Joined: 11/30/2008
Posts: 650
Location: a little city in the middle of nowhere
I just realised that for any finite amount of money, you will lose money using this strategy. if you have an infinite amount of money, why bet? you already have enough money to burn.
It's been well proven that no betting strategy can beat roulette, unless you note down every number in order to figure out if there is a wheel bias or if the croupier has a signature throw.
Actually, doubling your bet every time if you lose is actually a winning strategy if you are betting on a 50/50 game.
Didn't roulette have a couple of spots where the bank always wins and everyone loses? That makes it a non-50/50 game. And of course the mentioned bet limits are there precisely to make it a losing strategy in the long run. They say that in Las Vegas if you have a pattern like the Martingale system, what they tell you is "welcome".
There's one game, though, which does have a winning pattern, but it's pretty complicated: Blackjack. This is the only winning pattern which is both legal and will get you banned from all casinos in the US if you use it. (Not that many people are able to do it, as it's so complicated.)
(OTOH the method works only in classic blackjack where one set of n decks is always played through before a new shuffling. Many casinos nowadays prevent the strategy by shuffling all used cards back into the deck right away, which completely defeats the winning strategy.)
There's one game, though, which does have a winning pattern, but it's pretty complicated: Blackjack. This is the only winning pattern which is both legal and will get you banned from all casinos in the US if you use it. (Not that many people are able to do it, as it's so complicated.
(OTOH the method works only in classic blackjack where one set of n decks is always played through before a new shuffling. Many casinos nowadays prevent the strategy by shuffling all used cards back into the deck right away, which completely defeats the winning strategy.)
You're refering to card counting, the movie 21 gave a huge number of myths about card counting. Mainly making out you needed to be genius to do it. Concentration is the hardest part of doing it, but most people will use cancellation methods to count faster. In Atlantic city there are not allowed to ban players for being card counters so they get round it by shuffling earlier. It can be beat using shuffle tracking methods but it means being at the table for a lot longer.
In Atlantic city they are not allowed to ban players for being card counters so they get round it by shuffling earlier.
Technically, counting cards at blackjack is not illegal, but it's also not illegal for a casino to throw your card-counting ass off their property and tell you to never come back. They can't have you arrested for counting cards, but they could have you arrested for trespassing after they kick you out.
Here's my interpretation of the Monty Hall problem, if the host picks a door at random:
There's a 2/3 chance you picked a goat on your first try. When the host picks a random door, there is a 1/2 chance that he picks a goat (making switching a good strategy) and a 1/2 chance that he picks the car (so you have no chance of winning).
There is a 1/3 chance that your first choice was a car. In this case, staying where you are is certainly the best choice.
So, there is a 1/3 chance that you will be in a situation where you should switch, a 1/3 chance that you will be in a situation where you should stay, and a 1/3 chance that you have no chance of winning. Therefore, it doesn't matter whether or not you switch.
One analysis I read said to try exaggerating the numbers in the problem. So say there were 1000 doors, and 999 had a goat behind them, and when you first picked a door, the host opens 998 other doors that had a goat. Now, do you really think you picked the one door out of a thousand correctly on your first try? Obviously the odds are that you didn't, so you have a 99.9% chance of winning by switching to the other door.
put yourself in my rocketpack if that poochie is one outrageous dude
Is it just me, or you are deforming the Monty Hall problems ? The host will always remove a goat. Else the statistics isn't the same.
From wikipedia :
The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it.
Joined: 3/11/2004
Posts: 1058
Location: Reykjavík, Ísland
To understand more clearly why you should always switch in the Monty Hall game:
Instead of only 3 doors, imagine that you have one THOUSAND doors, 999 of which contain goats, and one contains a car. You choose ONE door. Now the game show host opens 998 of the other doors, all of which contain goats. Now you are left wiht two choices. Your original choice, or the one other door not opened by the host. NOW what do you think, should you switch? The actual Monty Hall problem is in principle exactly the same, but with fewer doors so the advantage of switching is much less pronounced. (and thus less obvious).
I have another probability problem. It may be very trivial, but atm I just can't think of a solution for it.
Let's say, we have two bowls of capsules. One contains red ones, the other the exact same amount of blue ones. Both, the red and blue ones have numbers from one to "x = amount of capsules in total / 2" on them. Now we mix all of them up into a bigger bowl and pick x capsules at random. The other half of them gets flushed down the toilet. Some kind of wizard then refills the bowls with new capsules (x red and x blue ones), but you keep the capsules you already have.
How high is the probability for each step that we already have a collection of capsules from 1-x? How high that we have all blue ones from 1-x? How high that we already have all of them?
What if the wizard used some additional magic? We only pick one capsule at a time from the mixed up bowl. Whenever we do that the capsule with the corresponding number in different color disappears instantly. How does this affect the solution?
