Easy math puzzles

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September 15th, 2020 at 6:44:32 PM permalink

Quote: Wizard
Quote: gordonm888
(1) How many bridge deals (13 cards out of 52) on average will it take before a player has held all 52 cards (in his own hand) at least once?

16.4116...

Solved it, hopefully correctly, with a Markov chain.

CORRECT answer for the first part

My source says the answer is:

545928847452258491492389829320450626160841197170151499543576764239417023255451861085626905883156684405195107023780479317372271274509508567967387626740665018015009371549436149407597300658313676235368117130572713596038893838341598601195428485082034721825135468987/
33263651815411301455093132853409898491173712201961171863009501862172850150290107365581703955244599875291300187598652081762922197224670489260819393711347009189515918586716637050967309053192156474313569986447499658535026680206545307025594664981127461443909937520 ≈ 16.412 .

Not sure why this answer uses about 250 digits in the numerator and denominator and then rounds it to so few digits. Weird, huh?

Given a Markov chain calculation, I understand about formulating the transition matrix and then starting with an initial state vector/matrix and repeatedly multiplying it with the transition matrix. I just haven't really attempted it yet.

Wizard, did you use a spreadsheet or a computational program to do this?

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

ThatDonGuy

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September 15th, 2020 at 7:30:49 PM permalink

Quote: gordonm888
Quote: Wizard
Quote: gordonm888
(1) How many bridge deals (13 cards out of 52) on average will it take before a player has held all 52 cards (in his own hand) at least once?

16.4116...

Solved it, hopefully correctly, with a Markov chain.

CORRECT answer for the first part

My source says the answer is:

545928847452258491492389829320450626160841197170151499543576764239417023255451861085626905883156684405195107023780479317372271274509508567967387626740665018015009371549436149407597300658313676235368117130572713596038893838341598601195428485082034721825135468987/
33263651815411301455093132853409898491173712201961171863009501862172850150290107365581703955244599875291300187598652081762922197224670489260819393711347009189515918586716637050967309053192156474313569986447499658535026680206545307025594664981127461443909937520 ≈ 16.412 .

Not sure why this answer uses about 250 digits in the numerator and denominator and then rounds it to so few digits. Weird, huh?

Given a Markov chain calculation, I understand about formulating the transition matrix and then starting with an initial state vector/matrix and repeatedly multiplying it with the transition matrix. I just haven't really attempted it yet.

Wizard, did you use a spreadsheet or a computational program to do this?

That's what I get as well

I did it computationally, using a vector of rational numbers and working from the expected number needed with 52 already drawn (hint: zero) down to the number needed with 13. The solution is 1 + the number needed with 13, as the first deal will always result in exactly 13 cards having been dealt.

After all, what's a Markov chain other than a series of simultaneous equations?

charliepatrick

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September 15th, 2020 at 7:39:23 PM permalink

In the end I ran a simulation and then a simple program looping through all the months using nested loops (do a=1 to 12, do b=1 to 12 etc perms=daysinmonth[a]*...).
Slightly worrying is the simulation gave slightly different answers!

Simulation

Parms: No of trials:1000000000 Time:3:0:59:776
Summary Trials 1000000000 Wins 572657896 Losses 427342104
Parms: No of trials:1000000000 Time:3:5:40:425

Parms: No of trials:1000000000 Time:3:31:15:942
Summary Trials 1000000000 Wins 572637546 Losses 427362454
Parms: No of trials:1000000000 Time:3:36:4:144

Calculation: Summary Trials 17748900625 Wins 10164157800 Losses 7584742825.

The calculation agrees with

No of	No of	No of	Ways	Days	Days	Days	Days	Contribution
31-day	30-day	28-day		first	second	third	fourth
months	months	months		person	person	person	person
4			1	217	186	155	124	775 757 640
3	1		4	217	186	155	120	3 002 932 800
3		1	4	217	186	155	28	700 684 320
2	2		6	217	186	120	90	2 615 457 600
2	1	1	12	217	186	120	28	1 627 395 840
1	3		4	217	120	90	60	562 464 000
1	2	1	12	217	120	90	28	787 449 600
	4		1	120	90	60	30	19 440 000
	3	1	4	120	90	60	28	72 576 000
							Total	10 164 157 800
				365	365	365	365	17 748 900 625
							Prob	57.266 408%

Wizard
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Wizard

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September 15th, 2020 at 7:43:05 PM permalink

I used a spreadsheet. It doesn't give an exact answer, but after 116 hands the probability of seeing every card is 0.999999999999836, resulting in an expected number of hands of 16.41217418, which should be accurate to that many decimal places.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

charliepatrick

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September 15th, 2020 at 8:03:07 PM permalink

For the bridge hands an idea I was thinking of using is similar spreadsheet method to working out EVs at Blackjack, i.e. using recursion.

With 52 found, your expected number of hands to go is zero : f(52) = 0 .

