nmacgre
nmacgre
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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
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JB
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March 1st, 2011 at 2:00:47 PM permalink
10.5 units
nmacgre
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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
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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
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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
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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
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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
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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
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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
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March 1st, 2011 at 5:10:19 PM permalink
It's quite amazing how much time i can waste by not reading the question carefully. Ugh.
RaleighCraps
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March 1st, 2011 at 5:24:46 PM permalink
In school, I was always one of the first ones done with every test. I didn't get the highest scores, but I bet I had the highest scoring avg per minutes spent taking the tests.
So one day the teacher hands us all a surprise test and says this is a timed test, and speed plus following directions counts most. SWEET!

Teacher says go, and flip the test over. Typical instruction nonsense about reading carefully, answer accurately, and take the time to read all the questions first, before you begin answering......blah blah.
I jump right in and am on a heater. Questions are simple and I am flying. Get to page 2 and sense some people are really struggling, as they don't seem to be answering any questions. Pretty soon hardly anyone is answering questions.... Uh oh. What did I miss?

Last question says, "Once you have read this question, you are done. You do not need to answer any other questions" So, had I followed the instructions, and read through all the questions first............
Always borrow money from a pessimist; They don't expect to get paid back ! Be yourself and speak your thoughts. Those who matter won't mind, and those that mind, don't matter!
thecesspit
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March 1st, 2011 at 5:27:36 PM permalink
I reckon I scored 5-10% higher on almost every test I did at school by reading through the questions first and planning my order of attack...

I'd divide my time on "so long per mark" with time to check answers at end. If a question took longer than the marks were worth, I'd stop it (unless almost done) and move onwards. I'd almost always do one of the low scoring questions first, then do the big mark questions before going back to the short answers.
"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
RaleighCraps
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March 1st, 2011 at 5:32:06 PM permalink
Quote: thecesspit

I reckon I scored 5-10% higher on almost every test I did at school by reading through the questions first and planning my order of attack...

I'd divide my time on "so long per mark" with time to check answers at end. If a question took longer than the marks were worth, I'd stop it (unless almost done) and move onwards. I'd almost always do one of the low scoring questions first, then do the big mark questions before going back to the short answers.



Too much work for me. I just wanted the damn things over with, and 93 paid the same as 100, so the extra effort wasn't worth it..... :-)
Always borrow money from a pessimist; They don't expect to get paid back ! Be yourself and speak your thoughts. Those who matter won't mind, and those that mind, don't matter!
spinswizard
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July 28th, 2020 at 3:19:48 AM permalink
Intuitively you can check that the possible outcomes are the following:
2 probab: 1/4
2+3=5 probab: 1/4*1/3
2+5=7 probab: 1/4*1/3
2+9=11 probab: 1/4*1/3
2+3+5=10 probab: 2*1/4*1/3*1/2 = 1/4*1/3
2+3+9=14 probab: 2*1/4*1/3*1/2 = 1/4*1/3
2+5+9=16 probab: 2*1/4*1/3*1/2 = 1/4*1/3
2+3+5+9=19 probab: 3*1/4*2*1/3*1/2 = 1/4

And then from here since 2 and 19 have the same probability to occur we can take their average 21/2=10.5 with probab 1/2
And in the same way for the others 5+6=21, 7+14=21, 11+10=21 => avg = 10.5 with probab 1/2.

I am summing them 2 by 2, just to show that the distribiution is symmetric and i think its nice to see quite visually why that the EV is just (2(collect) + 19(sum of all) )/2, like the user gog stated before me :)

and then by induction this is true for a game with n buckets and even repeated values of the prizes :)
ThatDonGuy
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July 28th, 2020 at 8:20:17 AM permalink
Quote: RaleighCraps

Last question says, "Once you have read this question, you are done. You do not need to answer any other questions" So, had I followed the instructions, and read through all the questions first............


I have seen a number of these. Usually, the first two instructions are, "Read everything before doing anything," and, "Write your name at the top of the paper," and the last instruction is, "Now that you have read all of the instructions, do only the first two." Usually, one of the other instructions is something like, "If you are the first one to read this instruction, say out loud, 'I am the best at following instructions.'"

I wouldn't be surprised if someone did a variation on this where one of the instructions in the middle is, "Ignore the final instruction."
charliepatrick
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July 29th, 2020 at 2:13:43 AM permalink
It's the first time I've seen this puzzle.

One neat way of solving it is consider the buckets placed in a row, and then going through all the permutations.

Person one makes their picks from left to right, they will pick any prizes to the left of the "Collect" bucket and the "Collect" bucket.
Person two goes from right to left, they will pick any prizes to the right of the "Collect" bucket and the "Collect" bucket.
Together they pick all the prize buckets and 2 x "Collect" = 3+5+9+2+2 = 21.

If they work through all the permutations, every time the total of the two will be 21, and each person will have covered all the permutations. Hence the average for two people is 21, so the average for one person is 10.5.


On the "ignore the final instruction" there's a puzzle you probably know where it starts with "You are driving a bus and at the first stop 15 people get on...."
spinswizard
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August 2nd, 2020 at 1:45:59 AM permalink
Hii I came up with a pretty neat formula for calculating the EV w.r.t. any number of buckets, including multiple "collect"-s. It goes like this:
Sum of all buckets (incl. all collects) - S
Count of collects - h
Value of Collect - c

(S + c) / (h +1) ;

I've come to it via the general formula with the permutations you suggested, which is the following
h *( SUM i from 0 to n-h ((S-c(h-1)) *Binomial [n-h-1; i-1] + c* Binomial [n-h-1; i]) * i! * (n-1-i)! /n! )

Its a bit long but the end result is so satisfactory! Maybe you will be interested to see :)

https://www.wolframalpha.com/input/?i=SUM+h*%28%28S-c*%28h-1%29%29+*+Binomial%5Bn-h-1%2C+j-1%5D+%2B+c*+Binomial%5Bn-h-1%2C+j%5D%29+*+j%21+*+%28n-1-j%29%21+%2Fn%21+0+to+n-h+
Last edited by: spinswizard on Aug 2, 2020
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