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Q: Red Ball Payout ( No Answer,   1 Comment )
Subject: Red Ball Payout
Category: Science > Math
Asked by: ignorantbliss-ga
List Price: $5.00
Posted: 18 Oct 2006 11:10 PDT
Expires: 20 Oct 2006 08:30 PDT
Question ID: 774751
There are 12 white balls and 1 red ball in an urn. I draw one ball at
a time. If the white ball comes up, I remove the white ball and repeat
the trial.  if the red ball
comes up, I replace all the removed white balls and repeat the trial
with 12 white and 1 red again.

Every trial where white shows up I pay you 60 cetns. Every trial where
red ball shows up I pay out 70 cetns.

Over the long term what is my average payout per trial?
There is no answer at this time.

Subject: Re: Red Ball Payout
From: jack_of_few_trades-ga on 18 Oct 2006 12:58 PDT
For simplicity sake, 
trial = drawing 1 random ball from the urn
round = the trials including when there are 13 balls until the Red ball is drawn.  

Looking at the odds from the beginning of the round, there is exactly
a 1 in 13 chance that the Red ball will be chosen in any given trial.
Trial 1, of course there are 13 balls and 1 is red.
There is a 12/13 chance that you will get to Trial 2, so 12/13 * 1/12
(because now there is 1 red ball of 12 balls total)... 12/13 * 1/12 =
If you do the math, you will find that every trial is 1/13.

On average, the round will end on trial 6.5 because that is 1/2 of
13.(theoretically of course since there is no 6.5th trial)

That theoretical round would include 5.5 White balls and 1 Red ball.  

So the payout would be (5.5 * $.60 + 1 * $.70) / 6.5 = $.615

I think you may have mis-stated the question a bit (perhaps payout vs
getting paid), if so I think you have enough of the method listed
above so you can figure out the math for other scenarios as needed.

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