If he thinks Grunty's graph proves his point he should probably look at a graph that plots how many more/less tails than 50% have been thrown for increasing coin tosses. The diversion should be printed in numbers, not in percentage. The resulting curves should look pretty random for good rngs. >_>
If he's so sure he's right he should try and get rich with roulette.
The Monty Hall problem works only if the host knows the correct door and deliberately opens an incorrect one every time. In this case the host is giving some information to the contestant. If you simulate the situation many times, then the result is indeed that switching is advantageous.
A more interesting question is what happens if the host does not know the correct door, and opens a door at random. If the randomly opened door happened to be incorrect, is it still advantageous to switch?
If you simulate this many times (assuming that if the randomly opened door was the correct one, the contestant immediately loses), then the result is that switching doesn't change the chances of winning in average.
However, if the game is played only one time, and the randomly opened door was incorrect, is it advantageous to switch?
However, if the game is played only one time, and the randomly opened door was incorrect, is it advantageous to switch?
Yes, because it then becomes the exact same problem. Switching is advantageous in both cases (whether the host does or does not know). If the host doesn't know, that only affects your chance of not even getting to switch. If the door opened is incorrect, switching is advantageous. What makes you think the host's knowledge has any effect on that chance?
If he selects at random from the other 2 doors, the game master will knock you out in 50% of the cases where you'd normally win by switching. That limits the problem to only 4 cases in which you even have a choice. In 2 of them you get the car by not switching, in the other 2 by switching. Chances of winning are 33.33% no matter what.
As the amount of available doors (with goats in them) increases, it stays the same. Switching pays off for the friendly game master and it doesn't matter for the random one. If they open more than one door at a time, it still stays that way. Not sure what happens if there are more cars though.
However, if the game is played only one time, and the randomly opened door was incorrect, is it advantageous to switch?
Yes, because it then becomes the exact same problem. Switching is advantageous in both cases (whether the host does or does not know). If the host doesn't know, that only affects your chance of not even getting to switch. If the door opened is incorrect, switching is advantageous. What makes you think the host's knowledge has any effect on that chance?
It has effect on multiple runs: If the host knows the correct door and always deliberately opens an incorrect one, then switching is advantageous in average, when the game is played many times. In this case there is information being passed from the host to the contestant. Basically the host is saying "the 66% probability of these two doors is actually in this one door". Switching gives a 2/3 chance of winning in average.
However, if the host does not know, and opens a door at random, then in multiple runs there is no advantage in switching: The contestant wins with a probability of 1/3 regardless of whether he switches or not. (The reason is that in 1/3 of the runs in average he automatically loses when the host opens the correct door, thus removing that third from the total.)
These always work when averaging the results of multiple runs (the more runs, the more the average result approaches 2/3 or 1/3 respectively).
The interesting question is, however, if switching makes a difference with only one run, assuming the host does not know, and opened an incorrect door at random.
Joined: 5/1/2004
Posts: 4096
Location: Rio, Brazil
Why does it matter if he knows or not? You still have 1/3 chance of having picked the wrong one at first. So, if out of the 2 other doors remaining he revealed one that doesn't have the prize, then the prize has 2/3 chance to be on the door you haven't picked.
If the host opens the door with the prize by accident what happens? Start over?
If the host opens the door with the prize by accident what happens? Start over?
No, that means you've lost. Theoretically you could chose one of the two remaining goats and lose either way. It basically means game over.
If he picks the door at random there are 6 cases.
1: You pick goat1, he opens goat2.
2: You pick goat1, he opens car. Game over.
3: You pick goat2, he opens goat1.
4: You pick goat2, he opens car. Game over.
5: You pick car, he opens goat1.
6: You pick car, he opens goat2.
There are 4 cases in which you can decide if you want to switch (and still have an influence at the restult). The rest should be obvious.
a) In 2 of these 4 cases switching gets you the car. (1 and 3)
b) In 2 of these 4 cases not switching gets you the car. (5 and 6)
c) In the 2 remaining cases switching and not switching both get you a goat. In the original problem switching gives you the car in these two cases as well because the game master mustn't open the car in that one. So cases c) lead to a) instead for the original one.
Why does it matter if he knows or not? You still have 1/3 chance of having picked the wrong one at first. So, if out of the 2 other doors remaining he revealed one that doesn't have the prize, then the prize has 2/3 chance to be on the door you haven't picked.
How do you prove that switching is advantageous? If you run the experiment many times, trying to calculate the probability, you will end up seeing a 1/3 winning probability.
(Of course if we only count those runs where the host picked the wrong door, and discard the ones where the host picks the right door, then we will see a 2/3 probability of winning by switching. However, which one of these two probabilities is the correct one for the one single case?)
Joined: 5/1/2004
Posts: 4096
Location: Rio, Brazil
The point of the game is that the host shows you a door that doesn't have the prize.
But if your new problem involves removing a door from the game without opening it, then it still is better to switch to the remaining door, because:
1 - in case he removed the door with a prize, you'll lose anyway, if you switch or not doesn't matter.
