## Poll

19 votes (48.71%) | |||

13 votes (33.33%) | |||

6 votes (15.38%) | |||

2 votes (5.12%) | |||

12 votes (30.76%) | |||

3 votes (7.69%) | |||

6 votes (15.38%) | |||

5 votes (12.82%) | |||

11 votes (28.2%) | |||

8 votes (20.51%) |

**39 members have voted**

Quote:ThatDonGuyHere'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

2 .25

3 .234375

4 .21875

5 .205078

10 .160179

50 .078029

100 .055791

500 .025175

1k .017821

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

Quote:Ace2My 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

link to original post

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?

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

Quote:ThatDonGuyHere'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

Quote:Ace2Do I get a beer ?

link to original post

Wizard, give Ace2 one of my accumulated beers at the next WoV gathering

Quote:ThatDonGuy

Wizard, give Ace2 one of my accumulated beers at the next WoV gathering

link to original post

I would be proud to.

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?