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Guest Message by DevFuse

Staying alive with an eight-sided die

Best Answer CaptainEd, 30 January 2013 - 06:35 PM

Spoiler for Here's an argument

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16 replies to this topic

#11 CaptainEd

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Posted 30 January 2013 - 05:21 PM

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I think the same words would apply equally well if the die had sides reading 0,1,1,1, rather than 0,1,2,3, yet I feel the probability of going bust should be different. Perhaps that's the trick that I'm falling for.

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#12 CaptainEd

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Posted 30 January 2013 - 06:35 PM   Best Answer

Spoiler for Here's an argument

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#13 bushindo

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Posted 30 January 2013 - 06:46 PM

You glue two small square pyramids together at their bases to form an eight-sided object that becomes a fair die. You mark the opposite pairs of faces with 0 1 2 and 3.

You begin the game with \$1. You roll the die; the number that shows is your new stake. That is, with equal probability you lose you dollar, you keep your dollar, you double your dollar or you triple your dollar.

A minute later, you bet each of your dollars, if any remain after the first roll, with another roll of the die; one roll for each dollar that you have, and collect your winnings, if any. After another minute passes, each of your dollars, if any, suffers the fate of another roll. To be clear, in all cases each dollar is wagered individually. As the minutes turn into years, you eventually become rich, or you go bust.

What is the probability that you go bust?

I like the conditions that Prime gave in his last game. You can simulate the game if you like. But a submitted solution must comprise an answer and a method, both of which are correct. Enjoy!

Here's an approach

Spoiler for

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#14 bonanova

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Posted 30 January 2013 - 06:46 PM

CaptainEd, Good solve.
You're on a roll.
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Vidi vici veni.

#15 bonanova

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Posted 30 January 2013 - 06:51 PM

You glue two small square pyramids together at their bases to form an eight-sided object that becomes a fair die. You mark the opposite pairs of faces with 0 1 2 and 3.

You begin the game with \$1. You roll the die; the number that shows is your new stake. That is, with equal probability you lose you dollar, you keep your dollar, you double your dollar or you triple your dollar.

A minute later, you bet each of your dollars, if any remain after the first roll, with another roll of the die; one roll for each dollar that you have, and collect your winnings, if any. After another minute passes, each of your dollars, if any, suffers the fate of another roll. To be clear, in all cases each dollar is wagered individually. As the minutes turn into years, you eventually become rich, or you go bust.

What is the probability that you go bust?

I like the conditions that Prime gave in his last game. You can simulate the game if you like. But a submitted solution must comprise an answer and a method, both of which are correct. Enjoy!

Here's an approach
Spoiler for
I think you have it.
The Ap=p is the key and A can be the first part of your matrix.
Nice solve.
Cap'n beat you by only a few minutes. ;(
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Vidi vici veni.

#16 bushindo

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Posted 30 January 2013 - 07:03 PM

Spoiler for Here's an argument

Good solve, CaptainEd. Nice and compact solution =)

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#17 CaptainEd

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Posted 30 January 2013 - 07:19 PM

Spoiler for Here's an argument

Good solve, CaptainEd. Nice and compact solution =)

Thank you both so much!

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