Somehow I'm really confused over this atm.
Probability, after having chosen X capsules, that we have each number represented:
First pick doesn't matter
Second pick must be anything except what you picked first (2x-2 options)
Third pick must be anything except what you picked in the first two times (2x-4 options)
Et cetera.
So you have a 1 in PI(2x-2n) chance. (PI is like Sigma, except it does multiplication instead of addition).
If you constrain it to be just one color, then for each pick, you must avoid x capsules, so your odds are:
First pick: x in 2x
Second pick: x-1 in (2x - 1)
Third pick: x-2 in (2x - 2)
Et cetera
Total is PI((x - n) / (2x - n))
If, every time you pick a capsule from the bowl, the corresponding differently-colored capsule is removed, then your odds of getting every number are 1. Your odds of getting every number in the same color are improved, because you now must simply make a 50% toss for each number that you get the right color. Thus your odds are 1 in 2^(x-1) (first pick's color doesn't matter).
Pyrel - an open-source rewrite of the Angband roguelike game in Python.
Probability, after having chosen X capsules, that we have each number represented:
First pick doesn't matter
Second pick must be anything except what you picked first (2x-2 options)
Third pick must be anything except what you picked in the first two times (2x-4 options)
Why do you subtract 2 and 4 instead of 1 and 2?
Also thanks.
Edit: I figured it out. Thanks again.
Hey guys, interesting problem. Did you know the contestant actually has a higher chance of winning if he switches after the host removes a door? Very counter intuitive, but it's true. There's a wiki page on it somewhere, maybe someone knows what it's called.
Scratch that.
To get all numbers: (x is the total amount of capsules / 2 again)
100% for the first one, 2x-1 is the amount of capsules left, there's one that doesn't improve our status among them. 1-1/(2x-1) should be the odds for improving our status, shouldn't it?
For the third pick we have two possibilities. We have a 1/(2x-1) probability that our next pick has a 100% chance of improving our status. That is if we've picked the same number as in the first one with our second pick.
Or we could have a 1-2/(2x-2) chance if we've picked a new one. Because the amount is 2x-2 now and there are 2 bad numbers.
This is so confusing.
Hey guys, interesting problem. Did you know the contestant actually has a higher chance of winning if he switches after the host removes a door? Very counter intuitive, but it's true. There's a wiki page on it somewhere, maybe someone knows what it's called.
Wow, did you see that movie "21"?
put yourself in my rocketpack if that poochie is one outrageous dude
How about this one:
Let's assume that in a country the probability of a newborn child to be either a boy or a girl is exactly 50%. Naturally this means that about 50% of the population will be male and the rest female.
To limit overpopulation, the country passes a law that after a boy has been born to a woman, she can no longer have any further children. There's no limit how many girls she can have before her first son is born.
Question: Does this change the male/female ratio of the population as time passes, and if so, how much?
Joined: 11/30/2008
Posts: 650
Location: a little city in the middle of nowhere
I would have to say that it remains 50/50. the probability of your first born being any gender is 50%, if you had a girl first, then the probability of the second child's gender is still 50/50 and so on. It only affects the total number of children born.
#!/usr/bin/python
import random
numMale = 0
numFemale = 0
for i in xrange(1,100000):
while True:
if random.random() < .5:
numFemale += 1
else:
numMale += 1
break
print numMale,numFemale
Printed "99999 99600" on my first run, and since intuitively the number of female children would go up, so much for intuition. :)
Pyrel - an open-source rewrite of the Angband roguelike game in Python.
Sidenote:
>Let's assume that in a country the probability of a newborn child to be either a boy or a girl is exactly 50%. Naturally this means that about 50% of the population will be male and the rest female.
No, only if males and females die at a similar rate. Which they don't do in most countries (especially not those at war).
It's indeed curious how the first intuition is to think that the amount of females would raise compared to males in these conditions, but that isn't so.
There's a relatively simple way of seeing why the ratio keeps at about 50/50 even in this situation:
Let's assume a sample of 1000 women who give birth. About 500 will be boys and 500 will be girls. So the ratio has remained equal.
Now let's assume that some of the women who gave birth to a girl want to have a second child. Let's say, for example, that 400 of them want to do so (it doesn't really matter how many, it's still the same). From them about 200 will have boys and 200 girls. Now there will be 500+200=700 boys and 500+200=700 girls in total. Still the ratio is preserved.
Let's assume 100 of the women who gave birth to a second girl want a third child. Again, about 50 boys and 50 girls will be born, in which case we will have 700+50=750 boys and 700+50=750 girls, the ratio still remaining equal. And so on.
Perhaps a bit surprisingly, and a bit contrary to intuition, the ratio will indeed keep at about 50/50, no matter how many women give birth to how many children. It doesn't even matter if they decide to stop having children even if they only have female children.
