Hedge fund interview question

[@verkuilb]

Presh, why are you including the win/loss result of the third set in this assessment at all?? There is NEVER a need to assess the win/loss results of that set. We don’t care who wins or loses the third set-we only care whether it occurs.

[@lousyoldpotato3587]

Wish I had scrolled down farther in the comments before I posted, you were way ahead of me on this one.

[@Mrtranquil01]

Wondering the same. What's the reason?

[@benstevens44]

@@Mrtranquil01 Well, the video would have been quite a bit shorter if that hadn't been factored in, that might have something to do with it.

[@gled9880]

The easy way to calculate the probability of winning in three sets is to subtract the probability of winning in two sets from 1.

[@skylark.kraken]

I didn’t even consider considering the 3rd match so when it appeared it confused me, we only care about the conditions leading up to it being 2 or 3, so there was no reason to include the 3rd match even for completeness as it’s entirely irrelevant

[@LuisSergioCarneiro]

I think the solution might be simpler. The probability of finishing in 2 sets is equal to the probability of the winner of the second set being the same player that won the first one. If sets are independent and players were equally likely to win each set, that probability would be 50%. However, if one player has already won one set, we might think this was due to him being the best one of them, so I would bet he will win the second set as well. Therefore I would bet on a 2-set match.

[@Steve.M]

I agree. The question, as stated, doesn’t require expressions for all the different probabilities. Start by considering two players that are exactly equally matched, and it’s 50/50 whether the match will go to three sets. Then suppose that one player is slightly better than the other. It must then be slightly more likely than 50/50 that that player will win the first two sets.

[@blahsomethingclever]

My conclusion also, came to that almost instantly. Players aren't evenly matched.

[@anandgautam3057]

Came to the comments to say exactly this, but am already late :) I would love to see an edit from Presh as these nice way of understanding solutions are amazing.

[@juanmanuelmunozhernandez7032]

I agree, but the second part of the answer may not convince some interviewers because it relies on Bayesian reasoning. If I still want to make the heuristic point, which I'm happy with, I would rephrase it like this: if one player is better than the other (p>1/2 wlog), then by definition this superiority will show up (ending in 2 sets) more often than not (3 sets).

[@SimonLaudati]

Just thinking the same. If we start with the assumption that probability is stable for all sets it’s fairly logical that 2 sets will be more likely

[@eventhorizon853]

It is beyond pointless to look at the 3rd set. If the outcome of the first two isn't WW or LL, there will always be a 3rd set, so it should simply be 1-Pr(2 sets)

[@leif1075]

Right ots 50/50 since if all.prpbsvikties are ewually likely then you have two possible.putcomes for each so 2 put of 4 is 50mpercent..QED

[@oliviervancantfort5327]

You don't understand. It is absolutely necessary... in order to triple the video length 😂

[@waltlock8805]

@@leif1075 It's not 50/50 though. The two players are never said to be equal in ability. That's what makes ending in two more likely.

[@nigh_anxiety]

@@leif1075 Its not 50 percent. We're not flipping coins.

[@jollyjoker6340]

It's pointless to write any equations here whatsoever. If the chance of a player winning a set, defined at 1:03, "is fixed and independent for each set", the player more likely to win the first set is by definition more likely to win the second.

[@Business_Memo]

I think there's an easier way to approach it. There are 4 possible outcomes to the first two sets: WW, LL, WL, LW. In the case that W=L=0.5, WW=LL=WL=LW=0.25. The match ends in two games 50% of the time (WW+LL), so for players of equal skill, it's just as likely to end in 2 games as it is in 3. If one player is better than the other, it's more likely to end in 2 games. This is because WW and LL must sum to a greater quantity than WL and LW. For example, W=0.9 and L=0.1, then WW=0.81 and LL=0.01, so WW+LL=0.82. The off-diagonals of WL and LW sum to 0.18 (because the probability set must sum to 1). Therefore, if the players are NOT matched in skill it's more likely to end in two games rather than 3.

