Pascal’s Probability Puzzle
In my political science classes, the following exercise is given to the students to decide on as a group:
“Harry and Tom are engaged in a contest. The first to win six coin-tosses will get $800 dollars. Harry is ahead 5-3 when the contest is stopped. It is up to you, the class, to decide how that $800 gets distributed to Harry and Tom.”
For this particular assignment, there is no right or wrong answer to this problem; it is what the class decides as a group on how to award the $800. What seems fair to you, considering the game is based entirely on luck? Of the twenty or so times this exercise has been given to a class, the allocation of $500-$300 has been the decision in every class except one (over 95%)! This is revealing in that it demonstrates our human tendency to look for patterns that make our decision-making easier. (I have already discussed how students often look for patterns on the multiple-choice math problems in lieu of solving them.) In this case, the 5-3 score likely made an impression that influenced the decision; perhaps a form of availability bias.
A variation of this puzzle was posed to mathematician/philosopher Blaise Pascal in 1654, and the “solution” has been attributed to kickstarting probability theory.
A fair decision to this problem could be based on the probability that Harry would win the game from this point. For simplicity we can assume that Harry will be “heads” and Tom will be “tails,” since the probability of tossing either is the same. The key to calculating the probabilities is understanding that the only way Tom could win is if he can get three tails in a row (TTT). Therefore, we have to determine what all the possibilities are with these final three tosses (even though Harry could win with the very next toss). Then we could use the formula for the probability that Harry will win:
ρ = (# of desired outcomes) / (# of total possible outcomes)
There are 8 total combinations that could occur on the remaining three tosses. Focusing on heads, how many ways could we get one head out of three tosses?
HTT THT TTH
How many ways can we get two heads from three tosses? (The same as one tail out of three tosses.)
HHT HTH THH
The other remaining combinations are HHH and TTT. And we’ll get the same 8 combinations if we focused on tails. Using the Fundamental Counting Principle, we can also calculate that for three independent events, (2 choices) x (2 choices) x (2 choices) = 8 total possibilities.
Now we see that there are 7 combinations of the final three tosses where Harry would win, with only 1 combination, TTT, where Tom would win. Therefore, the probability that Harry would win the contest is 7/8, and a fair distribution would be $700 for Harry and $100 for Tom.
We can also calculate this using the formula for a binomial probability distribution:
ρ = nCr (pr)(qn-r)
where nCr is the combination of n objects taken r at a time, p is the probability of a head (1/2) and q is the probability of a “not-head” (1/2). Combinations are discussed in the Misc. Math module, and pages 191-192 of the Math Skills Instruction Manual. The formula for calculating nCr = n! / [(n-r)! (r!)]
Reminder: the ! is for factorial. For example, 5! = (5)(4)(3)(2)(1)
First look at the probability of one head in three tosses. This is a combination of 3 tosses taken one head at a time, and the probability is calculated by:
(3C1) (1/2)1 (1/2)2 =
[(3! / (3-1)!(1!)] (1/2) (1/4) = 3/8
Then look at the probability of two heads in three tosses:
(3C2) (1/2)2 (1/2)1 =
[(3!) / (3-2)! (2!)] (1/4) (1/2) = 3/8
And the probability of three heads in three tosses:
(3C3) (1/2)3 (1/2)0 =
[(3!) / (3-3)! (3!)] (1/8) (1) = 1/8
Adding these three probabilities, we get: 3/8 + 3/8 + 1/8 = 7/8
Coming from the other direction, we could also calculate Tom's chance of winning a bit more easily, with (1/2)(1/2)(1/2) = 1/8; thus Harry would be 7/8.
RKL