# Sunday Times Teaser 3148 – Quiz Probabilities

*by Edmund Marshall*

#### Published Sunday January 22 2023 (link)

Each of four contending couples in a quiz game has equal probability of elimination at the end of each of the first three rounds, one couple going after each round. In the fourth round, the remaining couple has a constant probability p, less than ½, of winning the jackpot, which consists of £1000 in the first game; if the jackpot is not won, it is added to the £1000 donated in the next game. Each couple may enter three successive games of the quiz, except that any couple having played for the jackpot in the fourth round of any game then withdraws altogether, being replaced by a new couple in the next game.

If the probability that a couple, competing from the first game, wins £2000 is 7/96, what is the value of p as a fraction?

I believe that this teaser has a flaw. If it is interpreted as the author intended, it has no solution.

Consequently I don’t have a full Python solution. But here is a program that paves the way to a discussion of possible solutions by giving the equations for the probability of winning the various possible amounts in terms of the probability \(p\) described in the teaser.

Please note that this code uses the latest version of my polynomial library which I have just updated to provide features that help with this teaser.

Here is its output (edited):

If we set \(P[p]\) to be \(q/96\) for a win of £2000 and simplify the appropriate polynomial, we obtain the cubic equation \[27p^3+9p^2-36p+2q=0\]

For the teaser \(q\) is 7, for which the polynomial has one negative real root and two complex ones, none of which can solve the teaser. In practice it seems likely that the probability \(p\) set by the author in the teaser is a rational fraction so it makes sense to consider other \(q\) values to see if any provide solutions of this form. When this is done two possible solutions emerge: \(p=1/3\) when \(q = 5\) and \(p=2/3\) when \(q=6\). The teaser hence has solution when the teaser’s probability of \(7/96\) is replaced by \(5/96\).

Jim Randell (on his S2T2 site) has suggested that “£2000” could be interpreted as “£2000 or more”. When we use the polynomial for this value and set it to \(7/96\) it simplifies to give a quadratic equation:\[9p^2-9p+2=0\]with solutions of \(p = 1/3\) and \(p = 2/3\) the first of which meets the teaser constraints.

Of these I prefer the former since it is nice to see a solution involving a cubic instead of the much more common quadratics.

Here is a linear approach enumerating many of the 512 possible outcomes of the three games and calculating the probabilities of winning £2000 or of winning £2000 or £3000.

Forwarned that there is no solution for ‘7/96’, a solution is considered found if the denominator of the probability = 96.