[@beng4186]

This was exactly my reasoning too.

[@kristianemilkjrgaard531]

You would want p(W) to be a variable between 0.5 and 1.0 (without loss of generality) to prove it, right? p(WW) = p(W)^2 p(LL) = p(L)^2 = (1-p(W))^2 = 1 + p(W)^2 - 2p(W) p(WW) + p(LL) = p(W)^2 + 1 + p(W)^2 - 2p(W) = 1 + 2p(W)^2 - 2p(W) Letting p(W)=x, the function is now f(x) = 2x^2 - 2x + 1 and simple analysis tells us that f(x)>= 0.5 always

[@tonyennis1787]

What happens when you ask the interviewer if the players are equally strong, and the answer is "I don't know"?

[@beng4186]

@@tonyennis1787 It's a compound answer. You don't need to ask the interviewer a question. If p=0.5 then P(2)=P(3). Otherwise, P(2)>P(3).

[@BlurbFish]

@@tonyennis1787 Nothing changes, as you're betting on 2 sets no matter what the interviewer says. If the players are not exactly equal, then the match is more likely to end in 2 sets. the two players are *exactly* equal, then the match is as likely to end in 2 sets as it is to end in 3 sets. You don't need to know anything about the players, because betting on 2 sets is either as good as or better than the alternative.

[@randomxnp]

After the first set there are two outcomes: if the same player wins it is won in two sets. If the other player wins then it is won in three sets. So if equally skilled the chance is 50:50. However the chances are the first winner is better, which means that player is more likely to win the next match.

[@jay-jaykay3707]

thats not how that works...the "better" player winning 2 in a row only has a chance of >50% if that player is a 71%+ favourite per set, otherwise their chances of winning 2 in a row are less than 50%...only if we take the chances for the worse player winning 2 in a row into account we come to the conclusion of the video...

[@paulnieuwkamp8067]

@@jay-jaykay3707 "winning twice in a row" is a different probability than "I've already won once, what's the chance I'll win again?" After winning the first set (which is the premise set by randomxnp ) you don't need to be "a 71%+ favourite per set" to have a greater than 50% chance to win again; every tiny sliver you are better than the other player shifts the percentage in your advantage.

[@randomxnp]

@@jay-jaykay3707 You are misunderstanding, and in the process missing one possibility that puts the probability of the same player winning the first two above 50%. I did not say that the probability of the better player winning both was more than 50%. I said that (given the players are not equally skilled) the probability that the winner of the first is the better player is more than 50%. This means that there is greater than 50% chance the second match is won by the same player as the first match. What you are missing is the possibility that the less skilled player wins both matches. If you add that to the probability that the more skilled player wins both you get more than 50%. How about this to explain. Assume the second match is replaced by a coin toss. I am sure you would agree that the chance of the player winning the first match winning the coin toss is 50:50. Replace the coin toss with a second match. The more skilled player has more than a 50% chance in each match. So there is more than a 50% chance that the winner of the first match is someone who can beat the odds on the coin, and less than 50% chance that the winner has worse odds than the coin. So the odds are improved compared to the coin toss more often than they are reduced, meaning that the odds must be better than 50% that the same person wins both.

[@tonyennis1787]

I think your "however" statement is interesting. Suppose you ask the interviewer, "Are the players equally skilled?". If they say "no" then the answer is trivial - 2 sets are more likely than 3. If they say "yes" then the answer is also trivial, but it's 50/50, so 2 or 3 sets is an equal bet. BUT if the answer is "I don't know" then what you said is very important. Absent other information, when the first player wins the first game, there is a better chance that they are more skilled, which means that there is a greater than 50/50 chance they win the next match. So "I don't know" means 2 sets are more likely than 3.

