Two dice can be tricky

2009-11-26

In the 17th century French salons, one of the favourite pastimes was gambling. One of the essayists of those times, Chevalier de Méré was playing the following game: roll a single die 4 times and bet of getting a 6. His reasoning was the following: if the probability of getting 6 for one roll is 1/6, then for 4 rolls it would be 4/6, so in the long run he would win more than he would lose.

When nobody wanted to play with him anymore because he was winning, he changed his game a little: roll 2 dice 24 times and bet on getting a double 6. I this case his reasoning was: the chance of getting a double 6 for one roll (of 2 dice) is 1/36. So by throwing 24 time the probability would be 24/36 = 4/6 so he would still win (although it would take a longer time) but the game being different he could convince people to play again with him. To his total surprise he started to lose in this game. Since he could not explain this he asked one of his friends, Blaise Pascal to help him with the mathematics.

So let’s see the correct way to estimate the probabilities in this (simple) case.

Game 1

First of all it is not correct to add probabilities in this case. If we have 2 events A and B then:

P(A \text{ or } B) = P(A) + P(B) - P(A \text { and } B)

P(A and B) is 0 only if the events are mutually exclusive, that is there is no basic outcome that is common for A and B. In our case if event A is: get 6 on the first roll and event B is: get 6 on the second roll, there is a basic outcome: get 6 on the first roll and get 6 on the second roll, which is common to both A and B, so that means we cannot ignore P(A and B) in the formula above. This reasoning obviously applies to the remaining die rolls.

The correct way to calculate the probability is the following:

P(\text {getting at least one 6 in 4 die rolls}) =
\text{  }1 - P(\text{not getting 6 on any roll})

where

P(\text{not getting 6 on any die roll}) =
P(\text{(not getting 6 on the first roll) AND}
\text{(not getting 6 on the second roll) AND ... so on})

For 2 events:

P(A \text{ and } B) = P(A)P(B)

only if the 2 events are independent that is A happening has no influence over B happening. In our case that means that getting 6 on the first roll has no influence over getting 6 on the second roll which is true.

So the correct probability for this game is:

P(win) = 1 - \left(\frac{5}{6}\right) \left(\frac{5}{6}\right) \left(\frac{5}{6}\right) \left(\frac{5}{6}\right)= 1 - \left(\frac{5}{6}\right)^4 = 0.51775

Thus, Chevalier de Méré had almost 52% chance of winning (instead of his estimate of 66.6%) so we can say he was lucky in his probability estimation.

The expected value is a useful measure of the winnings/losses in the long run. Assuming that each participant is paying 1 dollar per game the expected value is:

E = 2 \cdot P(win) + (-2) \cdot P(lose) = 0.071

Game 2

As seen from the previous game, in this case the correct probability is:

P(win) = 1 - \left(\frac{35}{36}\right)^{24} = 0.4914

and the expected value:

E = 2 \cdot P(win) + (-2) \cdot P(lose) = -0.035

As indicated by practice (presumably) Chevalier de Méré would lose on this second game. If there’s a lesson we can learn from this is don’t estimate probability if you don’t know the rules, you will get it wrong even for simple 2 dice games.

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