With 51 found, it takes you 1/p(51) hands to get out of the state (where p(n) is the chance of finding at least one card in state n - in this case the last card) + 1* f(52).

With 50 found, it takes you 1/p(50) to get out, then P(s51|s50 finds a card)*f(51) + P(s52|s50 finds a card)*f(52) more hands.

With fewer found, it's the sum of how long do you have to wait for a decision ( again 1/p(n) ), and then add SUMPRODUCT (probabilities of going to each new state, f(new state) ).

The tricky part is working out the probabilties, given you've found at least one new card, of how many you might actually have found.

I think when this was a roulette puzzle, working out how long it took to roll all 37 numbers, you could only go up one state so it was fairly easy. In this case you can go up multiple states.

Ace2

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September 15th, 2020 at 11:03:15 PM permalink

Quote: Gialmere
A sort of sequel to the famous birthday problem...

What is the probability that in a group of 4 people, at least 2 of them are born in the same month?

Assume a 365 day year (i.e. 28 days in February) but, apart from this caveat, use the correct probabilities for each of the months.

Where
a = 28/365
b= 120/365
c=217/365
d=30/365
e=31/365

n= a(1-a)^3 + b(1-d)^3 + c(1-e)^3 = 0.770101

4(1-n) - 6(a^2 + bd + ce) + 4(a^3 + bd^2 + ce^2) - (a^4 + bd^3 + ce^3) = 44.646% probability

It�s all about making that GTA

Gialmere

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September 16th, 2020 at 6:43:55 AM permalink

Quote: Ace2

Where
a = 28/365
b= 120/365
c=217/365
d=30/365
e=31/365

n= a(1-a)^3 + b(1-d)^3 + c(1-e)^3 = 0.770101

4(1-n) - 6(a^2 + bd + ce) + 4(a^3 + bd^2 + ce^2) - (a^4 + bd^3 + ce^3) = 44.646% probability

Also very close, but no.

Have you tried 22 tonight? I said 22.

ThatDonGuy

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September 16th, 2020 at 7:14:59 AM permalink

Quote: gordonm888
(2) How many Pai Gow Poker hands (7 cards out of 53) on average will it take before a player has held all 53 cards at least once?

1 / 32.92568564

Specifically:
629,097,238,224,941,789,116,137,284,618,328,573
,253,267,781,017,644,372,314,522,497,141,028,041,898,951,883,234,392,254,059,720,888,441,444,593,624,346,465,280,910,908
,992,267,124,905,045,814,131,513,624,902,821,633,667,261,105,718,956,288,311,603,040,542,103,677,201,489,748,650,413,559
/
19,106,579,741,561,884,505,224,318,707,257,710
,891,622,671,725,425,855,998,089,223,811,780,836,135,833,611,417,577,161,108,975,083,498,545,215,815,129,315,281,867,166
,023,015,450,873,297,597,254,359,533,368,618,569,986,012,443,663,161,895,129,820,324,319,062,323,201,573,486,955,270,120

ChesterDog

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September 16th, 2020 at 9:10:28 AM permalink

Quote: Gialmere
...

What is the probability that in a group of 4 people, at least 2 of them are born in the same month?

Assume a 365 day year (i.e. 28 days in February) but, apart from this caveat, use the correct probabilities for each of the months.

I get approximately 0.428431201777603.

Edit: Now, I agree with you others. (In calculating the probability of four different months, I had neglected one combination: September-April-June-November.)
Now, my Excel gives: 0.427335922672112.

Last edited by: ChesterDog on Sep 16, 2020

charliepatrick

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September 16th, 2020 at 11:23:27 AM permalink

Quote: Gialmere
Quote: gordonm888
0.427335923
I agree with Charlie's approach and calculations except that the total number of combinations should be 365^4 because the four people can be born on the same day.

Close, but no cigar.

I've tried yet another approach, not the actual coding as I used a spreadsheet, but the logic is as follows...

...yes there's a better way using perms, but this is bullet proof...

DO I=1 TO 12
 DO j=1 TO 12
  DO k=1 TO 12
   DO l=1 TO 12
    determine whether any months are the same
    perms = DAYSINMONTH(i) * DAYSINMONTH(j) * DAYSINMONTH(k) * DAYSINMONTH(l)
    add perms to "diff" or "same"
ENDDO ENDDO ENDDO ENDDO
print diff same

   57.266 408%      42.733 592%
10 164 157 800    7 584 742 825   17 748 900 625

So I'm wondering whether I've misunderstood the question.

gordonm888
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September 16th, 2020 at 12:02:18 PM permalink

Quote: ThatDonGuy
Quote: gordonm888
(2) How many Pai Gow Poker hands (7 cards out of 53) on average will it take before a player has held all 53 cards at least once?