[@randomxnp]

@@tonyennis1787 except that there is no way to know if the players are, indeed, equally skilled, let alone equally skilled on that day in those conditions. Therefore from the point of view of anyone considering the problem the probability of two sets is always greater than 50%.

[@simonmarcstevenson]

The math is often easier to use p and q instead of p and 1-p, with the additional assumption that p + q = 1. Now, the probability of winning in 2 sets is p^2 + q^2. The probability of winning in 3 sets is p.q.p + q.p.p + q.p.q + p.q.q = 2p.q.(p+q) = 2pq. Then we ask.- which is more likely p^2 + q^2 or 2pq. Subtract 2pq from both sides. The question is - which is bigger - p^2 - 2pq + q^2 or 0. Note that the left side is (p - q)^2 which is non-negative, so p^2 - 2pq + q^2 >= 0, or p^2 + q^2 >= 2pq

[@deathwingdeathwing5246]

exactly, this is way easier

[@grumpylibrarian]

The usefulness of this as an interview question is that it contains a trap. It's very easy to approach this problem by considering a 50/50 chance of a given player winning, as we have no data to suggest otherwise. If we calculate the first two games with 50/50 odds of a given player winning, we see 4 outcomes, two of which has that player winning both or losing both. So it appears from a "back of the napkin" approach that the probability of 2 vs 3 matches to get a best-of-three is itself a coin toss. The fallacy was in assuming the 50/50 chance. We "have no data to suggest otherwise," but we have no data to suggest that this is in fact the case, either. Once we realize that the probability of perfectly-matched opponents is less than 1 (in fact extremely less than 1), then betting on 2 rounds being decisive is going to pay out in the long run. The less equally matched the opponents are, the more likely that 2 rounds will be decisive. So this appears to be an excellent interview question for this type of job. I don't know if it really is or if that's just a modern legend, but it seems cool.

[@stigcc]

Agree, it has layers.

[@wwoods66]

"The less equally matched the opponents are, the more likely that 2 rounds will be decisive." If the match is me vs. Serena Williams, there's a good chance the match will end after **one** set, if there's a mercy rule.

[@richbuckingham]

You're actually missing the real trap. You always *should* consider it a 50/50 chance at first, not considering that first would be a very impractical approach, then not telling the interviewer the simplicity behind this one possibility would be a huge miss. The real trap is someone not being able to eloquently explain this and over complicating with math what is actually a very simple problem to solve.

[@Connor-c5t]

So the probability of 2 sets is >= 50% and the probability of 3 sets is

[@stigcc]

@ If the probability of winning the second set increases after losing the first set, you can have that three sets have higher than 50% chance

[@malvoliosf]

You are way overthinking this. Just realize that “ending in 2 sets” versus “ending in 3 sets” is logically equivalent to “the outcome of the first two sets are the same” versus “the outcome of the first two sets are the different.” Or to put it another way, what are the odds that the player who wins the first set will also win the second? Look at from a Bayesian perspective: *the only piece of information you have* is that the player who won the first set... won a set. Maybe he’s the better player, maybe he cheated, maybe the other player dropped dead of a heart attack. Maybe it’s simply a rule that that player always wins, he is Kim Jong Un or whatever. It is also possible that it was a fluke, that the underdog won in a never-to-repeated Cinderella story. Whatever, doesn’t matter. The only piece of information you have is that the player who won the first set won a set. The underpinning of probability - of all reasoning - is that events in the future tends to resemble the events of the past. And you follow this reasoning instinctively. If you walked into a after the first set and had the opportunity to for or against the winner of the first set to win the second, lacking any other information about the players or the game, you would of course bet in favor of the winner winning again.

