Probability that 6 Peas will SinkDate: 1/23/96 at 14:36:11 From: Anonymous Subject: probability We need to know the probability that all six of these durn peas will be floating on top of the water before they are all six sunk down to the bottom of the jug. Conditions: 1 jug of magic math water and six special peas. Start: four peas floating and two peas resting on the bottom. Each second, peas can move: 0, 1, or 2 peas can go top to bottom with equal probability for each case. 0, 1, or 2 peas can go bottom to top with equal probability for each case If there is exactly one pea in either location, 0, or 1 pea moves with equal probability When all peas are in either location, then game is done. Date: 6/1/96 at 17:7:10 From: Doctor Gary Subject: Re: probability Let's call the probability of all the peas going to the top before they all go to the bottom "t" and the probability of all the peas going to the bottom before they all go to the top "b". Eventually, one or the other will happen, so t + b = 1. Since all we care about is which happens first, and not how long it will take to happen, we can safely ignore those so-called "moves" in which no peas change location. We can calculate t as a sum: There are only four possible non-null first moves, each with probability .25. 1. If the first non-null move is 2 from bottom to top, the game is over. This means that t will be AT LEAST .25. 2. If the first non-null move is 1 from top to bottom, we have 3 peas on top and 3 on the bottom. If moves are independent of one another, the chances are now even that all the peas will be on the top before they are all on the bottom, and vice versa. Add another .125 to our calculation of t. 3. If the first non-null move is 2 from top to bottom, then there will be 4 peas on the bottom and 2 on the top. FROM THIS POINT ON, the probability that the peas will all be on the bottom before they are on the top is the same as it was at the beginning of our game that they would all be on the top before they were all on the bottom, namely t. The probability that they will all be on the top before they are all on the bottom is b, which is equal to (1-t). This means that our computation of t will include a term of .25(1-t). 4. If the first non-null move is 1 from bottom to top, then there are only two possibilitites for the SECOND non-null move, each with probability .5. The second such move must either end the game with all the peas on the top (add .125 to our calculation of t) or restore the original situation with 4 on top and 2 on the bottom (add .125t to our calculation of t). t can now be expressed as the sum of: .250 + .125 + .250(1-t) + .125 + .125t which equals: .750 - .125t Now we can solve the equation: t = .750 - .125t 8t = 6 - t 9t = 6 t = 2/3 -Doctor Gary, The Math Forum |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/