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Roulette Gambler's Fallacy thought experiment.

Started by RouletteFanatic, June 13, 2010, 02:45:05 PM

0 Members and 1 Guest are viewing this topic.

RouletteFanatic

Lets say a guy plays a number a day on a no limit roulette table.

He simply waits for a specific number to appear say for example 7 and then bet 1 unit on 7, if it doesn't hit. He finishes.

If it hits, he take his profit and original 1 dollar(total 36) and bet on 7 again. If it doesn't he finishes, it it hits again, he takes all his money (again) which would be 1296 total and bet on 7 again, it it doesn't hit, he finishes. If it hits he has $46,656 and walks away a happy man.

The thing is what is the probability that he walk away with $46,656? According to gambler's fallacy theory that each spin is independent of each spin, the probability is calculated from the moment he started betting. So if he is playing on an european roulette, it would be (1/37)^3, since he bets on number 7 three consecutive times. That amounts to an average of approximately once in 50'653 times.

However, his betting sequence is already FIXED before he even started betting, meaning he can profit $46,656 (original $1 bet included just for the sake of it) only after 4 consecutive 7s appear, as he waits for it to appear first before betting on it 3 times.

Everyone knows that a number appearing 3 times and a number appearing 4 times is different. According to Gambler's Fallacy theory the odds are once in 50'653 times, but then again for this gambler to get $46,656 he needs a particular number to appear 4 times in a row! which is (1/37)^4.

I know the future is uncertain and no one can predict what spins are coming out next, but lets say you are the creator and you have a set of 1,000,000 actual spins from past results that you are going to make appear on a Roulette wheel. For certain (or close to), 3 times in a row is going to appear more then 4 times in a row for a specific number. And if that particular gambler with that playing style plays on this wheel, he is going to need that 4 times in a row so the odds are (1/37)^4.

Maybe I'm severely misunderstood but this whole thing is very paradoxical and you should more or least get the gist of what Im saying and maybe guide and explain the correct reasoning for the above?

please discuss.

Edit: I  realized that by choosing any number the first probability will still be (37/37) * (1/37)^3. Therefore I changed it to him waiting for a specific number to come out, I overlooked this part.

RouletteFanatic

95 views no answers? Just voice out your opinions guys, don't matter if right or wrong..

Edit:

I got a simple solution for this, the answer is that this gambler would win in (1/37)^4 spins, since 4 7s in a row needs to appear. Though the moment he starts betting he would have a (1/37)^3 chance to win.

Shorty

You're right, he has a 1/50,653 chance of winning.

Waiting for a specific number makes no difference, it just decreases your betting opportunity.

Shorty

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