Mahjong - probability

[bossteck]

Hi,

Yes... you got me. I stand corrected. I did some further thinking:
Here's my second attempt.

For sake of simplicity, let's say we bet on the number six.

Total number of outcomes for 3 die game = 6 x 6 x 6 = 216
Case1: Total number of outcomes with 3 sixes = (1 x 1 x 1) = 1
Case2: Total number of outcomes with 2 sixes = (1 x 1 x 5) x 3 = 15
Case3: Total number of outcomes with 1 six = (1 x 5 x 5) x 3 = 75

So, from the point of view of the person betting, chance of winning something from his bet is 1/2, BUT wait...

Chance of winning 3 times his bet is 1/216
Chances of winning 2 times his bet is 15/216
Chances of winning 1 time his bet is 75/216 and last but not least
Chances of losing is 108/216

[timebomb]

Hmm, tough. At first glance, I would immediately say it should be the same for both. But thinking about it further, the big hospital needs an extra 10 boys and 10 less girls to achieve the ratio but for the small hospital, it needs only 2 of each.

Let's say there's a 3rd hospital that handles only 10 births a day. For it to hit 60:40, it takes only 6 boys and 4 girls This is very likely because the smaller the number, the more likely the ratio is to fluctuate.

So my final answer is "the small hospital".

Loh K L

[bossteck]

For the hospital problem, my guess is, it is the same.
Timebomb, you see it as needing 10 more and 10 less, I see it as having a larger margin of error.

Cheers

[timebomb]

Speaking of boys and girls, this other puzzle from Marilyn vos Savant will really get you. It looks like the simplest question in the world but think carefully before answering. Like I've said before, things always aren't what they seem.

Here's the question:

A couple has 2 children. One is a boy. What is the probability the other is a girl?

Loh K L

[bossteck]

At the rate we're going... we are going to need to have a new folder

Here's my solution:

4 possible outcomes

Boy, Boy - 25%
Boy, Girl - 25%
Girl, Boy - 25%
Girl, Girl - 25%

Given that one's a boy... the girl, girl case is now invalid.
So 3 cases are left.

Boy, Boy - 33.3%
Girl, Boy - 33.3%
Boy, Girl - 33.3%

Cases 2 and 3 are the desired outcomes... hence 66.6%. Correct or not?

Cheers

[msmurf]

Cases 2 and 3 are the desired outcomes... hence 66.6%. Correct or not?

Cheers

yo eric,

my answer also 2/3. right?

[checkerboard]

The answer to the hospital question is the small hospital.

When you have a larger sample size, you tend closer to the true probability (by law of large numbers). It is easier to observe deviations from natural probability with small samples.

Applying this to the gambling question: with more gamblers, the natural outcome (that the bets are more spread out) will occur more often than not, so the gangsters will win, when there are many gamblers and also, when they play the game over a long time (when the beneficial odds of them winning because of spread out bets also plays out over the many games), the gangsters earn more money.

[checkerboard]

Actually, to the Boy:Girl question, I think the answer is still 0.5.

The probability of the second child being a boy or girl is independent of the first.

I will view it this way:
Think of coin toss. Say we toss the a coin.
First time: Head
What is the probability of the second toss being a tail?
Still half isn't it?

If you think it is 2/3, I will like to play this game with you. I will flip a coin once, then you bet on the next flip, and according to you, you will be betting the opposite of what I flip each time. I think the chance of you winning is still half isnt it. Just like you bet BIG/SMALL, if the first opening is BIG, you do not always bet SMALL the next time thinking it has higher chance of winning isn't it?

When people think it is 2/3, they may be thinking the next outcome has a higher chance of being an outcome that will bring the sample probability closer to the natural probability. But I think in this case, for instance coin toss, each toss is an independent event. So it is half, just like i think the probability of the baby being girl is still half.

P(one girl|one boy) = P(one girl)P(one boy|one girl)/[P(one girl)P(one boy|one girl)+P(one boy)P(one girl|one boy)] = (3/4)*(1/2)/{3/4*1/2+3/4*1/2)} = 1/2
You learn nothing from the fact that one is boy, because the next one is still an independent event.

As again, open to discussion.

[bossteck]

Think you interpreted the questions as...

If the first child is a boy, what is the probablity that the second is a girl.

But if you read the question again, there's no mentioned of first or second child, it merely states that if one of them is a boy.

Trick question?

Cheers

[timebomb]

If the question had been - "The older child is a boy. What's the probability of the second child being a girl?", then Checkerboard's answer would definitely be the right one. But according to Marilyn, Eric and Koh are right. If a couple has 2 children and one is a boy, the probability of the other one being a girl is 2/3.

The key lies in reading the question carefully. It's more of a case of interpreting the meaning of the question rather than just thinking about the math. So it's more of a language puzzle than a mathematical one.

Because if for instance, you bump into an old friend and he tells you, "I have 2 kids. One is a boy". You can quite safely assume the other has to be a girl. If not, he would have said, "I have 2 kids. Both are boys".

Loh K L

[checkerboard]

er, actually, i made no assumptions on sequences if you read carefully. I think the confusion lies in he use of the word "second". What I meant was the other child. In fact, whether it is the first or second child if we consider sequences, there is no difference, the answer is still half, because they are independent events.