[@templarknight7]

your scenario actually changes the probabilities now since you're not betting on a game ending in 2 sets but on an individual to win one more set. If player A and B have equal odds of winning, the probabilities remain the same. However, if player A has 70% chance of winning a set and player B has 30% chance of winning a set, the probability that the game ends in 2 sets is 58%. Odds of player A winning the first 2 sets is 49% while it's 9% for player B. If player A wins set 1, the odds of the game ending in 2 sets is 70%. if player B wins set 1, the odds of the game ending in 2 sets is 30%. So you see, it would be unwise to bet on the same outcome without any information; your scenario is no different from betting on who will win the first set.

[@malvoliosf]

​@@templarknight7 The second scenario where you show up late and can bet on the current leader winning? Yes, it’s a different game, with different odds, but my point is, you would instantly do the right thing: bet according to the only piece of information you have.

[@lpslpslpslpslpslps]

This is convoluted. Why do we care about the outcome of the third game? If it goes to a third game, its a 3 game win. So we just need to know the probability of WW + LL compared to the probability of WL + LW.

[@ottokopp770]

In fact, this makes his approach wrong and just happens to come to the right conclusion. Assuming 50% win rates, the likelihood is equal, but any win rate above 50% means 2-sets is more likely due to momentum and the winning player getting the initial serve. (Which explains the

[@nigh_anxiety]

@@ottokopp770 Winning or losing does not determine the next serve. It always alternates between games. The player who serves to start the second set is the player who received in the last game of the previous set, regardless of whether they won or lost. If the previous set went to tie-breaks, the player who received first in the tie break serves to start the next set. His approach is correct for two reasons: 1) In an interview, you may get asked to prove it absolutely mathematically, and 2) The full approach can be expanded to answer any scenario, so you can use the technique to answer any more complicated probability question they may go against gut intuition.

[@lpslpslpslpslpslps]

@@nigh_anxiety but, again, you don't need to know the result of the third set to rigorously prove the result mathematically. You simply need to know the results of the first two sets.

[@victorkaplansky]

It is obvious that Pr(3 sets) = 1 - Pr(2 sets). This observation could allow you to shorten your calculations. Actually I've done it in my mind, so this is good inteview question.

[@michalp79]

no, it is not helpfull

[@cannot-handle-handles]

Absolutely! Another way to avoid the inefficient calculations for Pr(3 sets) would be to show that Pr(2 sets) is always at least 50%.

[@verkuilb]

Another solution, which I think is far more intuitive: do it graphically: Put the outcome of set 1 on the x axis, with the probability of player 1 winning being p, and the probability of player 2 winning being 1-p. Do the same for set 2 on the y axis. Your 1x1 square is now broken into four rectangles: one with player 1 winning both sets having are p^2, one with player 2 winning both sets having area (1-p)^2, and two each player winning one set, each with area p(1-p). If either player wins both sets, the match ends in two sets, so the probability of that is the sum of the areas of those two squares. If they split the first two sets, there will be a third set, and the probability of that is the sum of the areas of those two rectangles. Now the simple comparison of those two sums (either algebraically, or by simple visual comparison of the areas) reveals that the probability of two sets is always greater than the probability of three, unless p is exactly 0.5, when all four rectangles become .5 x .5 squares. There is NEVER a need to assess the win/loss results of the third set. We don’t care who wins or loses that set-we only care whether it occurs.

[@kpberry11]

One clean way to solve this is to note that one player's probability of winning is 0.5 + d and the other's is 0.5 - d for some non-negative d. The probability that one wins the first match and the other wins the second, thus requiring a third match, is 2 * (0.5 + d) * (0.5 - d) = 0.5 - 2d^2. Since d is non-negative, the probability of requiring a third match is less than or equal to 0.5, so it's better to bet that a third match is not required.

[@Bob94390]

This was a cute and efficient change in variables, thank you.

[@yurenchu]

That's how I calculated it. But d doesn't need to be non-negative.

[@Jhfm1793]

I assumed he meant d^2 is non negative.

[@yurenchu]

@@Jhfm1793 He wrote "[...] one player's probability of winning is 0.5 + d and the other's is 0.5 - d _for some non-negative d._ " But the stipulation "non-negative" was unnecessary.

