Samuel Pepys's Problem

[Kenshin]

Samuel Pepys, the famous keeper of diaries, was an earnest gambler and once posed to Isaac Newton the following gambling problem.

In a game of dice, one man has six dice and is required to score at least one Ace (a 1 showing on the face of a die), while the second player has twelve dice and needs to score two aces or more. Does one player have an advatage over the other?
If not, why not and if so, which player and why?

Could some one help me out, with working out so i can understand?

 

[Opalg]

What Pepys asked was this:

"A - has 6 dice in a Box , with which he is to fling a 6.
B - has in another Box 12 Dice, with which he is to fling 2 Sixes.
C - has in another Box 18 Dice, with which he is to fling 3 Sixes.
Q: whether B & C have not as easy a Taske as A, at even luck?"

Newton gave this answer: "If the question be thus stated, it appears by an easy computation that the expectation of A is greater than that of B or C; that is, the task of A is the easiest. And the reason is because A has all the chances of sixes on his dyes for his expectation, but B and C have not all the chances on theirs. For when B throws a single six or C but one or two sixes, they miss of their expectations."

Newton's answer (that A has the best chances) is correct, but the logic of his answer is not at all convincing by modern standards. You can find the exact solution at [url]http://mathworld.wolfram.com/Newton-PepysProblem.html[/url], but it's not that simple. I don't know whether that counts as "working out so i can understand."

 

[soroban]

Hello, Kenshin!

In a game of dice, one man has six dice and is required to score at least one Ace,
while the second player has twelve dice and needs to score two Aces or more.
Does one player have an advatage over the other?
If not, why not and if so, which player and why?

We know that: \(\displaystyle \(\text{Ace}) \,=\,\frac{1}{6}\;\;\;P(\text{other})\,=\,\frac{5}{6}\)


Player \(\displaystyle A\) has six dice and must roll at least one Ace.

The opposite of "at least one Ace" is: "no Aces".

The probability that he gets no Aces is: \(\displaystyle \:\left(\frac{5}{6}\right)^6 \:=\:0.334897977\)
. . Hence: \(\displaystyle \(\text{at least one Ace}) \:=\:1\,-\,0.335\:=\:\L\fbox{0.665}\)


Player \(\displaystyle B\) has twelve dice and must roll at least two Aces.

The opposite of "at least two Aces" is: "0 Ace or 1 Ace".

. . The probability that he gets no Aces is: \(\displaystyle \:\left(\frac{5}{6}\right)^{12} \:=\:0.112156655\)
. . The probability t hat he gets one Ace is: \(\displaystyle \:12\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^{11} \:=\:0.269175671\)

So: \(\displaystyle \(\text{0 Ace or 1 Ace}) \:=\:0.112156655\,+\,0.269175971\:=\:0.381332626\)

. . Hence: \(\displaystyle \(\text{at least 2 Aces}) \:=\:1\,-\,0.381\:=\:\L\fbox{0.619}\)


Player \(\displaystyle \L A\) has the advantage.