Poll

8 votes (50%)
6 votes (37.5%)
3 votes (18.75%)
2 votes (12.5%)
6 votes (37.5%)
2 votes (12.5%)
3 votes (18.75%)
2 votes (12.5%)
8 votes (50%)
5 votes (31.25%)

16 members have voted

ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
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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
gordonm888
Joined: Feb 18, 2015
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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, were dead. And a thousand thousand slimy things lived on, and so did I.
ssho88
ssho88
Joined: Oct 16, 2011
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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
Gialmere
Joined: Nov 26, 2018
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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.
ssho88
ssho88
Joined: Oct 16, 2011
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Thanks for this post from:
Gialmeregordonm888
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
Gialmere
Joined: Nov 26, 2018
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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.
Ace2
Ace2
Joined: Oct 2, 2017
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September 17th, 2020 at 9:22:18 PM permalink
Thereís 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?
Itís all about making that GTA
ssho88
ssho88
Joined: Oct 16, 2011
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September 17th, 2020 at 9:41:35 PM permalink
Quote: Ace2

Thereís 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
Ace2
Joined: Oct 2, 2017
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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)
Itís all about making that GTA
Gialmere
Gialmere
Joined: Nov 26, 2018
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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.

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