throw must be T (otherwise Bill would win and the game would end)

(ii) If the sequence HH appears, the next two throws must be TT (in order for the game to continue)

(iii) If we have the sequence TT. the process can start all over again as if nothing had happened. This is indicated by an asterisk in the diagram below.

Still working on it

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

That's the method I used of finding the probability as well.

Why did the vector cross the road?

It wanted to be normal.

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**Maelwys****Member**- Registered: 2007-02-02
- Posts: 161

I did something pretty similar to what Mathsyperson described, and I got ...

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**Fruityloop****Member**- Registered: 2009-05-18
- Posts: 135

This post is a few years old, but I find this problem fascinating...

It reminds me a little bit of the game of c r a p s where you have to roll your 'point' before a 7 in order to win. You have a infinite number of possible sequence of events depending on when the 'point' or 7 appears. Using a nice diagram like Jane's...

Ben wins with the far branch on the right H-T-H or with the branch ending on the bottom left H-H-T-H.

Bill wins with the branch ending on the middle left H-H-H.

So the probability of Ben winning is (1/8+1/16)=3/16 and Bill's chances are (1/8)=2/16.

So, Ben's chances are 3/16 and Bill's chances are 2/16.

So, Ben is favored 3 to 2 with a fair coin. Ben has a 60% chance of winning and Bill has only a 40% chance of winning.

We want to find the probability of heads which will give an even chance to both parties.

Let the probability of heads be x and the probability of tails be 1-x.

Bill must win with three heads and Ben can win with either two heads and a tail or three heads and a tail so...

which gives the solution as

or about 61.8% chance of the coin being heads.

*Last edited by Fruityloop (2009-08-31 18:12:05)*

**The eclipses from Algol come further apart in time when the Earth is moving away from Algol and closer together in time when the Earth is moving towards Algol, thereby proving that the speed of light is variable and that Einstein's Special Theory of Relativity is wrong.**

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**quittyqat****Member**- Registered: 2009-04-08
- Posts: 1,213

What if the special coin has a tail on both sides? Then they both have no chance of winning.

I'll be here at least once every month. XP

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