Why not think about it in this way. No sequence case:
Mrs Ong and Mrs Tan each has one kid.
They bring them to the park. Mrs Ong kid is a boy. What is the probability that Mrs Tan kid is a girl?

Sequence case:
Your wife has given birth to your first child. It is a boy. Two years later, you have a second child.
What is the probability that the second child is a girl?

Let's see you answer for both. Mine is the same for both cases. Half.

[bossteck]

Sequence case:
Your wife has given birth to your first child. It is a boy. Two years later, you have a second child.
What is the probability that the second child is a girl?

Let's see you answer for both. Mine is the same for both cases. Half.

My answer for both your questions are 0.5, just like you. But

The sequenced case you have presented already has an added restriction. That is:

First Child is a boy, so there're only 2 possible cases:

[First child, second child]
CaseA: Boy, Girl
CaseB: Boy, Boy

However, the earlier questions was, one of the child is a boy.

[First child, second child]
Case1: Boy, Girl
Case2: Boy, Boy
Case3: Girl, Boy <- this is the case which is not possible in your question but possible in the one presented by Timebomb.

[checkerboard]

Eric,

Maybe this is where the difference is:

There are only 3 possible outcomes. Not 4.

1) 2 Boys
2) 2 Girls
3) 1 Boy and 1 Girl.

You know there is 1 Boy already. So Outcome 2 is out.
Therefore, probability of the other baby being a girl is 0.5

[checkerboard]

Bossteck, you are right, the sequence does have a restriction.

Timebomb question, is a non-sequence one.

I am not saying that the questions are the same.
I am just trying to illustrate that independent events probability do not depend on sequences.

[timebomb]

Guys, remember the counter-example that Checkerboard raised, the one about a pistol with 3 bullets randomly inserted within the 6 chambers and six persons taking turns at it? Are you aware that such a game is known as "Russian Roulette"? I don't know if anyone in the world actually plays such a game but it was made famous in the movie "The Deer Hunter". In Russian Roulette, the game involves 2 players and there's only 1 bullet in the pistol. Spectators lay their bets on who blows his own head off first.

There's this very interesting puzzle in one of my books that is a variation of "Russian Roulette". This is the puzzle:

Imagine you are involved in a game of Russian Roulette with another player But instead of just one bullet, there are 2 in the pistol. The 2 bullets are inserted in successive chambers. In other words, they are just next to each other. The pistol has 6 chambers. Your opponent goes first. He takes up the gun, puts it to his head and pulls the trigger. "Click", he's one lucky fellow; he clicked on an empty chamber. It's now your turn. You are allowed to spin the barrel again if you wish.

Here's the question: Should you spin the barrel or you should simply take up the gun and trigger it?

Loh K L

[bossteck]

Wow... tough one.
At the moment, all I can muster is that there are 5 different arrangments.

4/5 chance of surviving if it is spun? Not even sure about this.

[checkerboard]

I think the answer is No. We should not spin the barrel.

There are 5 ways in which the bullets can be placed. The second bullet is "irrelevant". You survive if the first bullet is placed in chamber 1,3,5. So, when the game starts, you are at an advantageous situation being the second player, since you only die if the first bullet is at the 2,4 slots. (i.e, Prob of surviving = 2/5 for you and 3/5 for Player 1)

Now, player 1 shoots, he survived. We know combination 1,2 is out. There is 4 more possible outcome of which in 2 you will die. Now, your prob is 1/2. If you think that by respinning the barrel you have a higher chance of survival at 2/5, then you are possibly wrong i think. Because, once you re-spin, you have effectively re-set the game with you as Player 1, because it is your turn to shoot. Therefore the probability of you dying is 3/5, which is greater than 1/2 if you do not spin .

Hence, I think you should not spin again.

[timebomb]

Hence, I think you should not spin again.

Checkerboard, your answer is correct but the logic behind it is wrong. If there are 6 chambers in a gun and 2 of the chambers are loaded, the probability of the player dying is 2/6 which is equal to 1/3 which is 33.3&#37;. For this calculation, we can ignore the fact that the loaded chambers are next to each other but if you're the next player to take your turn with the gun, you have to take this into consideration.

It could be you misunderstood the puzzle so I'm going to lay it out here for you. The gun has 6 chambers. 2 are loaded with bullets. The loaded chambers are next to each other. So, if E represents an empty chamber and L is a loaded one, it will look like this:

E1, E2, E3, E4, L5, L6

or it could look like this:

E4, L5, L6, E1, E2, E3

Loh K L

[XnSdVd]

You people scare me.. lol =)

[checkerboard]

Ok, consider six scenarios, I missed out the case of L6,L1 which is possible.

Adding this case, the prob of player 2 dying is 1/3. Player 1 prob of dying is 2/3. Given that Player 1 shoots and survive, this eliminates 2 scenarios, 1,2 and 6,1.
So same argument. Now player 2 prob of dying is now 1/2. If he respin the wheel, he is effectively player 1 where Prob of dying is 2/3. So he should not re-spin the wheel.

Exact same argument applies adding in the last scenario.