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## Comments

789Subscriber1,048Pro1)So here it would mean if we say "ok you flip two coins, look at them, and tell me nothing if they are both tails otherwise say something" then you can quickly see it's1/3looking at the 4 total combos being (TT/HT/TH/HH). However I'll write out the parallel solution since the next one is a bit more confusing, you can compare it to this one.-Each coin has 2 possibilities

-There are 4 total combos

-3/4 of the total combos are considered (i.e., qualify for "one of them is heads")

-Only one of the three is valid (i.e., qualify for "both heads")

3)Extracting the same logic, then we say "ok let's take a stranger with 2 children, and you tell me something if and only if one of the children is a boy born on Tuesday, otherwise you say nothing", then it's13/27because:-Each children has 14 possibilities (7 days times 2 sexes)

-There are 14x14 = 196 total combos

-27/196 are considered (i.e., qualify for "at least 1 is a boy born on Tuesday"). This is found by 1-(13/14)^2, since (13/14)^2 don't qualify.

-13 of the 27 are valid (i.e., qualify for "both are boys"). We have the first boy being ThursdayBoy times 7 combos for the second, then vice versa and subtract the one double counted combo where both are ThursdayBoy.

2)is the obviously the same as question 1) if we're assuming each child can only be a boy or a girl.428Subscriber.

1/2, 1/2 and 13/27

Pretty unintuitive how the Tuesday changes it from .33 to .48

428Subscriber1,592Subscriberthe Tuesday info has no bearing on the calculations.

Mrs Smith either has a boy and a girl, or two boys.

if we didn't know the sex of one, then she could have B/G, G/G, B/B or G/B

but once we know the sex of one - she can only have B/G or B/B

unless we're counting hermaphrodites...

1,048ProLike you said yourself, B/G and G/B are different entities. The reason it's slightly under 50% in this case is because the scenario of *both* boys being born on Tuesday is only counts 1 time.

1,592Subscriberit's like saying your Villain is wearing a blue shirt, what is the probability he's got AA?

one thing has nothing to do with the other!

1,048ProAnyway the question is intentionally layered to not be trivial, but if you think of a simpler version of the same problem, it may be more obvious. Let's say that Mrs Smith has 1 child. I will ask Mrs Smith the gender/birthday, and I will relay the information to you if and only if it's a boy who is born on any day *other* than Christmas. Otherwise I won't tell you anything.

I come back and tell you nothing. What are the chances her child is a boy?

1,592Subscriberyou're blowing my mind, here.

if we didn't know the sex of one child (or side of a coin) - then there's more combinations to consider. but once we know one, it seems that it no longer should be considered in the possibilities.

Mrs Smith has 2 kids. what are the odds that BOTH are boys - is a very different question than Mrs Smith has 2 kids, one is a boy - what are the odds that the other is a boy.

The other can only be a boy or girl. 50/50. The sex of the first doesn't influence the sex/gender of the other. The date of birth also doesn't effect the calculations as the question doesn't ask for probability of when the other might be a boy.

same holds true with coins.

I've got 8,455 coins in which 8,454 are heads - what's the probability that the 8,455th is heads? That last coin's probability is independent of the other coins. Again, only 2 possibilities - so 50/50.

Q3 - didn't say anything about "I'll tell you something if..." like your X-mas example.

It just said one boy was born on a Tuesday. That info is as helpful as saying one is a boy who likes to play with dolls. Maybe the question was worded incorrectly - but the probabilities seem to all be the same.

What's the probability that I'm wrong against Don Ding?

428Subscriber184Subscriber1,048ProWhat do you think the answer to 1) is?

Note that the problem is not quite the same as "Mrs Smith has 2 kids, one is a boy - what are the odds that the other is a boy.", which you're right the answer would be 50/50. Because in that scenario, there is no differentiation between G/B or B/G combos whereas there is in the OP's problem (which, as the very first thing I posted in this thread, I believe is slightly ambiguous but the most correct interpretation)

184Subscriber1,048ProTuesdayBoy/WednesdayBoy

WednesdayBoy/TuesdayBoy

whereas there's only 1 of

Tuesdayboy/Tuesdayboy

702SubscriberBoth born on Tuesday:

Tues/Tues

There are two ways for one boy to be born on Wednesday and one on Tuesday:

Wed/Tues

Tues/Wed

Remember that each child is not interchangeable. For ease let's say the first born is always blond and the second born is always has brown hair.

Blond born on Tuesday, Brown born on Wednesday is not the same as Blond-Wednesday, Brown-Tuesday, hence two options not one.

1,048Pro184Subscriber6,078AdministratorLeadProLet's say there is 1 car behind one of three doors A, B and C. We choose what's behind door B. Then we are told that door A does not contain the prize. Will changing our decision to door C increase the odds that we win?

What about someone at random who does not know the results of anything choosing door A and its empty. Does changing our selection to door C make a difference?

1,592SubscriberQ3 never mentioned anything about "you tell me something if and only if one of the children is a boy born on Tuesday, otherwise you say nothing"

so there are no circumstances provided that would affect the result I'm with Roger - they're all 50/50.

2 coins can be H/H, H/T, T/H & T/T

with one being H - then you're left with H/H or H/T... those are your only options!... 50/50

since the value of one is already known (the first H) - the result is only affected by one of two possibilities from the other coin... T/H and T/T are never possibilities because we've already said that coin #1 was and is only H Before any info revealed - we have a 33% chance of being correct

After info provided - our odds increase to 50/50

technically - changing our choice does not give us better odds

some would argue that by switching, we've gained the equity given away with the door opened - so we only have a 33% chance of winning if we stay but a 66% chance of winning if we switch. But I don't buy that logic.

when a door is opened and two doors are left - we can "choose" to stay with our original selection or choose to switch. either way, we've got the same probability because the problem has now changed to choosing one of two doors. the newly opened door does not provide any additional info as to whether either selection is better.

860Member