March 1st, 2011 at 1:50:26 PM
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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?

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?

March 1st, 2011 at 2:00:47 PM
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10.5 units

March 1st, 2011 at 2:27:55 PM
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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

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

March 1st, 2011 at 2:51:07 PM
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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.

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.

March 1st, 2011 at 2:53:47 PM
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Quote:nmacgreYes. 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 |

March 1st, 2011 at 3:05:47 PM
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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

Therefore total EV = 0.5(3+5+9)+2 = 10.5

March 1st, 2011 at 3:20:42 PM
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Quote:SOOPOOThe 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

March 1st, 2011 at 3:32:41 PM
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Quote:gogProbability 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.

March 1st, 2011 at 4:31:51 PM
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Quote:gogProbability 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