1 / 32.92568564

Specifically:
629,097,238,224,941,789,116,137,284,618,328,573
,253,267,781,017,644,372,314,522,497,141,028,041,898,951,883,234,392,254,059,720,888,441,444,593,624,346,465,280,910,908
,992,267,124,905,045,814,131,513,624,902,821,633,667,261,105,718,956,288,311,603,040,542,103,677,201,489,748,650,413,559
/
19,106,579,741,561,884,505,224,318,707,257,710
,891,622,671,725,425,855,998,089,223,811,780,836,135,833,611,417,577,161,108,975,083,498,545,215,815,129,315,281,867,166
,023,015,450,873,297,597,254,359,533,368,618,569,986,012,443,663,161,895,129,820,324,319,062,323,201,573,486,955,270,120

CORRECT (I will graciously assume you meant the inverse of your answer.)

How do you calculate that many digits?

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

ThatDonGuy

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September 16th, 2020 at 12:06:48 PM permalink

Quote: gordonm888
How do you calculate that many digits?

I created a "Rational" library for .NET that does arithmetic on rational numbers, with arbitrary-precision integers ("BigIntegers") for the numerator and denominator.

gordonm888
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September 16th, 2020 at 12:25:21 PM permalink

Quote: charliepatrick
Quote: Gialmere
Quote: gordonm888
0.427335923
I agree with Charlie's approach and calculations except that the total number of combinations should be 365^4 because the four people can be born on the same day.

Close, but no cigar.
I've tried yet another approach, not the actual coding as I used a spreadsheet, but the logic is as follows...
...yes there's a better way using perms, but this is bullet proof...
DO I=1 TO 12
 DO j=1 TO 12
  DO k=1 TO 12
   DO l=1 TO 12
    determine whether any months are the same
    perms = DAYSINMONTH(i) * DAYSINMONTH(j) * DAYSINMONTH(k) * DAYSINMONTH(l)
    add perms to "diff" or "same"
ENDDO ENDDO ENDDO ENDDO
print diff same
   57.266 408%      42.733 592%
10 164 157 800    7 584 742 825   17 748 900 625
So I'm wondering whether I've misunderstood the question.

And, if you ignored the length of the months and simply assumed they were 1/12 of a year the easy-to-calculate combinatorial answer is 42.708 333%. And given the actual distribution of month lengths, -and especially the effect of February being only 28 days - it makes sense that the our answer of 42.733 592% is slightly higher than 42.708 333%.

MathAnswersMatter

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

ThatDonGuy

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September 16th, 2020 at 12:28:39 PM permalink

For the birthdays problem, I was getting what everybody else was getting, until...

I did a brute force search, and got 42.7391782%

For some reason, the method used was overcounting the number of times there were four different months by 991,440.
The number of sets of people with different months are lower than from Charlie's counts as follows, where (A, B, C) is A people born in 31-day months, B people born in 30-day months, and C people born in 28-day-months:
(4, 0, 0): 124 x 155 x 12
(3, 1, 0): 4 x 120 x 155 x 6
(3, 0, 1): 4 x 28 x 155 x 6
(2, 2, 0): 6 x 120 x 90 x 2
(2, 1, 1): 12 x 28 x 120 x 2
The others are the same

Last edited by: ThatDonGuy on Sep 16, 2020

charliepatrick

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September 16th, 2020 at 2:03:03 PM permalink

Another way to look at the scenario where everyone is born in a 31-day month is to consider it as two parts.
(i) Each person could be born on any of 31 days in their month. That gives 31*31*31*31.
(ii) Looking for times where the months are different, the first person can be any of the 7 months. The second can be any of 6 etc. That gives 7*6*5*4.
31*31*31*31*7*6*5*4 = 775757640.

gordonm888
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September 16th, 2020 at 2:43:34 PM permalink

Quote: ThatDonGuy
For the birthdays problem, I was getting what everybody else was getting, until...

I did a brute force search, and got 42.7391782%

For some reason, the method used was overcounting the number of times there were four different months by 991,440.
The number of sets of people with different months are lower than from Charlie's counts as follows, where (A, B, C) is A people born in 31-day months, B people born in 30-day months, and C people born in 28-day-months:
(4, 0, 0): 124 x 155 x 12
(3, 1, 0): 4 x 120 x 155 x 6
(3, 0, 1): 4 x 28 x 155 x 6
(2, 2, 0): 6 x 120 x 90 x 2
(2, 1, 1): 12 x 28 x 120 x 2
The others are the same

I admit, I am stumped. I have spent about an hour studying this post and Charlie's calculations and I simply can't reconcile them. I see no problem with Charlie's approach and calculations. I am human, maybe I have a blind spot.

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

Gialmere

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September 16th, 2020 at 5:17:29 PM permalink

Here is the official Birthday solve from the puzzle site. I'll let you guys debate the finer points.
(Note that I was too lazy to code and reconvert the two tables.)

The answer is .43122136 or 43.12%.

