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ChesterDog
ChesterDog
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July 28th, 2022 at 4:12:01 AM permalink
Quote: ThatDonGuy

Here's my problem of the day:

500 men each toss a Ronald Reagan Presidential Dollar Coin.
Simultaneously, 500 women each toss a Sacagawea Dollar Coin.
What is the probability that the number of Reagan coins that come up heads is exactly one more than the number of Sacagawea coins that come up heads?
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I used a spreadsheet to get about 2.51747%.
charliepatrick
charliepatrick
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July 28th, 2022 at 5:46:06 AM permalink


2 .25
3 .234375
4 .21875
5 .205078
10 .160179
50 .078029
100 .055791
500 .025175
1k .017821
Ace2
Ace2
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July 28th, 2022 at 12:24:34 PM permalink
My kinda problem

c(1000,499) / 2^1000 =~ 0.025174669

A very good approximation is (500π)^(-.5) =~ 0.025231325. Works well for any n over about 30

This works because: if you look at the binomial distribution for p=0.5 for any n, take the products of all the individual probabilities (groups of 2) and then sum them, it comes out to c(2n, n-1) / 2^(2n). Hope I said that right. I just discovered this by trying some low n's and saw that the product-sums all fit into Pascal's triangle

The approximation works due to application of Sterling's formula and cancellation of terms for the probability that a result exactly matches expectations when p=.05. In this case, 499/100 is close enough to 500/1000. Negligible difference

Last edited by: Ace2 on Jul 28, 2022
Itís all about making that GTA
Wizard
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Wizard
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July 28th, 2022 at 12:51:26 PM permalink
Quote: Ace2

My kinda problem

c(1000,499) / 2^1000 =~ 0.025174669

A very good approximation is (500π)^(-.5) =~ 0.025231325. Works well for any n over about 30

This works because: if you look at the binomial distribution for p=0.5 for any n, take the products of all the individual probabilities (groups of 2) and then sum them, it comes out to c(2n, n-1) / 2^(2n). Hope I said that right. The approximation works due to application of Sterling's formula and cancellation of terms for the probability that a result exactly matches expectations when p=.05. In this case, 499/100 is close enough to 500/1000


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That's very interesting. However, I didn't understand why that approximation works. I don't dispute it does, of course. Can you elaborate on how pi fits into it?
It's not whether you win or lose; it's whether or not you had a good bet.
Ace2
Ace2
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July 28th, 2022 at 1:35:08 PM permalink
Let's say you want to know the probability of getting exactly 50 heads in 100 flips. The exact answer would be given by c(100.50) / 2^100

Stirling's approximation says n! =~ (n/e)^n * (2πn)^.5

c(100,50) = 100! / (50!)^2. Use Stirling for those factorials, divide by the 2^100 and the terms cancel out nicely to (50π)^(-.5) = 0.07978 vs the exact answer 0.07958

The problem at hand is answered exactly with: c(1000,499) / 2^1000.<a> This is almost the same as the probability of getting 500 heads in 1000 flips, which would be c(1000,500) / 2^1000 =~ (500π)^(-.5)

Incidentally, Stirling's approximation is already accurate within 1% for 9!, though you wouldn't need to approximate a figure that low

<a>I think so anyway. Answer hasn't been confirmed yet
Last edited by: Ace2 on Jul 28, 2022
Itís all about making that GTA
ThatDonGuy
ThatDonGuy
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Thanks for this post from:
ChesterDogcharliepatrick
July 28th, 2022 at 4:40:52 PM permalink
Quote: ThatDonGuy

Here's my problem of the day:

500 men each toss a Ronald Reagan Presidential Dollar Coin.
Simultaneously, 500 women each toss a Sacagawea Dollar Coin.
What is the probability that the number of Reagan coins that come up heads is exactly one more than the number of Sacagawea coins that come up heads?
link to original post



Let N be the number of Reagan coins that come up heads.
This means there are N-1 Sacagawea coins that come up heads, and 500-(N-1) = 501-N Sacagawea coins that come up tails.
Reagan heads + Sacagawea tails = 501.
Thus, each set of tosses where the Reagan heads = 1 + the Sacagawea heads maps to a selection of 501 of the 1000 persons; each man in the selection tossed a head, and each woman tossed a tail. Since the probabilty of tossing a tail = the probability of tossing a head, each of the 2^1000 possible sets of tosses is equally likely.

The solution is C(1000, 501) / 2^1000, or about 1 / 39.7224689 (0.025174669)
Note that this is also C(1001, 499) / 2^1000

Ace2
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July 28th, 2022 at 6:03:22 PM permalink
Do I get a beer ?
Itís all about making that GTA
ThatDonGuy
ThatDonGuy
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July 28th, 2022 at 8:18:31 PM permalink
Quote: Ace2

Do I get a beer ?
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Wizard, give Ace2 one of my accumulated beers at the next WoV gathering
Wizard
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Wizard
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July 29th, 2022 at 2:50:22 AM permalink
Quote: ThatDonGuy


Wizard, give Ace2 one of my accumulated beers at the next WoV gathering
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I would be proud to.
It's not whether you win or lose; it's whether or not you had a good bet.
Gialmere
Gialmere
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August 1st, 2022 at 8:00:03 AM permalink
It's easy Monday. Here are two easy puzzles...

This one is an easy classic.



A local newspaper is made of 16 large sheets of paper folded in half and has 64 pages altogether. The first sheet contains pages 1, 2, 63, 64.

If you pick up a sheet containing page number 45. What are the other pages that the sheet contains?

What's a simple formula to quickly get the answer?



There are three hockey pucks on the ice in a rink. A player shoots one of them between the two others. He continues shooting one between the other two but never shoots the same puck twice in a row.

What's the minimum number of shots it'll take him to return the pucks to their original positions on the ice?
Have you tried 22 tonight? I said 22.

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