# Purple Jelly Beans & Probability

I have a question about probability, hope someone could help me.

Imagine we had a jar of 80 different color jelly beans (don't know if their are 80 different colors in the world, but it doesn't matter, this is all theoretical) and someone randomly chose 20 jelly beans from the jar.

I am told to choose 1 color  & if my color matches any one of the twenty chosen jelly beans I win a jar of jelly beans.

The chance of me choosing the right color is 1 in 4.

Now, assuming I had insider information that the "purple"  jelly bean had a chance of 1 in 3 of being chosen (don't ask me why, it doesn't really matter. Let us just assume this is the case. Maybe the purple / green jelly bean is 3 times the size of the others. But, honestly, the reason doesn't matter. Just assume that this is the case.)

Now, let's further assume that I had additional insider knowledge, that, again for some unknown reason, the "purple" jelly bean had a probability of 1 in 6,  every 5 draws,  instead of 1 in 3.

In other words, 4/5 draws, the purple jelly bean has a probability of showing up 1 in 3. But for 1/5 draws it has a probability of 1 in 6.

Assuming I could choose when to play, when the 5th draw comes up I would naturally avoid playing that game, since my overall chances of winning is 1/6.  I would rather only play when I know the probability of choosing purple is 1/3.

My question is:

Assuming I played 4/5 games and avoided playing the one game with the chance of 1/6 -  by how much did I enhance my overall chance of winning by not playing this game?

And what would be the formula to calculate this?

P.S

As I was writing it occurred to me that this would be the correct formula

1/3 times 4 = 4/12

1/6 times 2 = 2/12

=  6/24

= 1/4

So by avoiding playing that game I would increase my overall chances to 1/3 which is 25% increase?

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Colors. :-) –  Brian M. Scott Dec 24 '12 at 12:21

If you play five consecutive games, your expected number of wins is $4\cdot\frac13+\frac16=\frac32$. If you play four consecutive games and sit out the fifth, your expected number of wins is only $4\cdot\frac13=\frac43$, which is smaller, not larger. To put it differently, if you play $3000=5\cdot600$ consecutive games, on average you can expect to win about $\frac32\cdot600=900$ games; if you set out every fifth game, however, you can expect to win only about $\frac43\cdot600=800$ games. In other words, your expected number of wins has decreased by about

$$\frac{900-800}{900}=\frac19\approx0.11111\;,$$

or a bit over $11$%.

Or perhaps you prefer to look at your probability of winning at least one game. If you play $5$ games in a row, your probability of losing all $5$ is $\left(\frac23\right)^4\left(\frac56\right)=\frac{80}{486}=\frac{40}{243}\approx0.16461$, so with probability $1-\frac{40}{243}=\frac{203}{243}\approx0.83539$ you’ll win at least one game of the $5$. If you sit out the fifth game, your probability of winning no games is $\left(\frac23\right)^4\cdot1=\frac{16}{81}\approx0.19753$, since you can’t win the game that you don’t play, and your probability of winning at least one game is $1-\frac{16}{81}=\frac{65}{81}\approx0.80247$: it has decreased by about

$$\frac{\frac{203}{243}-\frac{65}{81}}{\frac{203}{243}}\approx0.03941\;,$$

or about $3.941$%.

Added: If you pay $\$1$to play a game, and a win is worth$\$2$, your expected winnings if you play $5$ consecutive games is

$$\frac32\cdot2-5=-2$$ dollars, since you expect on average to win $\frac32$ games. If you sit out the fifth game, your expected profit is $$\frac43\cdot2-4=-\frac43$$ dollars. You’ve reduced your expected loss from $2$ to $\frac43$ dollars, so you’ve reduced it by $$\frac{2-\frac43}2=\frac13$$ dollars, or a little over $33$%.

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Thanks for reply but I'm looking at it from another respective. Say, for instance I had to pay 1 dollar to play a game. And the box of jelly beans (if I win) is worth 2 dollars -- Amd I have to play either 4 or 5 games. By avoiding playing the 5th -- how much would I save? –  Mike Dec 24 '12 at 12:57
That’s a somewhat different question from the one that you asked. I’ll add that possibility to my answer, using those specific dollar amounts. –  Brian M. Scott Dec 24 '12 at 13:00