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Strategy and Expected Value of Dice Game

Date: 09/03/2005 at 13:19:57
From: Matt
Subject: Dice game 

You have the option to throw a die up to three times.  You will earn
the face value of the die.  You have the option to stop after each
throw and walk away with the money earned. 

For example, if I throw the die once and a 2 shows up, I earn 2 
dollars or throw again.  If a 3 comes up after the second throw, I
earn 3 dollars and if I want to stop I can or I can throw a third
time.  The earnings are not additive, so my total earning is just the
3 dollars I got on the last roll. 

What is the expected payoff of this game? 

I know if I throw it once, the expected value is 3.5.  I just don't
know how to incorporate the stopping option. 

I was thinking of approaching this with a risk aversion idea, meaning
that if the throw gives 4 or above, the user should stop.



Date: 09/04/2005 at 01:59:47
From: Doctor Julien
Subject: Re: Dice game

Hey Matt,

Thanks for writing in.  I love these kinds of questions!

Here's how I would approach it.  Let's say that we've rolled two
consecutive ones.  We have one throw left.  What is our expected payoff?

As you correctly stated, it would be 3.5.  Let's call that the payoff
at stage 3.

So that tells us that if our second roll is 1, 2, or 3, we should
reroll.  Otherwise we should stop.  So our stage two payoff will be
the average of 4,5, and 6 half the time, and half the time it will be
the payoff for stage 3.

Stage 2 payoff = 1/2 [(4+5+6)/3] + 1/2 [3.5]
               = 1/2 [5] + 1/2 [3.5]
               = 1/2 [5 + 3.5]
               = 1/2 [8.5]
               = 4.25

With me so far?  Now stop for a moment and think what you would do for
the next step in this problem.  When you think you have an answer,
read on!

What we want to do is now calculate the function for stage 1 based on
stage 2.  If we roll a 5 or a 6 on the first throw, we'll want to keep
that.  Otherwise, we'll go to stage two where our expected payoff is 4.25.

Stage 1 payoff = 2/6 [(5+6)/2] + 4/6 [4.25]
               = 2/6 [5.5] + 4/6 [4.25]
               = 11/6 + 17/6
               = 28/6
               = 4.666...

So that's our expected payoff!  Can you see how to generalize this for
any number of throws?
  
- Doctor Julien, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 09/04/2005 at 15:55:32
From: Matt
Subject: Dice game 

Thank you for your quick reply.  I think my answer was wrong in the 
sense that I was assuming a constant risk aversion strategy.  Meaning 
I was assuming that at each throw I would replay only if I was getting
a 1, 2 or 3.  But your approach is more accurate. 

Thanks again. 
 
Matt
Associated Topics:
High School Probability

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