r/mathriddles • u/Shin_Cow • 12d ago
Easy A harder opponent twice, or an easier opponent twice?
You play three tennis matches. You have a 70% chance of beating your teacher in any match and a 30% chance of beating a champion. Assume the results of all three matches are independent, and the win probabilities stay fixed.
Your goal is to win at least two consecutive matches. You can choose one of these orders:
Teacher — Champion — Teacher
Champion — Teacher — Champion
Which order gives the higher probability of achieving your goal, and why? Winning all three also counts as success.
Bonus: if the two win probabilities are any 0 < q < p < 1, is the same choice always better?
Creator disclosure: I run Think in Odds, a daily probability puzzle site. This is a released puzzle version; optional hints and a worked explanation are here: https://thinkinodds.com/puzzles/tennis?utm_source=reddit-mathriddles&utm_medium=reddit&utm_campaign=r_mathriddles_first_players_oct2026
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u/QuagMath 12d ago
Intuitively, you have one chance to beat the person in the middle and two chances to beat the other person, and you have to beat each at least once so you’d rather play the harder person twice (though without the math I wasn’t sure this was always true).
Explicitly, for the teacher in the middle we get p(1-(1-q)^2 ) = pq(2-q). Similarly for the champ in the middle we get pq(2-p). (2-q)>(2-p), so it’s always better for the teacher in the middle.
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u/DodgerWalker 12d ago
To win two in a row, you must win the middle game. You also need to win either the first game or last game. Chance of winning at least one of two games is 1 - P(losing both).
If the teacher is the middle game then your chance of winning is: .7 * (1 - .72 ) [chance of losing both games to the champion is .7*.7] = .357
If the champion is in the middle, then your chance of winning is: .3 * (1 -.32 ) = .273.
In general, you want to face the opponent you're more likely to beat in game 2 since that is the one game that is a must win, while you could survive a loss in either game 1 or 3.
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u/Shin_Cow 12d ago
Thanks for the solutions! The “must win the middle game” argument and the inclusion–exclusion argument are two useful ways to see why the result holds for any 0 < q < p < 1. The general proofs above solve the bonus too. I’m marking this solved.
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u/Konkichi21 12d ago
Doing the math, the chance is the probability of beating the first two plus the probability of beating the last two minus the double-counted chance of getting all three. In an ABA structure, that's AB + AB - AAB. In order to maximize this, the ABs are the same regardless of which is two, so we want to minimize the AAB, requiring the A be the harder one (adding the lower chance twice), so the CTC has a better chance.
Intuitively, CTC is better because you're basically playing two games against A and one against B and need to win one against each. CTC makes the one with only one (which you need to win) the easier one, and gives you a second chance against the harder opponent.
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u/jaminfine 12d ago
The key is seeing that the middle match must be won in order to have two consecutive wins. So, it should be the easy match in the middle. That gives you effectively two chances to win the harder match.
To calculate the probability, first go by the chance we win the middle match. In the TCT case, that's a 30% chance. Now, to get the chance we win at least one of the edge matches, we can take the complement of the chance we win 0 of them. 0.3x0.3= 0.09. So the chance of winning at least 1 is 91%. Finally multiple that by the 30% from earlier and get 27.3% of two consecutive wins.
In the CTC case, same logic. Start with 70% middle match. Chance of 0 edge match is .7x.7=49%, so 51% chance to win at least 1. This gives us about 35%. Better odds.
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u/bitternerd_95 11d ago
Whichever way you go it is clear that you need to beat each person at least once. It is clearly in your best interests to play the champion twice to maximize your chance of getting a win against the better opponent.
An interesting variant would be this: you play N games against k<N opponents of different skill levels. You need to beat each opponent at least once. You must decide in advance how many times to play each opponent. How do you distribute the games among opponents to maximize your chances of success
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u/Shin_Cow 11d ago
Nice variant! Assuming all matches are independent, each opponent has a fixed win probability
p_iwith0 < p_i < 1, and there is no consecutive-win requirement:If you assign
n_imatches to opponenti, your success probability is the product of1 - (1 - p_i)^n_i. Start with one match per opponent. Allocate each remaining match to whoever has the largest multiplier(1 - (1 - p_i)^(n_i+1)) / (1 - (1 - p_i)^n_i). This greedy rule is optimal: taking logs makes the objective a sum, and each opponent’s successive gains decrease. So you select the largest remaining gain at every step. It needn’t mean putting every extra match against the hardest opponent; the diminishing returns matter.
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u/Needless-To-Say 10d ago edited 10d ago
Intuitively
If you are more likely to lose to the Champion then having the champion in the middle significantly reduces your chances of 2 in a row.
Putting the teacher in the middle increases your chances of 2 in a row.
Head Math says the chances of winning 2 in a row with the teacher in the middle is just over 35%. While with the champion in the middle, the must win spot, never gets more than 30%
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u/Both-University3955 10d ago
You're better playing the champion twice unituitively. Proof : i read it in Nonplussed
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u/kalmakka 12d ago
P(winning at least two consecutive) = P(winning first two) + P(winning last two) - P(winning all three).
Since P(wining first two) = P(wining last two) = pq in both cases, you should pick the strategy that minimizes P(winning all three), i.e. facing the champion twice.