# Complex Probability: The real chances of getting a 6, I'm not sure

Below is the link for the mission.

https://app.dataquest.io/m/379/solving-complex-probability-problems/7/combining-formulas

I’m confused about the use of the formula below:

When applied on:

Because:

On a six-sided die, I have Ω={1,2,3,4,5,6}. How can the chances of getting a value that is not 6 be lower than the chances of getting a 6 (1-0.4823=0,51)?

I mean, for every time I throw the die, I have only one positive result in 6.

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Note that the die is thrown four times and the probability that is referred to is after 4 throws. It is likely that majority of the time the die does not land on `6`. However, it is highly likely of getting a `6` at least once in that `4` throws, as compared with not landing a `6` at all.

Hope this clarifies!

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Let’s pretend that it is a bet. If I had to bet all my coins on:

• Getting a six from a six-sided die in one throw.

The advice is that it is less likely to happen or rarely would happen.

However, if the rules changed to:

• Getting a six from a six-sided die in four throws.

The advice is that it is slightly more likely to happen mathematically speaking because as the numbers of throws increase, the probability of getting six increases too.

Is it almost correct?

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Yes @italo.moises:

You can think of it in this way. Remember that the die must be a 6-sided fair die for the probabilities to be so.

The probability of getting 6 stays the same (hence `5/6 **4` --i.e. prob of not getting `6` stays at `5/6` for the 4 throws) but it is more likely theorectically speaking to get a `6`. Like for example using some software or pseudo-random number generator.

Hope this helps!

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Thank you very much. I’ll remember that when I get into a casino .

HAHAHAHAHA @italo.moises. Do you mind marking my reply as the solution if it helped you? Thanks!