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Q: Red Ball Payout ( No Answer,   1 Comment ) Question
 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?```  ```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 = 1/13 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.``` 