For each of the seven months containing 31 days, the probability of being born in that month is 31/365.
For each of the four months containing 30 days, the probability of being born in that month is 30/365.
For February which contains 28 days, the probability of being born in that month is 28/365.

We must figure out the probability that all four are born in different months, and then subtract that number from 1 to get the probability that at least two are born in the same month.

We must first determine the number of combinations of four DIFFERENT months. Using 12C4, we get 495 different combinations of 12 months taken 4 at a time. Luckily, we can break this down into nine categories since many of those 495 combinations would have the same probability:
JAN-FEB-MAR-APR would have the same probability as JAN-FEB-SEP-DEC because they each have two months with 31 days, 1 month with 30 days, and 1 month with 28 days.

So, here are the nine categories:

Category 1st Month 2nd Month 3rd Month 4th Month Number of Combinations
#1 31 31 31 31 7C4 = 35
#2 30 30 30 30 4C4 = 1
#3 31 31 31 30 7C3 x 4 = 140
#4 31 31 31 28 7C3 x 1 = 35
#5 31 31 30 30 7C2 x 4C2 = 21 x 6 = 126
#6 31 31 30 28 7C2 x 4 x 1 = 21 x 4 x 1 = 84
#7 31 30 30 30 7 x 4C3 = 7 x 4 = 28
#8 31 30 30 28 7 x 4C2 x 1 = 42
#9 30 30 30 28 4C3 x 1 = 4
Total 495

Now the next table shows the probability that all four people have different birth months for each of the nine categories:

Category 1st Month 2nd Month 3rd Month 4th Month Probability that all are different
#1 31 31 31 31 (31/365)4 = .00005203
#2 30 30 30 30 (30/365)4 = .00004564
#3 31 31 31 30 (31/365)3 x 30/365 = .00005036
#4 31 31 31 28 (31/365)3 x 28/365 = .00004700
#5 31 31 30 30 (31/365)2 x (30/365)2 = .00004873
#6 31 31 30 28 (31/365)2 x 30/365 x 28/365 = .00004548
#7 31 30 30 30 31/365 x (30/365)3 = .00004716
#8 31 30 30 28 31/365 x (30/365)2 x 28/365 = .00004014
#9 30 30 30 28 (30/365)3 x 28/365 = .00004259

Now, we must combine the two tables to obtain the sum of all 495 combinations:
Sum = 35 x .00005203 + 1 x .00004564 + 140 x .00005036 +
35 x .00004700 + 126 x .00004873 + 84 x .00004548 +
28 x .00004716 + 42 x .00004014 + 4 x .00004259
= .02369911

However, this only represents the nine combinations. We have computed the probabilities for each combination of four months. But we have ignored the possibility of the four people having different birth months within the same category.
For example:

Abe could be born in JAN, Ben in FEB, Carol in MAR, and David in APR; or
Abe could be born in JAN, Ben in FEB, Carol in APR, and David in MAR; or
Abe could be born in JAN, Ben in MAR, Carol in FEB, and David in APR; or
Abe could be born in JAN, Ben in MAR, Carol in APR, and David in FEB; or

Abe could be born in JAN, Ben in APR, Carol in FEB, and David in MAR; or
Abe could be born in JAN, Ben in APR, Carol in MAR, and David in FEB; or
Abe could be born in FEB, Ben in JAN, Carol in MAR, and David in APR; or
Abe could be born in FEB, Ben in JAN, Carol in APR, and David in MAR; or

Abe could be born in FEB, Ben in MAR, Carol in JAN, and David in APR; or
Abe could be born in FEB, Ben in MAR, Carol in APR, and David in JAN; or
Abe could be born in FEB, Ben in APR, Carol in MAR, and David in JAN; or
Abe could be born in FEB, Ben in APR, Carol in JAN, and David in APR; or

Abe could be born in MAR, Ben in JAN, Carol in FEB, and David in APR; or
Abe could be born in MAR, Ben in JAN, Carol in APR, and David in FEB; or
Abe could be born in MAR, Ben in FEB, Carol in JAN, and David in APR; or
Abe could be born in MAR, Ben in FEB, Carol in APR, and David in JAN; or

Abe could be born in MAR, Ben in APR, Carol in JAN, and David in FEB; or
Abe could be born in MAR, Ben in APR, Carol in FEB, and David in JAN; or
Abe could be born in APR, Ben in JAN, Carol in FEB, and David in MAR; or
Abe could be born in APR, Ben in JAN, Carol in MAR, and David in FEB; or

Abe could be born in APR, Ben in FEB, Carol in MAR, and David in JAN; or
Abe could be born in APR, Ben in FEB, Carol in JAN, and David in MAR; or
Abe could be born in APR, Ben in MAR, Carol in JAN, and David in FEB; or
Abe could be born in APR, Ben in MAR, Carol in FEB, and David in JAN; or

So, for each category of four months, there are 24 permutations!
So, we must multiply our probability sum by 24 to get 24 x .02369911 = .56877864.
This means that the probability that four people will have different birth months is .56877864.