[@kenfreeman7780]

The real world example might be skewed due to the way the brackets are set up ( high seed vs low seed). It might be interesting to see if two set wins is as prevalent in the later rounds.

[@Bob94390]

If a job applicant had answered this simple question with a lot of mathematics like shown here, I would not have hired that person. The solution to the question is extremely simple, and relies on considering only the second set: If the winner of the second set is the same as the winner of the first set, then it is a two set match. If the two players are equally good, then this will occur in 50 % of the cases, and you can just as well bet on a 2 set match. However, if the two players are NOT equally good, it is more likely that it was the winner of the first set will win the second set, so then you should definitely bet on a 2 set match. This assumes that the payout (odds) are the same for the two cases.

[@isambo400]

Uhh if you’re hiring someone to do math though…?

[@peetiegonzalez1845]

You put it much more succinctly than I did, and I agree. I wouldn't hire someone who went the long way around considering irrelevant event spaces.

[@DanielSloane]

​@@isambo400 Then you have asked them a different question. This isn't a question that requires complex math at all.

[@yurenchu]

That's a wrong approach. If the weaker player happened to win the first set, then it'll be more likely that the game ends in three sets. So you need to weigh off the possibility of the stronger player winning the first two sets in a row _plus_ the possibility of the weaker player winning the first two sets in a row, *against* the possibility of the weaker player winning the first set and the stronger player winning the second set _plus_ the possibility of the stronger player winning the first set and the weaker player winning the second set. And to do that systematically, without making a mess that looks like arbitrary numbers juggling, one needs a sound mathematical basis. But you were right to comment about the payout.

[@rizka7945]

@@yurenchu No, you can figure it out intuitively. A simpler way to frame the exact same puzzle is to think of a coin. The coin may be fair, but is unlikely to be. It's probably more or less weighted but you don't know by how much. The coin is flipped twice. Would you rather guess that it falls the same side up both times? Or is it more likely that the result of the second flip differs from the result of the first flip? The answer is obvious enough.

[@Antoine.Comeau.Racing]

The probability that the winner of game 1 wins game 2 is 50% if they are equally good. If they are not equally good, it is more than 50%. Since it isn’t garantee that they are equally good, finishing in 2 is more likely.

[@andrewmakar2035]

The two matches aren't exactly equal. First serve is different between the first and second games. The degree of advantage this conveys could be significant.

[@Antoine.Comeau.Racing]

@@andrewmakar2035 interesting! So if they had similar skills it could be less than 50% chance to have 2 consecutive wins.

[@deerh2o]

No. The worse player can win the first set (or game as you put it) and not be favored for the second one (given the assumptions that the sets are independent).

[@Antoine.Comeau.Racing]

@@deerh2o It is possible yet infinitely unlikely that the two players are truly equally good. If there is a difference of skills, the results of the sets will be correlated if we ignore the potential benefits of serving first.

[@briankirschbaum5367]

@@andrewmakar2035 First serve of the second set is based on who was on return service during the last game of the previous set. Given each player is entitled to 6 service games if they hold, it really makes no difference.

[@romain.guillaume]

Without doing math here is my reasoning. No matter the result of the first set, we want to know if the second set will be the same winner as the second basically. If both players are equal, it should be a 50/50 chance. Otherwise, the best player should be more likely to win two sets in a row. In both cases, two sets is better, and works even better when players’ levels are uneven. After the vid: well that was it, yaay

[@austinnar4494]

Assuming each set is independent (which mind you may not be a valid assumption in a sports setting) the probability of finishing in 2 sets is at least 50%. For equally matched opponents, it is exactly 50%; as the skill gap widens, the probability goes up, reaching as high as 100% if one player will always beat another. Since we do not know the skill gap of the players, and we know that the match will conclude in 2 sets at least half of the time, we conclude that betting on 2 sets is best.

