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A person is throwing the dice until it get a six. The person got it on the tenth throw. The probability for not getting the six on the first 9 throws is:

$$\frac{5^9}{6^9}=0,1938$$

Expressed in percentage is 19,38%.

I calculate the decimal odds like this 100/(probability(%)). Which results 1:5.16.

It’s the math correct?

We tell the person the next two statement:

1.”There was 19.38% chance that you will not get a six in the first 10 throws”

2.”The odds of you non getting a six in the first 10 throws were 1:6.19”

Are the statement correct?

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Firstly, the probability you have calculated is completely correct.

Secondly, we would usually say "The odds of winning are $1$ in...", not "The odds of winning are ...%".

Thirdly, we would say "There is a ...% chance that you will win" instead of "There is a $1$ in ... chance that you will win", although the latter statement is true.

For this problem, statement $2$ is false:

There is a $19.38$% chance that you won't get a six in the first $10$ throws.

Because there is a $19.38$% chance that you won't get a six on any specific $10$ throws, for example last $10$ throws, middle $10$ throws,... so the statement above is true.

The odds of you not getting a six in the first $10$ throws were $1$ in $6.19$.

The only thing incorrect here is that the chance is actually $1$ in $5.16$, and yes, it is correct to say that the odds of ... is $1$ in $x$, even if $x$ is not an integer.

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  • $\begingroup$ Tnx. Regarding my statements I intended to point out to someone how unlucky he was to not get a six in let say in 20 throws and express the probability and the odds to emphasis the bad luck. $\endgroup$ – RoyenSolo Mar 16 '18 at 8:28

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