And finally, the probability that at least two of the four are born in
the same month is 1 - .56877864 = .43122136 or 43.12%.

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

Birthdays are good for you.

Statistics show that people who have more of them live longer.

Have you tried 22 tonight? I said 22.

ThatDonGuy

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September 16th, 2020 at 6:38:43 PM permalink

Quote: ThatDonGuy
For the birthdays problem, I was getting what everybody else was getting, until...

I did a brute force search, and got 42.7391782%

I found a glitch in my code (it was only counting 30 days for December) - when I corrected it,

all nine sets of numbers matched Charlie's, and the brute force answer is 42.733592267%.

...and, I found the error in the "official" answer:

Quote: Gialmere
Now the next table shows the probability that all four people have different birth months for each of the nine categories:

Category 1st Month 2nd Month 3rd Month 4th Month Probability that all are different
#1 31 31 31 31 (31/365)^4 = .00005203
#2 30 30 30 30 (30/365)^4 = .00004564
#3 31 31 31 30 (31/365)^3 x 30/365 = .00005036
#4 31 31 31 28 (31/365)^3 x 28/365 = .00004700
#5 31 31 30 30 (31/365)^2 x (30/365)^2 = .00004873
#6 31 31 30 28 (31/365)^2 x 30/365 x 28/365 = .00004548
#7 31 30 30 30 31/365 x (30/365)^3 = .00004716
#8 31 30 30 28 31/365 x (30/365)^2 x 28/365 = .00004014
#9 30 30 30 28 (30/365)^3 x 28/365 = .00004259

31/365 x (30/365)^2 x 28/365 = 0.00004401, not 0.00004014

Last edited by: ThatDonGuy on Sep 16, 2020

gordonm888
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September 16th, 2020 at 7:00:35 PM permalink

I think the official answer is wrong. Basically, in this step:

"Now, we must combine the two tables to obtain the sum of all 495 combinations:
Sum = 35 x .00005203 + 1 x .00004564 + 140 x .00005036 +
35 x .00004700 + 126 x .00004873 + 84 x .00004548 +
28 x .00004716 + 42 x .00004014 + 4 x .00004259
= .02369911"

The 8 digit fractions in that calculation were all rounded -and one of the rounded numbers was actually rounded up when it should have been rounded down. Those numbers were calculated in this step:

"Category 1st Month 2nd Month 3rd Month 4th Month Probability that all are different
#1 31 31 31 31 (31/365)4 = .00005203
#2 30 30 30 30 (30/365)4 = .00004564
#3 31 31 31 30 (31/365)3 x 30/365 = .00005036
#4 31 31 31 28 (31/365)3 x 28/365 = .00004700
#5 31 31 30 30 (31/365)2 x (30/365)2 = .00004873
#6 31 31 30 28 (31/365)2 x 30/365 x 28/365 = .00004548
#7 31 30 30 30 31/365 x (30/365)3 = .00004716
#8 31 30 30 28 31/365 x (30/365)2 x 28/365 = .00004014
#9 30 30 30 28 (30/365)3 x 28/365 = .00004259"

But the author should have displayed and used more digits. For example, the #3 product is actually .0000503541, which seems like a small difference from .00005036 but it is multiplied by (140 x24) to arrive at its contribution to the final answer. This roundoff in this one number alone causes an error in the final number that is much higher than the discrepancy we are trying to resolve.

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

ThatDonGuy

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September 16th, 2020 at 7:07:32 PM permalink

Quote: gordonm888
"Category 1st Month 2nd Month 3rd Month 4th Month Probability that all are different
#3 31 31 31 30 (31/365)3 x 30/365 = .00005036

But the author should have displayed and used more digits. For example, the #3 product is actually .0000503541, which seems like a small difference from .00005036 but it is multiplied by (140 x24) to arrive at its contribution to the final answer. This roundoff in this one number alone causes an error in the final number that is much higher than the discrepancy we are trying to resolve.

Er, even if you round 0.0000503541 to 0.00005035, the difference is only 0.00000001; multiplying it by 140 x 24 = 0.0000336, which does not make up the discrepancy.
There is another math error in there; see my post just above yours (click on the "Dontcha Just Hate Math Errors" spoiler button).

gordonm888
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September 16th, 2020 at 11:10:09 PM permalink

Quote: ThatDonGuy
Quote: gordonm888
"Category 1st Month 2nd Month 3rd Month 4th Month Probability that all are different
#3 31 31 31 30 (31/365)3 x 30/365 = .00005036

But the author should have displayed and used more digits. For example, the #3 product is actually .0000503541, which seems like a small difference from .00005036 but it is multiplied by (140 x24) to arrive at its contribution to the final answer. This roundoff in this one number alone causes an error in the final number that is much higher than the discrepancy we are trying to resolve.

