I was talking to this guy that loves gambling the other day, and he had this theory about how to win lots of money.

Using a roulette example (ignoring 0 for simplicity sake) he said that each time one colour follows a colour of the same type the probability of the ball landing on the opposing colour doubles. EG: 3 blacks in a row means that on the next spin theres a 1:8 probability that by betting red you would win.

I cant see how that would work. The way i see it is that you would still 2:1 odds (or perhaps only a tiny fraction less) as the result would be relevant to the point in time that you place the bet.

His argument was that it was due to the law of averages, which to him meant that he was right. AFAIK the law of averages was made up to measure the behavior of atoms when there was no accurate way of doing so and only really applies to orders of great magnitude.

Anyone know whats right here?

2. the right is - each next event does not depends on the previous one, so red probability is still even to black...

Thing changes when you talk about cards hoever - each card removed from the stack changes probability of the cards left... of course not in so drastic way how he wants to believe...

3. Ok, cool. I was pretty sure I must have been right. Shame I couldnt seem to convince the dude tho. I guess hes going to have to keep paying his stupid peoples tax then.

4. lottary is just another tax for people that do not know mathematics

5. Originally Posted by vart
lottary is just another tax for people that do not know mathematics
That's a good quote

6. Interestingly, I heard that there are more permutations resulting from shuffling Western playing cards than there are stars in the universe.

7. Good quote indeed, but it's "lottery".

"Dice have no memory."

You can try to explain it to your friend this way: if I flip a coin 10 times, probability says that the most likely outcome is 5 heads and 5 tails. The probability of the same result on all ten throws is 1:2^9 == 1:512; the probability of specifically heads or tails on all ten throws is 1:2^10 == 1:1024.
If I throw 9 heads, what's the probability that the tenth throw is tails? Your friend would think it's higher, but it's not. The probably is 1:2, just as for any individual throw. The thing is, if the tenth throw is heads, I've thrown 10 heads in a series, with a probability of 1:1024, BUT as I'm about to throw it for the tenth time, I've already thrown 9 heads, with a probability of 1:512. That was the hard part, so to say.

Or to put it another way: there are 1024 possible series of heads or tails for 10 throws, like hhhhhhhhhh, hthhttthtt, or something like that. Every specific series has a probability of 1:1024. In this sense, the series hhhhhhhhht (nine heads followed by a tail) is just as unlikely as hhhhhhhhhh (ten heads). Therefore, if we've reached the nine heads, the last throw must be equally likely heads or tails, because otherwise hhhhhhhhht would be more likely than hhhhhhhhhh.

Or: throwing one tail and 9 heads, regardless of order, has a probability of 10:1024 (there are ten possible series of throws that contain 1 head, each with a probability of 1:1024). However, having that tail exactly in the last place, that's not any less remarkable than throwing only heads.

Roulette is the same, except for the 0, which guarantees that the bank always wins in the long run.

8. Originally Posted by citizen
Interestingly, I heard that there are more permutations resulting from shuffling Western playing cards than there are stars in the universe.
If there are less than 10^68 stars in the universe, then that would indeed be correct.

Originally Posted by CornedBee
Roulette is the same, except for the 0, which guarantees that the bank always wins in the long run.
Not to mention that many casinos have roulette wheels with a 00 field, as well.

9. Originally Posted by mike_g
Can you ask the smart dude how much he actually win using this method?

He is right about one thing: if all first 9 turns are red/even, the tenth turn is more likely to be black/odd. If the payout is x2 and he had invested 1 dollar at each turn, he is very likely to win 2 dollars on the tenth turn, but he had to lose 9 dollars to get there.

and he forgot about the 00 where the house automatically wins.

tell the smart dude to play blackjack, at least there is the black jack that pays x1.5 principal, and the player has some control to make the odds even against the house.

--TING

10. Originally Posted by ting
Can you ask the smart dude how much he actually win using this method?

He is right about one thing: if all first 9 turns are red/even, the tenth turn is more likely to be black/odd.
Even if you like it more - the chance is the same

11. i think i got what the smart dude means. he has to seat besides the roulette and wait for probably 5 odd's to show up, then start betting money hoping that 6th turn on even because it's more probably according to the "law of average"

why does it sound logical & illogical at the same time? i learned probability before, but i want to disapprove this smart dude's "laws of average" based on predicate logic and reasoning. Besides i never believe probability theory until i studied it myself. the smart dude probably won't be convinced by probability alone.

i think the problem is that he neglects that averages not only refers to chance but also to the pattern; If his theory were true, then in a 10 roll bet, if all of the first five rolls are even, the next 5 rolls must be all odd to make up for the lost probability in the first 5 rolls. this sounds ridiculous, but that's what the smart dude's "laws of average" dicates.

--TING

12. The human mind wants to believe that in a 50:50 situation, the more you get one output, the higher the chance you will get the opposite output the next time around.

That is false, however. There is always a 50:50 chance on every iteration. It is just our notion to want to believe otherwise.

It is true, however, that although you always have a 50:50 chance of getting one or the other - you have less of a chance of getting a long string of the same thing.

For example, take a coin. If you have a 50&#37; chance of getting heads, and you flip the coin 5 times, then you have a:

0.5 * 0.5 * 0.5 * 0.5 * 0.5 = 0.03125

3.125% chance of getting 5 heads in a row. That does not change the fact that you have a 50% chance each individual toss.

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13. By the way, I don't want to create too large of a tangent, but in my statistics class my professor showed us a clip from a movie in which a guy kept flipping a coin 150 times and he got heads every single time.

Does anyone know what movie that is?

14. Well, there is nothing wrong in the laws of probabilities if I throw a perfect 6-sided dice 1^64 times and they all fall on 1. There is always the chance the next 1^64 * 6 throws will even out the result. Meanwhile my next throw has a 1:6 chance of being 1.

Or, if you want... if we somehow counted all dice ever thrown by mankind ever since they were invented to this day, the odds of the next throw being 1 would be 1:6.

15. >> Does anyone know what movie that is?
Rosencrantz and Guildenstern Are Dead? Fiction, though.

The gambler probably isn't hurt by his bad prob-abilities because the odds of it hitting black are still the same whether he picks it for good or bad reasons.

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Reminds me of another probability question (that I often ask in interviews):

On the game show Let's Make a Deal, the host is Monte Hall. Monte shows the final contestant three doors, 1, 2 and 3. Behind one of the doors is \$100,000 and behind the other two are donkeys.

Monte will ask the contestant to pick a door. Then, he will open a different door to reveal a donkey leaving two doors still closed (one of which is the door picked by the contestant). Monte then offers this deal to the contestant, "If you want to switch to the other closed door, you can do it for free." Once the contestant decides then the doors are opened and the contestant wins what is behind the one that was chosen.

If you're the contestant, should you switch doors? Should you stay on your door? Does it matter?