In a random experiment, a fair die is rolled until two fours are obtained in succession. The probability that the experiment will end in the fth throw of the die is equal to:

Guest Apr 24, 2019

#1**+3 **

It seems to me that this is going to be the same as rolling a dice 5 times and finding the probability that 4 is rolled any 2 times in a row.

call a 4 a success and any other a failure then for any individual roll P(S)=1/6 P(F)=5/6

44444 Any 5 fours 1 way (1/6)^5

4444 any 4 fours 5 ways 5*(1/6)^4*(5/6)

444 an 3 fours 5C3 ways = 10 ways 10*(1/6)^3*(5/6)^2

except 4*4*4 no but all other 3 fours ok. -1 way (1/6)^3*(5/6)^2

No fours 1 way (5/6)^5

P(4*4**)=P(4**4*)=P(4***4)=P(*4*4*)=P(*4**4)=P(**4*4) = (1/6)^2*(5/6)^3

\(\text {P(rolling 2 fours together in 5 rolls ) }\\ =\text{1-P(no fours)-P(1 four)-P(2 fours but not together) - P(3 fours but none together) } \\=1-(\frac{5}{6})^5- 5(\frac{1}{6})^1(\frac{5}{6})^4 - 6(\frac{1}{6})^2(\frac{5}{6})^3 - (\frac{1}{6})^3(\frac{5}{6})^2\)

1-(5/6)^5-5*(1/6)^1*(5/6)^4-6*(1/6)^2*(5/6)^3-(1/6)^3*(5/6)^2 = 0.0965792181069959 \(=\frac{751}{7776}\)

2 fours in the right places

44***

*44**

**44*

***44

P(2 fours in the right places) = 4*(1/6)^2(5/6)^3

OR

P(rolling 2 fours together)

=P(5 fours)+P(4 fours)+ P(3 fours) - P(4*4*4) +P(2 fours in the right places)

(1/6)^5+5*(1/6)^4*(5/6)^1+10*(1/6)^3*(5/6)^2-(1/6)^3*(5/6)^2+4*(1/6)^2(5/6)^3 = 0.0965792181069959 \(=\frac{751}{7776}\)

That is excellent.

I did it 2 different ways and got the same answer both times :)

Melody Apr 24, 2019

#2**+3 **

I interpret this question to mean the probability of the experiment ending exactly on the fifth throw. Melody has calculated the probability that it could end on any throw from the second to the fifth (I think!).

If my interpretation is correct, then the fourth and fifth throws must be 4s, and the third throw must be not4. This means the possible combinations for the first two throws are (1) 4 followed by not4, (2) not4 followed by 4, (3) not4 followed by not4.

Putting all this in probability terms we have p = (1/6^{2})(5/6)[ (1/6)(5/6) + (5/6)(1/6) + (5/6)(5/6) ]

or p = 175/7776 ≈ 0.0225

This is borne out by a MonteCarlo simulation (which also confirms Melody’s calculation using her interpretation of the question).

Alan Apr 24, 2019