Easy math puzzles

Sorry to step on the craps puzzle, but I think we can handle two at the same time.

On Jan 1 a hospital nursery has 3 boys and an unknown number of girls. On Jan 2 a baby is born and added to the nursery. On Jan 3 a random baby is selected from the nursery and it is a boy.

What is the probability the baby born on Jan 2 was a boy?

Quote: Ace2

Quote: charliepatrick

5/9 pays 6 to 4]

Out of curiosity, why did you use 6:4 instead of 3:2 ? I’ve also seen UK sportsbooks use 6:4.
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This comes from old racing odds used in the UK (US tracks tend to use 6/5 7/5 3/2 8/5 9/5 etc). Also we used to have Pounds, Shillings and Pence which is why some early prices were things like 100/6 100/7 100/8 and showed how many shillings you needed to put on to win £5 = 100/-. I'm guessing 1/8th (e.g. 13/8) were used as there was a half-crown (2/6) coin, and 6/4 used as it fitted into the Ev 11/10 5/4 6/4 7/4 2/1 9/4 5/2 11/4 3/1 100/30 7/2 4/1 simple pattern. (Some small greyhound tracks don't use 6/5 11/8 13/8 15/8 85/40 17/2.)

In the old days Starting Prices were from an "average" of "leading" on-course bookmakers. This was done via a group of people going round checking the prices and, as the race started, having a "huddle". It wasn't a mathematical average but a fair price that was available on a number of good bookmakers. They, nearly always, used the nearest odds from the approved list, although in some big race tracks I've known prices such as 95/40, 35/1, 7/5 to be used.

Nowadays, at UK racetracks, Starting Price is calculated using a computer which can directly read the bookmakers prices. I'm not sure whether they still use on course bookies or have moved to include off course ones, but it uses a sample of (say) 12 and takes the median (i.e. a price available on half of the sample). This means it's a price from a specific bookie, so you now do get intermediate prices such as 7/5 8/5 11/5 16/5.

Interesting. Incidentally, 6:4 is the true inverse of winning a 5 or 9 as in 6 ways to lose and 4 ways to win

My zero-edge craps version:

Point win on hard six pays 3:1.
Point win on five (2-3) pays 1:2

So 3.5% of resolutions are affected with two changes for a score of 0.07. The only caveat is bets should be made in even amounts due to the 1:2 payout on a 2-3 five point win, but the only adjustment would probably be for a $25 bettor to increase to $30.

Zero-edge can also be accomplished as:

Point win on hard six/eight pays 2:1.
Point win on five (2-3) pays 1:2

Though this has a higher score of 0.14 since a higher percentage of resolutions are affected and there are three changes instead of two

There are probably more ways to get to zero edge with minimal changes and practical payout ratios, though I don’t know of any method to find them besides trial and error. I believe there’s no way to get there with less than two changes since none of the individual outcomes’ probabilities are divisible by the 7 of 495

Wiz Idea #3

Point win on a 4 pays 3-2.
If point is a 3-3 and made with another 3-3, pays 11-10.
All other rules stay the same.

I know that second rule seems nearly worthless, but I had to find 0.0002525253 somewhere.

Wiz Idea #4

My best yet!

If point is established and made with a hard 6 or hard 8, then win is 19 to 5.

Quote: Wizard

Wiz Idea #3

Point win on a 4 pays 3-2.
If point is a 3-3 and made with another 3-3, pays 11-10.
All other rules stay the same.

I know that second rule seems nearly worthless, but I had to find 0.0002525253 somewhere.

Wiz Idea #4

My best yet!

If point is established and made with a hard 6 or hard 8, then win is 19 to 5.
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Great answers!

They both beat my score of 0.07. Your second version only affects about one half of one percent (1/198) of all resolutions

Good spot since it creates 1/11, so can be factored against 495=5*99.

Chance of coming out using hard 6 = 1/36
Chance of then winning with hard 6 = 1/11
Hence chance of either 6 or 8 = 1/198 = .5/99
House edge required = 7/495 = 1.4/99 (= 2.8/198)
Hence additional payout needs to be 1.4/.5 = 2.8
This means the actualy payout required = 3.8 = 19/5.

Deleted

Quote: Wizard

Sorry to step on the craps puzzle, but I think we can handle two at the same time.

On Jan 1 a hospital nursery has 3 boys and an unknown number of girls. On Jan 2 a baby is born and added to the nursery. On Jan 3 a random baby is selected from the nursery and it is a boy.

What is the probability the baby born on Jan 2 was a boy?
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Since no one else seems to have bothered trying to answer this one, I'll give it a shot:

Quote: Wizard

Sorry to step on the craps puzzle, but I think we can handle two at the same time.

On Jan 1 a hospital nursery has 3 boys and an unknown number of girls. On Jan 2 a baby is born and added to the nursery. On Jan 3 a random baby is selected from the nursery and it is a boy.

What is the probability the baby born on Jan 2 was a boy?
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Quote: ThatDonGuy

My Answer

1/2, assuming the laws of genetics still apply.

Other than that, I don't know where to start working on this one. Care to clue us in?

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Incorrect. I'll give a little more time to see if you wake up the problem.

Quote: Ace2

Show Spoiler

I get 4/7.

Let BB represent the probability a boy is born and a boy is chosen at random. Irrespective of the starting number of girls, the probability of BB is 4/3 of GB. We can eliminate BG and GG since we know a boy was chosen. So 4 / (4+3) =0.57 %

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I agree!

Addendum to hospital nursery answer: