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If I rolled a die 300 times and recorded each outcome, what is the chance of rolling at least one four?

I know that the answer will be very close to $1$, but I don't know if there is a formula for finding that exact value.

If I did this with two dice, then $P(4)=\frac{11}{36}$, which I only know how to work out if I draw a two-way table.

Any help is appreciated, thanks!

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    $\begingroup$ what is the chance of rolling one die and not getting a $4$ $\endgroup$
    – Will Jagy
    Dec 9, 2017 at 3:35

3 Answers 3

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Ask the complement question:

What is the probability that a 4 will not occur?

That, of course, is $\left(\frac56\right)^{300}$. So the probability of rolling at least one 4 is $$1-\left(\frac56\right)^{300}=1-1.76046×10^{-24}$$ It really is so close to 1 that I had to resort to just writing the difference out – the raw probability cannot be distinguished from 1 in 64-bit floating point.

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    $\begingroup$ 0.9999999999999999999999982395399029391309762257262419055102251211908921770069473469484227656987448800036480484026060865346688318593143029748068267104504629372594244497420023271959179818572087298048624982104315100456144229404158197674211459366490711346236493512966764819393858753624879494489514438991936456571054467798333829513964154167318124062601782710543919211588649621794421724543568061554829299705582265889185433781171602989870703404799041163188518625465303986151857705092478175719004297511914398783193404181826171373444375817743202126215836441456158428430047492421715130342804011797491906449991 $\endgroup$
    – wizzwizz4
    Dec 9, 2017 at 12:08
  • $\begingroup$ (08 rounded to 1) $\endgroup$
    – wizzwizz4
    Dec 9, 2017 at 12:08
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    $\begingroup$ @wizzwizz4 You must not care about accuracy with that frivolous rounding. The exact answer is 278852867695983428743551408716824211571664697829381258181828892993927916762314855747883045062452701904309604839405583020789789185327970692400334723465328185879400412956651706716567588739737136253812693643832872062715906971679128072751/278852867695983428743551899626170741344320007406576756809393190515479166707271366902794763772978174075895250849193986753985016903685127205588186040257189228351290693708134117612912814050283582240005547538013970502446610802398122213376 $\endgroup$ Dec 9, 2017 at 13:43
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    $\begingroup$ @JohnColeman It's hard to find the fraction solidus there. $$\frac{278852867695983428743551408716824211571664697829381258181828‌89299392791676231485‌57478830450624527019‌04309604839405583020‌78978918532797069240‌03347234653281858794‌00412956651706716567‌58873973713625381269‌36438328720627159069‌​71679128072751}{27885‌28676959834287435518‌99626170741344320007‌40657675680939319051‌54791667072713669027‌94763772978174075895‌25084919398675398501‌69036851272055881860‌40257189228351290693‌70813411761291281405‌02835822400055475380‌13970502446610802398‌​122213376}$$ $\endgroup$
    – wizzwizz4
    Dec 9, 2017 at 13:51
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    $\begingroup$ Hey dudes, digits are just mayonaise anyways. $\endgroup$ Dec 9, 2017 at 14:06
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Hint: What is the probability of NOT rolling a single 4?

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    $\begingroup$ Better as a comment $\endgroup$ Dec 9, 2017 at 3:38
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    $\begingroup$ I think this is a good hint and like it better than a full answer. It lets OP think about the problem. $\endgroup$ Dec 9, 2017 at 4:01
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    $\begingroup$ You see hints like this all the time on this website. $\endgroup$
    – Cornman
    Dec 9, 2017 at 4:03
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The easiest way to look at this is first to compute the probability that none of the outcomes is a $4$. So for each roll there would then be five possibilities out of six and for two goes this would be $\left(\frac 56\right)^2$ and the probability of at least one $4$ would be $1-\left(\frac 56\right)^2= 1 - \frac {25}{36} = \frac {11}{36}$.

Now apply this thinking to your main case.

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    $\begingroup$ and the probability of no 4 would be - wouldn't $1-(\frac56)^2$ be the probability of at least one 4? (Also, that's $\frac{11}{36}$, not $\frac{25}{36}$. $\endgroup$ Dec 9, 2017 at 5:40
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    $\begingroup$ @numbermaniac Fairly fundamental typo spotted - thanks $\endgroup$ Dec 9, 2017 at 9:23
  • $\begingroup$ And I shouldn't post when I am tired ... $\endgroup$ Dec 9, 2017 at 13:31

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