Chevalier de Méré's Problem Type Question

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?

Answer 1

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 $1-\dfrac{35^{24}}{36^{24}}$

Answer 2

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 $\dfrac{35}{36} \approx 0.97222$

Your second should be

• 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 $\dfrac{5}{6} \approx 0.83333$

Your next stage should to raise these to suitable powers

• 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.