nmacgre Joined: Aug 23, 2010
• Posts: 84
March 1st, 2011 at 1:50:26 PM permalink
Consider a game with four buckets. Three of them contain prize values (see table below) and one contains a “Collect” card with a prize value associated with it. The player is asked to select icons until the “Collect” prize is revealed (you cannot select from the same bucket twice). When the Collect icon is revealed, the player wins the sum of all prizes revealed.

Table of Prizes

Prize Prize Value
Prize 1 3
Prize 2 5
Prize 3 9
Collect 2

What's the expeced value of this game?
JB Joined: Oct 14, 2009
• Posts: 2088
March 1st, 2011 at 2:00:47 PM permalink
10.5 units
nmacgre Joined: Aug 23, 2010
• Posts: 84
March 1st, 2011 at 2:27:55 PM permalink
Yes. I used a table, How'd you do it?

Possibilities
1 punch Combinations Prob Pay EV
c 0.25 2 0.5
2 punch
1,c 0.083333333 5 0.416666667
2,c 0.083333333 7 0.583333333
3,c 0.083333333 11 0.916666667
3 punch
1,2,c 0.041666667 10 0.416666667
1,3,c 0.041666667 14 0.583333333
2,1,c 0.041666667 10 0.416666667
2,3,c 0.041666667 16 0.666666667
3,1,c 0.041666667 14 0.583333333
3,2,c 0.041666667 16 0.666666667
4 punch
1,2,3,c 0.041666667 19 0.791666667
1,3,2,c 0.041666667 19 0.791666667
2,1,3,c 0.041666667 19 0.791666667
2,3,1,c 0.041666667 19 0.791666667
3,1,2,c 0.041666667 19 0.791666667
3,2,1,c 0.041666667 19 0.791666667

EV 10.5
SOOPOO Joined: Aug 8, 2010
• Posts: 6820
March 1st, 2011 at 2:51:07 PM permalink
The EV is definitely going to be higher than 10.5. Too long to put my entire blob of paper calculations down, but just using the probabilities of getting collect in one of the first 8 tries yields an EV of 12.56. You also would have to add more for each time you did not hit collect until the 9th, 10th, etc. time. If I did it correctly, the collectible EV for round one is

1/4 (the chance of hitting collect) x 2 (the value of the collect bucket) = .5

For round two

3/4 x 1/4 ( the chance of hitting collect on exactly the second bucket) x {(3 + 5 + 9)/3 + 2} (the average value of bucket one, plus bucket 2, which must be 2 = 1.44

Round 3 1.88
Round 4 2.00
Round 5 1.95
Round 6 1.80
Round 7 1.60
Round 8 1.39

(all numbers rounded to hundreths).

Round 100 will be 3 to th 99th power divided by 4 to the 100th power, x [2 + 17/3 x 99}

I am sure my friend ME will be able to put this in better formula form, as it has been 30 years since I have used that part of my brain.
JB Joined: Oct 14, 2009
• Posts: 2088
March 1st, 2011 at 2:53:47 PM permalink
Quote: nmacgre

Yes. I used a table, How'd you do it?

Well, you always know that the 2 units from the Collect symbol will be awarded. Then there are either 0, 1, 2, or 3 of the other prizes. The mean of the non-Collect prizes is (3+5+9)/3 = 17/3, therefore:

# of Picks before Collect Average Prize Probability Return
0
2 + (0*17/3) = 02.000000
1/4 = 0.25
0.500000
1
2 + (1*17/3) = 07.666667
3/4 * 1/3 = 0.25
1.916667
2
2 + (2*17/3) = 13.333333
3/4 * 2/3 * 1/2 = 0.25
3.333333
3
2 + (3*17/3) = 19.000000
3/4 * 2/3 * 1/2 * 1/1 = 0.25
4.750000
Totals
1.00
10.500000
gog Joined: Jan 7, 2011
• Posts: 105
March 1st, 2011 at 3:05:47 PM permalink
Probability of any non-collect bucket being picked is 1/4 + (2/4)(1/3)+(2/4)(1/3)(1/2) = 0.5
Therefore total EV = 0.5(3+5+9)+2 = 10.5
thecesspit Joined: Apr 19, 2010
• Posts: 5936
March 1st, 2011 at 3:20:42 PM permalink
Quote: SOOPOO

The EV is definitely going to be higher than 10.5. Too long to put my entire blob of paper calculations down, but just using the probabilities of getting collect in one of the first 8 tries yields an EV of 12.56. You also would have to add more for each time you did not hit collect until the 9th, 10th, etc. time. If I did it correctly, the collectible EV for round one is
....

Each bucket is collectible only once. You can't pick more than 4 times.
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829
JB Joined: Oct 14, 2009
• Posts: 2088
March 1st, 2011 at 3:32:41 PM permalink
Quote: gog

Probability of any non-collect bucket being picked is 1/4 + (2/4)(1/3)+(2/4)(1/3)(1/2) = 0.5
Therefore total EV = 0.5(3+5+9)+2 = 10.5

Excellent! Simple and elegant.
MathExtremist Joined: Aug 31, 2010
• Posts: 6526
March 1st, 2011 at 4:31:51 PM permalink
Quote: gog

Probability of any non-collect bucket being picked is 1/4 + (2/4)(1/3)+(2/4)(1/3)(1/2) = 0.5
Therefore total EV = 0.5(3+5+9)+2 = 10.5

I usually do it the other way:

Avg # non-collect picks = 1.5
Avg non-collect pick value = 17/3
Total EV = 2 + 1.5*17/3 = 10.5
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
SOOPOO Joined: Aug 8, 2010