Er, even if you round 0.0000503541 to 0.00005035, the difference is only 0.00000001; multiplying it by 140 x 24 = 0.0000336, which does not make up the discrepancy.
There is another math error in there; see my post just above yours (click on the "Dontcha Just Hate Math Errors" spoiler button).

Yeah, I realized this after I posted it, but I am so over this issue that I didn't bother to correct it. I have spent too much time trying to reconcile your wrong calculation and Gialmere's erroneous "Official solution" and Charlie's original mistake. It was my turn to make an error.

Gilamere if you post another problem about people being born in the same month, physical violence may break out.

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

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September 17th, 2020 at 2:23:04 AM permalink

Quote: Gialmere
A sort of sequel to the famous birthday problem...

What is the probability that in a group of 4 people, at least 2 of them are born in the same month?

Assume a 365 day year (i.e. 28 days in February) but, apart from this caveat, use the correct probabilities for each of the months.

Here is my idea :-

A) Find the probability for all 4 people born in different month

1) Jan, Feb, March, April =====> Prob 1 = 31/365 * 28/365 * 31/365 * 30/365 * 4 ! = 19373760/365^4
2) Jan, Feb, March, May =====> Prob 2 =
.
.
.
495) Sept, Oct, Nov, Dec ====> Prob 495

B) The probability that in a group of 4 people, at least 2 of them are born in the same month = 1 - (Prob 1+ Prob 2 + . . .+ Prob 495)

LOL

Gialmere

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September 17th, 2020 at 8:14:09 AM permalink

Quote: gordonm888
Gilamere if you post another problem about people being born in the same month, physical violence may break out.

Heh. Okay, I'll retreat back to dice...

The game Jackpot Yahtzee is played using four 6-sided dice whose faces contain slot machine symbols.

Die 1: Orange, Orange, Bell, Bell, Cherry, Dollar

Die 2: Orange, Orange, Bell, Bell, Cherry, Dollar

Die 3: Orange, Bell, Dollar, Cherry, Cherry, Cherry

Die 4: Orange, Bell, Dollar, Cherry, Cherry, Cherry

The best roll you can get is one of each symbol since this allows you to put any of the four symbols on your payline rack for the turn.

Expressed as a fraction, what is the probability that you will roll one of each type of symbol on a single throw of the dice?

Have you tried 22 tonight? I said 22.

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September 17th, 2020 at 8:43:26 AM permalink

Jackpot Yahtzee

O = Orange, B= Bell, D = Dollar, C = Cherry

1) OBDC ===>Prob 1 = 2/6 * 2/6 * 1/6 * 3/6 = 12/1296
2) OBCD ===>Prob 2 = 2/6 * 2/6 * 3/6 * 1/6 = 12/1296
3) ODBC ===>Prob 3 = 2/6 * 1/6 * 1/6 * 3/6 = 6/1296
4) ODCB ===>Prob 4 = 2/6 * 1/6 * 3/6 * 1/6 = 6/1296
5) OCBD ===>Prob 5 = 2/6 * 1/6 * 1/6 * 1/6 = 2/1296
6) OCDB ===>Prob 6 = 2/6 * 1/6 * 1/6 * 1/6 = 2/1296
.
.
.
.
24)CDBO===>Prob 1 = 1/6 * 1/6 * 1/6 * 1/6 = 1/1296

Prob = Prob1 + Prob2 + . . . + Prob24 = 29/324

Last edited by: ssho88 on Sep 17, 2020

Gialmere

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September 17th, 2020 at 4:32:17 PM permalink

Quote: ssho88
Jackpot Yahtzee

O = Orange, B= Bell, D = Dollar, C = Cherry

1) OBDC ===>Prob 1 = 2/6 * 2/6 * 1/6 * 3/6 = 12/1296
2) OBCD ===>Prob 2 = 2/6 * 2/6 * 3/6 * 1/6 = 12/1296
3) ODBC ===>Prob 3 = 2/6 * 1/6 * 1/6 * 3/6 = 6/1296
4) ODCB ===>Prob 4 = 2/6 * 1/6 * 3/6 * 1/6 = 6/1296
5) OCBD ===>Prob 5 = 2/6 * 1/6 * 1/6 * 1/6 = 2/1296
6) OCDB ===>Prob 6 = 2/6 * 1/6 * 1/6 * 1/6 = 2/1296
.
.
.
.
24)CDBO===>Prob 1 = 1/6 * 1/6 * 1/6 * 1/6 = 1/1296

Prob = Prob1 + Prob2 + . . . + Prob24 = 29/324

Correct!
--------------------------

A gambler hits a huge slot jackpot.

"What are we going to spend 10 million dollars on?" asks his wife.

"31 black"

Have you tried 22 tonight? I said 22.

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September 17th, 2020 at 9:22:18 PM permalink

There�s a small off-strip casino that has the following game which costs one dollar to play.

