Plinko price is right game, help me figure math out

[furyshade]

i know the game uses a binomial distribution and im pretty sure it is the 1 8 29 56 70 56 29 8 1 level, but i cant figure out how to factor that in to the probability, anyone help?

http://en.wikipedia.org/wiki/Plinko

[oe39]

"The optimal strategy mathematically is to drop the chip down the center of the board, between the fourth and fifth pegs from the left in the top row. With this strategy, the odds of a chip landing in the $10,000 slot are 231 in 1024, or about 22.56%, and the odds of winning $50,000 are approximately 0.0584%." from your link

[furyshade]

yes, i know this, i can read, i want to know how you find that statistic

[tshort]

[ QUOTE ]
yes, i know this, i can read, i want to know how you find that statistic

[/ QUOTE ]

Using the binomial formula:

p=.5 , n=12, k=6

(12 choose 6) * (.5)^6 * (1-.5)^6 = .225585


Edit: Note, (.5)^k * (1-.5)^(n-k) is constant at any n

[furyshade]

hmm, where did you get the 12 choose 6, did i miss something? it seems there are 9 possible results so it should be 8 choose 4

[tshort]

[ QUOTE ]
hmm, where did you get the 12 choose 6, did i miss something? it seems there are 9 possible results so it should be 8 choose 4

[/ QUOTE ]

There are nine possible columns to the Plinko board, but it is restricted by walls. The player releases the plinko chip at row 0. The money is at row 12. The row 12 binomial would look as follows:

.0002 .0029 | .0161 .0537 .1208 .1934 .2256 .1934 .1208 .0537 .0161 | .0029 .0002

Notice that the .0002 and .0029 "columns" do not exist in the plinko board so those probabilities need to be added back into the columns that do exist. Where should that .0002 go?

It should be distributed amongst the $100, $500, and $1000 columns according n=4 binomial:

.125 | .375 .375 .125

Here, again, the .125 column is only theoretical and does not exist in the Plinko board because of the retaining walls. So, that .125 (.125 * .0002) should be distributed amongst the $100 and $500 evenly (or according to the n=2 binomial).

Similarly, the .0029 theoretical column should be distributed amongst the $100 and $500 evenly (or according to the n=2 binomial).

So, if you drop the plinko chip down the middle the chance of ending up in the columns is:

.0309 .0685 .1209 .1934 .2256 .1934 .1209 .0685 .0309

This gives the player an equity of $2,414 per chip ($2256 of which comes from the $10,000 column).

Notice if you drop it from one column off the middle your equity in the $10000 column drops by $322 while your equity in the $1000 looks to increase by maybe $15. Dropping your plinko chip in the far edge reduces your equity in the $10000 to $387. Without calculating it, I would guess your equity in dropping from an edge is no more than about $800 (probably less).

A simple script using recursion could compute the equities for dropping it from all the columns.

[tshort]

There is a mistake in my above post. Switch around the redistributions of .002930 and .0002 to get:

.0175 .0551 .1212 .1934 .2256

This is correct probability distribution for landing in the first five slots (mirrors for last 4 obviously).