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# Bin A has one white ball and four black b***s. Bin B has three b***s labeled \$1 and one ball labeled \$7 . Bin W has five b***s

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Bin A  has one white ball and four black b***s. Bin B  has three b***s labeled \$1 and one ball labeled \$7 . Bin W has five b***s labeled \$8 and one ball labeled \$500. A game is played as follows: a ball is randomly selected from bin A . If it is black, then a ball is randomly selected from bin B ;otherwise, if the original ball is white, then a ball is randomly selected from bin  W. You win the amount printed on the second ball selected. What is your expected win?

Sep 17, 2016

#6
+2489
+7

Here’s my logic.

This is a weighted average of probability.  The weights are proportional to the probability of each event.

The first event is the probability of either a black ball or a white ball.  Four of the five bálls are black and one is white. The black ball has a (4/5) 0.80 probability and the white ball has a (1/5) 0.20 probability of being drawn

IF a black ball (0.80 probability) is drawn, then bin “b” probabilities are in play.

The weights are calculated the same as above. There are four b***s – three are \$1 bálls and one is a \$7 ball. The probability of drawing a \$1 ball is (3/4) 0.75 and the probability of drawing a \$7 is (1/4) 0.25

0.75 * \$1 = 0.75

0.25 * 7\$ = 1.75

Total       =  2.50 average per event

\$2.50* 0.80 (Black ball Probability) = \$2.00

The logic for the white ball event is the same.

---------------------

Ugh!  CDD strikes again!!

I’m not posting it because I see from Sir Alan’s post that my numbers are wrong. I was using one \$500-ball and eight \$8-bálls. (I think I added 1 to the 8 instead of the five).

I added Sir Alan’s solution to Naus’ Tortoise and Hair AI formula for blonds, so it now works for red-heads.

-----------------------

Here's the fully repaired answer:

Black Ball probability 0.80

\$1 Ball drawn probability 0.75    Expected Value = 0.80 * 0.75 * 1 =  \$0.60

\$7 Ball drawn probability 0.25    Expected Value = 0.80 *0.25 * 7 =   \$1.40

Sum         \$2.00

White Ball probability 0.20

\$8 Ball drawn probability (5/6) 0.8333 Expected Value = 0.20 * 0.8333* 8 = \$1.33

\$500 Ball drawn probability (1/6) 0.1667 Expected Value = 0.20 * 0.1667 * 500 = 16.67

Sum         \$18.00

18.00+2.00 = \$20.00 Expected Value

Sep 18, 2016

#1
+2489
+7

(Assuming the player pays nothing to play)

Black Ball probability 0.80

\$1 Ball drawn probability 0.75    Expected Value = 0.80 * 0.75 * 1 =  \$0.60

\$7 Ball drawn probability 0.25    Expected Value = 0.80 *0.25 * 8 =   \$1.40

Sum         \$2.00

White Ball probability 0.20

\$8     Ball drawn probability 0.889    Expected Value = 0.20 * 0.8889 * 1    =    \$1.42

\$500 Ball drawn probability 0.111    Expected Value = 0.20 * 0.1111 * 500 =   \$11.11

Sum       \$12.53

Sum of expected values   2.00 + 12.53 = 14.53

Expected win per play \$14.53

-------------------------------------------------

I like to play the ponies. It’s a great venue to apply statistics and probability, and it is a lot of fun. It’s even more fun when you win more often.  A basic understanding of statistics helps to separate the whinnies from the neighs.

When the Troll-Master was tutoring me in statistics, he sent me links to two vids that parody the sport.

Classic Spike Jones with one of his many classical music parodies.

This one is contemporary and quite funny. Note that it has “adult” language. I’ve heard variations of this many times, but I never realized it was a horse race until I saw this.

Sep 18, 2016
#3
+2489
+6

I MUCKED this up when I edited it to make it neat. I can't blame it on a banana allergy.

The sums were/are correct for the correct values, so the answer stays the same

Here's the corrected version.

Black Ball probability 0.80

\$1 Ball drawn probability 0.75    Expected Value = 0.80 * 0.75 * 1 =  \$0.60

\$7 Ball drawn probability 0.25    Expected Value = 0.80 *0.25 * 7 =   \$1.40

Sum         \$2.00

White Ball probability 0.20

\$8     Ball drawn probability 0.889    Expected Value = 0.20 * 0.8889 * 8    =    \$1.42

\$500 Ball drawn probability 0.111    Expected Value = 0.20 * 0.1111 * 500 =   \$11.11

Sum       \$12.53

Sum of expected values   2.00 + 12.53 = 14.53

Expected win per play \$14.53

I'll explain the logic below.