The player bets on heads or tails and a fair coin is flipped. The game ends when the player wins a flip. He gets paid $1.00 if he wins on the first flip, $1.00/2 = $0.50 if he wins on the second, $1.00/3 = $0.33 if he wins on the third, etc.

For calculation purposes, assume payoff amounts are exact (no rounding)

What is the house edge?

It�s all about making that GTA

ssho88

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September 17th, 2020 at 9:41:35 PM permalink

Quote: Ace2
There�s a small off-strip casino that has the following game which costs one dollar to play.

The player bets on heads or tails and a fair coin is flipped. The game ends when the player wins a flip. He gets paid $1.00 if he wins on the first flip, $1.00/2 = $0.50 if he wins on the second, $1.00/3 = $0.33 if he wins on the third, etc.

For calculation purposes, assume payoff amounts are exact (no rounding)

What is the house edge?

HE = 1 - ( 0.5 + 0.5^2 *1/2+ 0.5^3 *1/3+ 0.5^4 *1/4+ 0.5^5 *1/5+ 0.5^6 *1/6 . . . . . . . ) =1-∑0.5^n/n = 1 - ln2 = 0.306852821738692,
where n = 1 to infinity.

Ace2

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September 18th, 2020 at 8:34:12 PM permalink

Quote: ssho88

HE = 1 - ( 0.5 + 0.5^2 *1/2+ 0.5^3 *1/3+ 0.5^4 *1/4+ 0.5^5 *1/5+ 0.5^6 *1/6 . . . . . . . ) =1-∑0.5^n/n = 1 - ln2 = 0.306852821738692,
where n = 1 to infinity.

Correct. I forgot about that series...I solved it by integrating (1/e^x)*(1-1/e^x)*1/x from zero to infinity, which is Ln(2)

It�s all about making that GTA

Gialmere

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September 19th, 2020 at 8:07:06 AM permalink

If you roll a pair of dice, the probability of getting midnight is 1/36.

How many times do you need to throw a pair of dice so that the chances of getting midnight is more than 50%?

Have you tried 22 tonight? I said 22.

rsactuary

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September 19th, 2020 at 8:15:51 AM permalink

Define midnight please?

Gialmere

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September 19th, 2020 at 8:18:00 AM permalink

Quote: rsactuary
Define midnight please?

12, double six, boxcars

Have you tried 22 tonight? I said 22.

rsactuary

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September 19th, 2020 at 8:23:55 AM permalink

Quote: Gialmere
12, double six, boxcars

Interesting, I've never heard that term before. thanks

rsactuary

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September 19th, 2020 at 8:29:44 AM permalink

There are different ways to interpret your question, I think. But based on my interpretation, I get 25.

unJon

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September 19th, 2020 at 8:53:30 AM permalink

Quote: Gialmere

If you roll a pair of dice, the probability of getting midnight is 1/36.

How many times do you need to throw a pair of dice so that the chances of getting midnight is more than 50%?

(35/36)^n = 0.5

n * ln(35/36) = ln(0.5)

n = 24.6

So 25 rolls

The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

Gialmere

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September 19th, 2020 at 12:10:35 PM permalink

Quote: rsactuary
There are different ways to interpret your question, I think. But based on my interpretation, I get 25.

Quote: unJon
(35/36)^n = 0.5

n * ln(35/36) = ln(0.5)

n = 24.6

So 25 rolls

Correct!

Very good. You'd think off hand it'd be 18 rolls, but it ain't.

Note that this solve is similar to the birthday puzzles.

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

Have you tried 22 tonight? I said 22.

ssho88

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September 19th, 2020 at 6:29:27 PM permalink

Quote: Gialmere

If you roll a pair of dice, the probability of getting midnight is 1/36.

How many times do you need to throw a pair of dice so that the chances of getting midnight is more than 50%?

I guess you should change your question to .... chances of getting AT LEAST a "midnight " is more than 50%.

Am I missing something?

Ace2

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September 19th, 2020 at 6:41:40 PM permalink

Seems some of you are splitting hairs.

What is ambiguous about the midnight problem ?

It�s all about making that GTA

Wizard
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September 19th, 2020 at 7:19:35 PM permalink

Quote: Ace2
Seems some of you are splitting hairs.

What is ambiguous about the midnight problem ?

I understood what was being asked from the wording.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

unJon

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September 19th, 2020 at 9:02:07 PM permalink

Quote: ssho88
Quote: Gialmere

If you roll a pair of dice, the probability of getting midnight is 1/36.

How many times do you need to throw a pair of dice so that the chances of getting midnight is more than 50%?

I guess you should change your question to .... chances of getting AT LEAST a "midnight " is more than 50%.

Am I missing something?

I think the problem works as written. Solve for how many rolls until you have a 50% chance of getting your first midnight, and see if you get a different answer . . .

The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

ssho88

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September 19th, 2020 at 9:43:34 PM permalink

Quote: Ace2
Seems some of you are splitting hairs.

What is ambiguous about the midnight problem ?