[@jay-jaykay3707]

1:32 thats where you went wrong, its a hedge fund question, those are finance guys, they dont bet on who wins, they bet on the better odds, so the correct "answer" would have been to ask for the betting odds...also, from the statistics as 85 of 122 matches were won in 2 sets this would imply that over all those matches on average there was one player who was a 81:19 favourite to win (which is interesting from a sportsbetting perspective, i doubt the odds being that clear very often, so underdog betting might be a winning strategy there), which would also mean that of those 85 2-set-wins 80 were won by that favourite and only 5 were won by the underdog...would be interesting to see if this held true...if not our assumption of fixed and independent win probabilities is likely wrong...

[@OMGclueless]

You don't need the betting odds to answer the question of what the expected chance of winning a bet is. Obviously they do matter eventually when choosing whether or not to place the bet, but it's normal in finance to calculate the expected returns first and then plug the price you're getting at the end to decide whether to buy.

[@klikkolee]

And who calculates the betting odds? This interview question is for a job where you need to be the one calculating those odds.

[@jay-jaykay3707]

@@OMGclueless but its wasnt the question of what the expected chance of winning the bet is, it was the question which outcome you would bet on...presh made the assumption that we want to bet on the more likely outcome, which is not necessarily true, from a financial or sportsbetting point of view we want to bet to make money longterm, so betting on an unlikely outcome that has huge payout to way overcompensate for the bad winning chances is absolutely a thing...

[@jay-jaykay3707]

@@klikkolee the guy at the sportsbetting company...and those guys dont calculate based on winning percentages, they balance the payout pots...the gambling company tries to never gamble themselves...and the person betting would rather go by statistics and not try to calculate winning chances with way too many unknown variables...

[@elSethro]

I chose 2 sets. Did not lay out formal equations, but my reasoning was as follows: If the 2 players are equally skilled and likely to win a given set, then, regardless of the outcome of the first set, there is a 50/50 chance that the winner of the second set will be the same as the first set (ending on 2 sets) or losing and going to 3 sets. HOWEVER, if one player is more skilled than the other, and therefore more likely to win a given set, then they are more likely to win the second set. If the probability for the 1st set was 50/50, then an advantage in the second set would only pay off 50% of the time, which would make the likelihood of 2 or 3 sets the same. But the skill advantage would carry over to all sets, so more often than not, the more skilled player has already won the first set, and a second victory ends the match after 2 sets. Don't know if that makes sense to anyone else, but that's my thinking.

[@robynrox]

Yes, not having watched the rest of the video yet, that was also my thinking.

[@ck3908]

yeah, bottom line, if that's an interview question, first ask if the players are equal ability or not. Then answer the question. My assumption was that players were equal skill. Clearly answer would be different if not and common sense dictates 2 sets without even doing the math and then do the math if needed.

[@robynrox]

@@ck3908 Why would the answer be different either way? If the players are of equal ability, each outcome is equally likely, and betting on two sets is not worse than betting on three sets; neither is it better. You could work through the logic and make the point. I suppose there's a chance that the fact that the events are not strictly independent might make three sets more likely than two sets where players are of equal skill because the player who won the first set would presumably feel more pressure to win the second set - perhaps that's the kind of insight the hedge-fund interviewer is looking for.

[@ck3908]

@@robynrox this question doesn't test anything but how one makes assumptions to be quite frank or if one is able to ask relevant questions. Remove winning sets and put in coin tosses so what we are we going to assume? one coin could be flawed or maybe flawed? I mean one would assume coin toss is fair.. Again, my instincts would be to immediately ask about assumptions. If none given then just say if equal probability 2 or 3 sets are same likelihood, if not equal abilities then 2 sets likely.

[@robynrox]

@@ck3908 I agree. It's all good. Happy new year!

