Could someone please assist me with this probability question which resulted from a game of Yahtzee we were playing in Melbourne, Australia on our holidays.

The object on this turn was to throw a “large straight” which is 5 numbers in sequence from 5 dice numbered 1 – 6. A player initially throws all 5 dice and then selects those dice they want to throw again for a further two more times. In this instance the player on their first throw, threw a 1,2,3,4 and 6.

Question:- What is the respective probabilities of gaining a straight if they were to –

a) put back say the 6 and try and throw a 5 on the two further throws or…..
b) put back the 1 and 6 and try and throw a 1 and 5 or 5 and 6 on the two further throws bearing in mind that if one of the numbers was a 5 on the second throw they could hold that number and try for a 1 or 6 on the third throw.

I would be most appreciative if someone could assist in showing me how to calculate the probabilities particularly in the second instance (b).

My name is Robert and I am not sure what level this question is and my interest is personal interest.

Kind regards




For a) consider the probability of failure - you throw a non-6 twice - it's (5/6)^2 or 25/36; thus the probability of success is 11/36.

For b) it might be easiest to see if you made a chart of all 36 possibilities from the two dice when thrown so as to count outcomes.

(1,1) ... (1,6)
(2,1) ... (2,6)
(6,1) ... (6,6)

You succeed if:

you throw (1,6) or (6,1) with the two dice - probability of this is 2/36;

you don't throw (1,6) or (6,1) but do throw a 1 or a 6 and then on the second throw, with the remaining die, you get the other - probability of this is (18/36)(1/6) = 3/36

you do neither of the above and then roll (1,6) or (6,1) on the second throw - probability of this is (16/36)(2/36);

in total p = 2/36 + 3/36 + 32/1296 = 212/1296 only a little more than 1/2 the chances if compared to a) which is 396/1296.