Chevalier de Méré's Problem Type Question

Is the following argument correct: A double six in a single turn in game B is 1/6 as likely as rolling a six in one turn in game A. But there are 6 times as many turns in game B as game A. Thus the two games are equally good bets.

Where game A is Pierre throws one die four times. He wins if at least once he rolls a six: .

and game B is he has 24 turns, and each time he throws two dice simultaneously. This times he wins if he rolls at least one “double six”

my working: P(not getting double 6)=$(36^1-35^1)/(26^1) = 1-(35/26)^1 = 0.3462 percent$

p(not getting a single 6) = $(6^1-5^1)=1/6=0.16%$

where do i go from here?

Your description is extremely unclear, so I'll just give you the probabilities assuming that we're dealing with fair dice...

• The probability that he loses game A is $\dfrac{5^4}{6^4}$

• The probability that he wins game A is $1-\dfrac{5^4}{6^4}$

• The probability that he loses game B is $\dfrac{35^{24}}{36^{24}}$

• The probability that he wins game B is $1-\dfrac{35^{24}}{36^{24}}$

• My apologies, the actual question was to prove whether the argument is correct or incorrect. – I hate math Sep 25 '14 at 8:06
• @Ihatemath: Do you mean prove or refute that "the two games are equally good bets"? Well, if you compare the 1st and the 3rd (or the 2nd and the 4th) in the answer above, you can clearly see that they are not equally good bets. – barak manos Sep 25 '14 at 8:14

Your first calculation should divide by $36$ not $26$, so be

• Probability of getting a double six is $\dfrac{36-35}{36}=\dfrac{1}{36} \approx 0.02778$
• Probability of not getting a double six is $\dfrac{35}{36} \approx 0.97222$

• Probability of getting a single six is $\dfrac{6-5}{6}=\dfrac{1}{6} \approx 0.16667$
• Probability of not getting a single six is $\dfrac{5}{6} \approx 0.83333$
• Probability of not getting a double six in $24$ attempts is $\left(\dfrac{35}{36}\right)^{24} \approx 0.5086$
• Probability of not getting a single six in $4$ attempts is $\left(\dfrac{5}{6}\right)^{4} \approx 0.4823$
Then, if it helps, subtract from $1$ to get the probability of success in the two games.