[@marcosolo6491]

Question is same as: Is a tie after two sets more likely than a win? If players are evenly matched, likelihood of tie is 50%. If one player is myself, likelihood tie is 0%. Thus, tie is LESS likely than a win. QED. Or, if you prefer more quantitative, probability of a tie is 2*p*(1-p) which has a maximum at p=1/2 and between 0 and 1/2 while 0

[@kamilrichert8446]

After the first match it's 1:0 for either player. Then either the winning player wins (2 sets) or not (3 sets). I don't understand where the difference happens so that it no longer is 50/50

[@MautozTech]

My dumbass's solution: the dude who won the first round is probably better because he won so probability of him winning second round is higher so competition is more likely to end in 2 sets

[@randomxnp]

No, that is a clever solution. It is the logic behind the maths presented in the video. It is far simpler than the maths and 100% correct.

[@BobJones-rs1sd]

@@randomxnp It is sort of "clever," but I wouldn't call it "100% correct," as it's incomplete logic. Certainly it works as a solution if one player has a GREAT advantage over the other. But it doesn't sufficiently explain cases where there's a smaller advantage. Consider if the probability of A winning over B is 90%. Then the probability of two consecutive wins (assuming independence, etc.) by A is 0.9*0.9 = 0.81, or 81%. Obviously in such a scenario where you have an 81% chance of two wins by the favored player, it makes sense to bet on "end in 2 sets." But now consider if the probability of A winning over B is only 60%. Then the probability of two consecutive wins by A is only 0.6*0.6 = 0.36 or 36%. In that case, the AA case doesn't account for more then half of scenarios, so it is uncertain -- from that logic alone -- that a 2-match case is favored. There, you'd have to at least consider a BB case (two wins by B), which would be 0.4*0.4 = 0.16, or 16%. Now, adding together 36+16 = 52% or a slight favoritism toward the 2-match solution. Alternatively, you could consider the probability of either AB or BA and you'd find they together have a 48% probability, making a 3rd match required only in slightly less than half of cases. To sum up: in cases where A is highly favored over B (or the reverse), the logic of the parent comment holds. But in cases where there's less favoritism (a win probability for one player less than 70.7%), the reason a 2-match solution is better is actually because the weakness of the second player makes all the scenarios involving the second player less likely. In particular, in the balanced case where players are literally 50/50 probability of winning, all outcomes (AA, BB, AB, BA) are equal. They'd each happen 25% of the time. But if the probability of B is less than 50%, then all the OTHER outcomes which involve B (i.e., BB, AB, BA) go down. And since AB and BA are the only ones leading to a 3rd match, if they are lower than 25% each, then they can't add up to 50% or more, meaning a 2-match outcome is favored. So, the logic is arguably backward mathematically. It's actually the disfavored player making his own wins less likely that makes mixed outcomes for the first two matches (AB or BA) less likely and therefore leading to fewer 3rd matches. Stated that way, I'd agree the logic would be 100% correct and a sufficient explanation. Considering consecutive wins by the favored player alone is only sufficient to consider mathematically if the win advantage is over 70%. (Technically more than sqrt(1/2).) Otherwise, you'd still need to add up the probability of AA and BB outcomes to assess whether a 2-match scenario is favored overall. (EDIT: Also, I'm giving the benefit of the doubt to the parent comment in making the assumption that he's really thinking about cases of the overall favored player winning twice. Otherwise, if it's really just, "The first dude who won is likely to continue winning" -- that is the so-called hot-hand fallacy and not valid in general statistically.)

[@f5673-t1h]

If one is better than the other, it's more likely that the better one wins both. (2 more likely than 3) If they're evenly matched, then one game goes to one of them, and it's about 50 percent chance that it ends with the next game. (2 equally likely as 3) So overall 2 is more likely.

[@67L48]

I read an article years ago discussing that relatively few matches go to three sets. I just looked it up and it’s around 35% (tennis magazine whose name I forget). I’d have said 2 sets, not because of theoretical modeling, but due to data. Data are the king.

