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**mrpace****Member**- Registered: 2012-08-16
- Posts: 54

Sam and Maria are playing a game where a dice is rolled. If it lands a 1 or a 2, Sam wins $10 off Maria. If the dice lands a 3,4,5 or 6, Maria wins $10 off Sam.

Seems like a great game for Maria right?

But what if Sam had an infinite amount of money, and Maria has a finite amount. ($100 for example.) What if these players played this game an infinite amount of times. (let's assume they are immortal and they never lose the dice.)

Does Maria eventually go broke?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

Hi;

This is a standard Gambler's ruin problem first expressed by Christian Huygen.

There are 2 possibilities. Maria can go broke or Maria can keep winning indefinitely and not go broke. Obviously Sam can never lose because he can not go broke.

The relevant formula is:

where i is the amount Maria starts with i = 10 bets. n = ∞ which means Maria never goes broke even in an infinite number of plays. p = 2 / 3.

Plugging in we get:

So Maria has a 99.90234375% chance of winning indefinitely and a .09765625% chance of going broke.

It has been known since the time of Huygen that if Maria and Sam both have the same percentage of winning then she is sure to go broke. But that is not the case here because Maria has a better chance of winning each bet.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**ElainaVW****Member**- Registered: 2013-04-29
- Posts: 291

Hello:

Huygen? I thought it was Pascal and Fermat.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

The exact chronology I do not know. Huygen reformulated Pascal - Fermat's work.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**ElainaVW****Member**- Registered: 2013-04-29
- Posts: 291

I see that now. So wonderful how they figured it out.

Is there a notebook?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

I will send you all the literature I have on this type problem.

Did you spot the small inaccuracy in the work given above. The one assumption that can cause a problem?

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**ElainaVW****Member**- Registered: 2013-04-29
- Posts: 291

The plugging in part was not the best idea.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

That is correct! Very good. M was nice to me there.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**ElainaVW****Member**- Registered: 2013-04-29
- Posts: 291

Later I need help on the solution to a problem that they were working on last night and I don't agree that their answer is right.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

Post the problem and who knows.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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