2 - in case he removed a door without a prize, the chances that the prize is in the other door are 2/3, and the chances that the prize is in your door are 1/3. This is fully explainable by the initial setup, where there were 3 doors and 1 prize. You only have a 1/3 chance of getting the correct door, leaving the bigger chance being that the prize is on a door you didn't pick. If the host removes your alternatives for which door to switch to (by leaving only 1 door), then the full 2/3 chances all go to the only door you didn't pick.
Prove this is trivial in this simple problem, because there are only 6 possibilities.
let's say prize is in door 1.
#1: pick door1, host removes door 2 -> switching = lose
#2: pick door1, host removes door 3 -> switching = lose
#3: pick door2, host removes door 1 -> switching = w/e (discarded)
#4: pick door2, host removes door 3 -> switching = win
#5: pick door3, host removes door 1 -> switching = w/e (discarded)
#6: pick door3, host removes door 2 -> switching = win
So, as you can see from this simple explanation (same as kuwaga's), what I just wrote was all wrong :) We can see here that in the case where it's possible for the host to remove the prize door when you've picked the wrong door at first, you lose 50% of the 2/3 odds you'd have to win if you picked a wrong door at first, totalling 1/3 chances to win by switching, 1/3 chances to lose by switching, and 1/3 chances to lose no matter what you do.
However, if the game is played only one time, and the randomly opened door was incorrect, is it advantageous to switch?
I don't get why if switching isn't advantageous generally (for the random game master), why would one single case be treated differently?
To me that's like asking the question of when you toss a coin chances are 50% for tails and 50% for head, but for one single toss, what's more likely to appear?
With the random one switching is the same as not switching, you can't get an advantage. No matter how often you play.
With the friendly one switching is advantegous. No matter how often you play.
Why would anything be different if the game is only played once?
If you don't know if it's the friendly game master or the random one, and only these two possibilities exist, then switching is advantageous.
If you know it's the random one and by chance he acts like the friendly one, then switching would be advantageous for that one game, but at that point we don't know if he did, so the point is moot. The probability that he acts like the friendly one for one try is of course exactly so low that it nullifies the advantage the friendly one would give to you. I don't really get your question.
I think there's another answer to my question of why it's still advantageous to switch even if the host chooses a random door (and it happens to be an incorrect one), even though in a long series of such events the probability of winning is 1/3:
In average in each third round the host happens to open the correct door, and thus your probability of winning in that round was 0.
In average in 2 out of three rounds the host happens to open an incorrect door, in which cases you have 2/3 of chances of winning by switching.
When we average these probabilities from a multitude of rounds, we end up having 1/3 of probability of winning (because the zero probability cancels one third of the other rounds). However, in each round where the host opened the wrong door, the probability of winning by switching was 2/3 in that round.
Thus switching is advantageous if you happen to be in such round.
However, in each round where the host opened the wrong door, the probability of winning by switching was 2/3 in that round.
No. I'm assuming by "wrong door" you mean goat. I'm also assuming you don't pressume we got a goat as our first door (so we could have either goat or car). If you had, then the probability of winning by switching would be 100% in those rounds.
So you say we got our random door, and we see the game master opening a goat. How great is the probability of winning by switching? You say 2/3.
I say, we are in one of 4 equally probable cases now. The only information we've got is that the game master randomly opened a goat. I'm going to refer to my post above now and say we are either in cases 1, 3, 5 or 6. Those are the ones were we get shown a goat by the game master. 2 of them win by switching, 2 of them lose. Makes it 1/2 and not 2/3.
So, I guess the statement of yours I've quoted has to be wrong.
The 2/3 probability comes up if we have 6 equally probable cases in which the game master shows us a goat (Actually there are 3 cases and one of them splits in 2, because he could select from one of two goats. So it's 4 cases, but they're not equally probable, so I'm counting the other two twice). That only happens if he always has to show us a goat. Switching wins in 4 of those 6 cases here, making it 2/3.
I could also say this (for the random host). We have a 2/3 chance that we get shown a goat (1/2 * 1/3 + 1/2 * 1/3 + 1/3). If we had a 2/3 chance of winning by switching now (like you basically said), the overall chance of winning would be 2/3 * 2/3 = 4/9
So our chance of winning would be > 1/3 and that contradicts the experiment you spoke of.
Joined: 6/2/2009
Posts: 1182
Location: Teresópolis - Rio de Janeiro - Brazil
I want to be a programmer too, so, let's see
what I can do with my amazing coding skills:
rem Dooty
program_begins;
make_global 1=1, 2=2;
{
draw_a_coin_on_screen;
ask_the_player_to_toss_the_coin;
}
if player_toss = 1
then show "heads"
else show "tails"
endif
if player_toss = 2
then show "tails"
else show "heads"
endif
end
Not bad eh? You can compile it
in any Commodore Amiga 500.
I am old enough to know better, but not enough to do it.