English is not my mother tongue, so the question is clear and correct but I can't understand it well, sorry about that.

ssho88

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September 19th, 2020 at 10:12:06 PM permalink

Quote: unJon
Quote: ssho88
Quote: Gialmere

If you roll a pair of dice, the probability of getting midnight is 1/36.

How many times do you need to throw a pair of dice so that the chances of getting midnight is more than 50%?

I guess you should change your question to .... chances of getting AT LEAST a "midnight " is more than 50%.

Am I missing something?

I think the problem works as written. Solve for how many rolls until you have a 50% chance of getting your first midnight, and see if you get a different answer . . .

I will solve it this way :-
1 - Prob(WITHOUT MIDNIGHT) > 0.5
1 - (35/36)^n > 0..5
(35/36)^n < 0.5
n*ln(35/36) < ln(0.5)
n*(-0.02817) < -0.69315
n> 24.61
n = 25

IF the 50% chance change to 40% chance, then :-
1 - Prob(WITHOUT MIDNIGHT) > 0.4 ------------------Formula 1
1 - (35/36)^n > 0..4
(35/36)^n < 0.6
n*ln(35/36) < ln(0.6)
n*(-0.02817) < -0.5108
n> 18.13
n = 19

IF Formula 1 is correct, then it means AT LEAST a "midnight" . . . . .

I am nobody, just for discussion purpose, NO OFFENCE.

Ace2

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September 19th, 2020 at 11:15:15 PM permalink

What is your native language, Sho88?

It�s all about making that GTA

ssho88

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September 19th, 2020 at 11:30:10 PM permalink

Quote: Ace2
What is your native language, Sho88?

Mandarin

unJon

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September 20th, 2020 at 7:03:17 AM permalink

Quote: ssho88
Quote: unJon
Quote: ssho88
Quote: Gialmere

If you roll a pair of dice, the probability of getting midnight is 1/36.

How many times do you need to throw a pair of dice so that the chances of getting midnight is more than 50%?

I guess you should change your question to .... chances of getting AT LEAST a "midnight " is more than 50%.

Am I missing something?

I think the problem works as written. Solve for how many rolls until you have a 50% chance of getting your first midnight, and see if you get a different answer . . .

I will solve it this way :-
1 - Prob(WITHOUT MIDNIGHT) > 0.5
1 - (35/36)^n > 0..5
(35/36)^n < 0.5
n*ln(35/36) < ln(0.5)
n*(-0.02817) < -0.69315
n> 24.61
n = 25

IF the 50% chance change to 40% chance, then :-
1 - Prob(WITHOUT MIDNIGHT) > 0.4 ------------------Formula 1
1 - (35/36)^n > 0..4
(35/36)^n < 0.6
n*ln(35/36) < ln(0.6)
n*(-0.02817) < -0.5108
n> 18.13
n = 19

IF Formula 1 is correct, then it means AT LEAST a "midnight" . . . . .

I am nobody, just for discussion purpose, NO OFFENCE.

Yes this is correct. But here I think getting at least one midnight is same as getting first midnight or getting midnight to a native English speaker. Implication is you stop after getting a midnight and what is the median number of rolls you stop after.

The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

Gialmere

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September 20th, 2020 at 9:16:59 AM permalink

If consonants are free, determine the various values of the five vowels A, E, I, O, and U from the following clues:

Audacious costs $260
Equivocation costs $340
Inimitable costs $255
Onomatopoeia costs $435
Unambiguous costs $225

Have you tried 22 tonight? I said 22.

ThatDonGuy

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September 20th, 2020 at 2:35:30 PM permalink

Quote: Gialmere
If consonants are free, determine the various values of the five vowels A, E, I, O, and U from the following clues:

Audacious costs $260
Equivocation costs $340
Inimitable costs $255
Onomatopoeia costs $435
Unambiguous costs $225

Simple case of five equations in five unknowns

A = 60
E = 75
I = 40
O = 50
U = 25

Gialmere

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September 20th, 2020 at 9:54:46 PM permalink

Quote: ThatDonGuy
Simple case of five equations in five unknowns

A = 60
E = 75
I = 40
O = 50
U = 25

Correct!
--------------------------------

I watch Jeopardy and Wheel of Fortune every night now which is weird because I don't remember turning 74 years old.

Of all the jobs made obsolete by technology, you�d think Vanna White�s would have been among the first.

Have you tried 22 tonight? I said 22.

Gialmere

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September 21st, 2020 at 8:24:42 AM permalink

Solve the KenDoku above.

Similar to Sudoku, the puzzle consists of a grid containing blocks surrounded by bold lines.

The object is to fill all of the empty squares so that the numbers 1 to 5 appear exactly once in each row and column.

The numbers in each block must produce the result shown in the top left corner of the block using addition. For example, the numbers in the three-square block in the upper left must add up to 9.

Have you tried 22 tonight? I said 22.

ThatDonGuy