GingerAle  Sep 18, 2016
#4
+118653
+5

Thanks Ginger, I am questioning your  0.8889  and your 0.1111

I think the first one is  5 eight dollar b***s out of 6 b***s = 5/6  = 0.83 repeater

and the second one is 1 five hundred dollar ball out of 6 b***s =1/6 =  0.16 repeater

Melody  Sep 18, 2016
#2
+118653
+5

Hi Ginger - it is nice to see you here

Tertre please do not double post like this !!!

You can repost if you want but link the repost to the original question so that it is not double answered.

Like this

http://web2.0calc.com/questions/bin-a-has-one-white-ball-and-four-black-b-s-bin-b-has-three-b-s-labeled-1-and-one-ball-labeled-7-bin-w-has-five-b-s#r2

If you do not know how to link it you can still copy in the address!

Now, my answer is on the other thread Ginger.

I admitted there and I will admit here as well that I am not very familiar with expected value questions but I cannot follow some of your logic Ginger.....

For instance in your first section you have a \$7 ball but there is no 7 in your calculation.....

Sep 18, 2016
#5
+33647
+6

Bin A  has one white ball and four black b***s. Bin B  has three b***s labeled \$1 and one ball labeled \$7 . Bin W has five b***s labeled \$8 and one ball labeled \$500. A game is played as follows: a ball is randomly selected from bin A . If it is black, then a ball is randomly selected from bin B ;otherwise, if the original ball is white, then a ball is randomly selected from bin  W. You win the amount printed on the second ball selected. What is your expected win?

E =  p(A= black)*(p(B=\$1)*\$1 + p(B=\$7)*\$7) + p(A=white)*(p(W=\$8)*\$8 + p(W=\$500)*\$500)

E =  4/5*(3/4*\$1 + 1/4*\$7) + 1/5*(5/6*\$8 + 1/6*\$500)

E = \$2 + \$108/6

E= \$2 + \$18

E = \$20

.

Sep 18, 2016
#6
+2489
+7

Here’s my logic.

This is a weighted average of probability.  The weights are proportional to the probability of each event.

The first event is the probability of either a black ball or a white ball.  Four of the five bálls are black and one is white. The black ball has a (4/5) 0.80 probability and the white ball has a (1/5) 0.20 probability of being drawn

IF a black ball (0.80 probability) is drawn, then bin “b” probabilities are in play.

The weights are calculated the same as above. There are four b***s – three are \$1 bálls and one is a \$7 ball. The probability of drawing a \$1 ball is (3/4) 0.75 and the probability of drawing a \$7 is (1/4) 0.25

0.75 * \$1 = 0.75

0.25 * 7\$ = 1.75

Total       =  2.50 average per event

\$2.50* 0.80 (Black ball Probability) = \$2.00

The logic for the white ball event is the same.

---------------------

Ugh!  CDD strikes again!!

I’m not posting it because I see from Sir Alan’s post that my numbers are wrong. I was using one \$500-ball and eight \$8-bálls. (I think I added 1 to the 8 instead of the five).

I added Sir Alan’s solution to Naus’ Tortoise and Hair AI formula for blonds, so it now works for red-heads.

-----------------------

Here's the fully repaired answer:

Black Ball probability 0.80

\$1 Ball drawn probability 0.75    Expected Value = 0.80 * 0.75 * 1 =  \$0.60

\$7 Ball drawn probability 0.25    Expected Value = 0.80 *0.25 * 7 =   \$1.40

Sum         \$2.00

White Ball probability 0.20

\$8 Ball drawn probability (5/6) 0.8333 Expected Value = 0.20 * 0.8333* 8 = \$1.33

\$500 Ball drawn probability (1/6) 0.1667 Expected Value = 0.20 * 0.1667 * 500 = 16.67

Sum         \$18.00

18.00+2.00 = \$20.00 Expected Value

GingerAle Sep 18, 2016
#7
+118653
0

Well at least we all agree now anyway :))

Ginger, I was looking for some hair dyes that could help you get the mix more consistently correct......

I don't think I achieved my goal but I did find this...

https://sntcollections.wordpress.com/2016/04/25/how-to-make-your-dyed-hair-last-longer-and-fresher/

This one is quite stunning !

Sep 20, 2016