Joined: 2/28/2006
Posts: 2275
Location: Milky Way -> Earth -> Brazil
FODA wrote:
We're releasing a patch that should make the first one a bit better.
I just haxored into your PCs and got it, sorry.
"Genuine self-esteem, however, consists not of causeless feelings, but of certain knowledge about yourself.
It rests on the conviction that you — by your choices, effort and actions — have made yourself into the
kind of person able to deal with reality. It is the conviction — based on the evidence of your own volitional
functioning — that you are fundamentally able to succeed in life and, therefore, are deserving of that success."
- Onkar Ghate
Bisqwit wrote:
This cute little fellow begs to differ.
Its antennas have the ability to directly influence your brain so that you're gonna think switching would be advantageous whenever you've picked the car first! Another ability is to force the game master to open the door with the car, even if he technically wouldn't be allowed to! The whole concept is bullet proof.
Its fur is actually made of micro glass fibres, effectively dispersing any sneaky capoeira laser attacks and then absorbing the energy via photosynthesis!
Oh now this will look good in an essay for university or something.
The chance of choosing the right door is 0, because laser-absorbing goats will brainwash you.
Warning! Believing this guy will cost you tons of money!
[URL=http://www.youtube.com/watch?v=gCHR3A4bMkc]This guy here[/URL] seems to agree with FODA's programmer friends. And he probably gets paid to pretend so.
Actually, this is called the [URL=http://en.wikipedia.org/wiki/Martingale]Martingale betting system[/URL]. There are no exploitable patterns.
Martingale would even lose if there was no 0 on the wheel because nobody has infinite cash.
Joined: 11/30/2008
Posts: 650
Location: a little city in the middle of nowhere
Actually, doubling your bet every time if you lose is actually a winning strategy if you are betting on a 50/50 game. It has nothing to do with the gamblers fallacy and is theoretically a good way to bet.
for example, take your first bet to be 5$, and say you lose about five or so rounds before you win again i.e.
Bet l/w $won
5 l -5
10 l -15
20 l -35
40 l -75
80 w 5
This is where you get double your money back if you win because there is a 50% chance of winning. Every time you win, you make the money given in your original bet, however, if the green comes up, you're in the shit if you've lost a few rounds. Generally, table minimums and maximums will prevent you from making any money with this strategy.
EDIT: oh and also how deep your pockets are. this strategy would work, if you had infinite money, in which case, why would you need to bet?
If you play that strategy for a long time (like say, to make a living), you're bound to reach a point where you'd have to bet such a high amount that you just can't because you don't own so much money. Table limits just make you reach that point earlier. That, and roulette isn't 50/50 to begin with.
Btw, stopping to play before you reach that point is also a failing strategy. If you do it once, you're likely to win. But if lots of people do it, there are some that lose a considerably huge amount of money before reaching their desired stopping point. The money you could lose by doing that is such a high amount, that it isn't profitable to risk playing the game. It's like risking 100$ for a below 2:1 chance to win 50$.
Joined: 11/30/2008
Posts: 650
Location: a little city in the middle of nowhere
You can bet on black or red. which is approximately 50/50, not including green.
EDIT: also it theoretically works for any other probability, not just 50/50.
You can bet on black or red. which is approximately 50/50, not including green.
EDIT: also it theoretically works for any other probability, not just 50/50.
That's the key word.
If you can bet infinite times, that's true for exactly 50/50.
But the the amount you have to bet increases exponentially, so that might be a problem. "...Well, I bet two solar systems full of money then! I can't keep losing forever!"
Somehow I feel like having some sort of rice dish now.
Regardless of any sort of table maximums, assuming exactly 50/50 this betting method does not improve your odds.
Bet 10 - Lose
Bet 20 - Lose
Bet 40 - Lose
Bet 80 - Lose
Bet 160 - Lose
Bet 320 - Win
You lost 10 + 20 + 40 + 80 + 160 = 310 and made 320. You made 10 dollars this round. No matter which round you win in you only make 10 bucks. In this case, there was a (1/2)^6 = 1/64 chance you lost 6 rounds in a row. You have to win 32 times to double your money, and there is a 32*(1/64) chance you will lose in the 6th round once.
This strategy assumes you had a maximum 630 dollars to start with. If you lose 6 bets in a row one time you just lost it all. If you were playing with 100,000+ dollars the problem is still the same. The odds of you losing 10-20 games in a row are considerably higher than one would like to think.
If you have infinite money you also have to play infinite games to double your money. 28 losses starting at 10 dollars reaches about 13 billion. You still have a 50% chance of losing after you've played enough rounds to double.
Time is money, so why waste your time playing an unnecessary amount of games for the same odds. =p
Its antennas have the ability to directly influence your brain so that you're gonna think switching would be advantageous whenever you've picked the car first! Another ability is to force the game master to open the door with the car, even if he technically wouldn't be allowed to! The whole concept is bullet proof.
Its fur is actually made of micro glass fibres, effectively dispersing any sneaky capoeira laser attacks and then absorbing the energy via photosynthesis!