[@abhisekmukherjee1811]

Here is a 2 liner here: Wlog, Player 1 wins both or loses both : p^2 + q^2 ( 2 set) Player 1 wins and loses once each in any order : 2pq ( 3 set) AM>=GM From the first case we have p^2+q^2>=2pq ( p+q=1)

[@BitcoinMotorist]

You didn't need to do all that complicated stuff. You only need to look at the second set. Many comments have already pointed out that you can ignore the 3rd set, but you can also ignore the 1st set. Someone will win the 1st set with 100% likelihood. We only care about the 2nd set. Assuming both players are equally good, there is a 50% chance the same player wins the 2nd set. So both a 2 set match and 3 set match are equally likely. But it's a reasonable assumption that the player who won the 1st set is better than the player who lost, so in most matches, a 2 set outcome is more likely. It took me 15 seconds to reason through this

[@wturber]

So I ran another set. This time I omitted the third flip reasoning that if I had either HT or TH this would be a three set match. If I had TT or HH, this would be a two set match. Whether H or T "won" shouldn't matter so no need to flip a third time. This time I had 32 matches won in two flips and 27 matches won in three flips. Combining the two runs I have 53 matches requiring two flips (sets) and 56 requiring three flips(sets). The more I test the more the results are converging on a 50/50 probability that the match would need two vs. three sets for a result. So if each player has an equal chance of winning any given set, the likelihood of two sets being required to complete a match appears to be the same as the likelihood that the match would require three sets.

[@RAFAELSILVA-by6dy]

I wanted to summarise the quick intuitive solution already described by several others. If we imagine the first set has been played. To bet on a three set match is effectively betting on the loser of the first set to win the second. With the information available, you have to bet on the winner of the first set to win the second, which is betting on a two-set match.

[@matheusGMN]

there's a much simpler way to solve this: it's conditional probability, thus the option with less events is always more likely to occur. I remember a ted-ed video which gave a pretty intuitve example: what is more likely, for a woman you meet to be a librarian or for a woman that you meet to be a librarian and have a graduation in math?, the fact the second option is conditional makes the first option more likely, because the only way for the second to be as likely as the first would be for all librarians to have a graduation in math. in this case, is it more likely to win in two sets or in 3 sets? well, obviously 2 sets, because there's only 2 conditional events, while in 3 sets there's 3 conditional events

[@daudnobel]

but if we put P = ½ that means the equation = 0, so Pr(2sets) = Pr(3sets) because it's like a coin flip, the chance are ½ head and ½ tail basically it depends on the winning chance of the player let say player A has much more points or way higher UTR than the player B, so the P > ½ but if the player A and player B has a close point or close UTR, the P will be close to ½ and that means more likely to go on the 3rd set the statistics on 2024 US open Women's Single shows much more won in 2 sets because almost all tennis tournaments use single elimination format, which is there are more matches between distant rating than close rating for example, you can see on the few first round will be much more won in 2 sets, and on the QF SF or final, it will be much more won in 3 sets

[@AnEnderNon]

I think it's simpler. The probability of either two wins or two losses in a row is x^2+(1-x)^2, where x is the probability of a win. Graphing this on desmos, it's an upwards facing parabola that is always above 0.5. Therefore I would bet the match ends in 2 sets.

[@JohnJones-pu4gi]

As many have noticed, pr(3)=1-pr(2) so all you need do is to show pr(2) >= 1/2 which is done by recouching p^2+(1-p)^2 as 2*(p-.5)^2+.5 qed

[@brians5232]

Before watching the video, I thought this is incredibly simple (it's either 50/50, or the more skilled player will win both, meaning 2 sets has a > probability).. then while watching the video I thought "maybe this is more complex than I originally thought.." then while reading comments I realized, "no, this is literally way simpler than the video is explaining it to be.."

[@Wise_That]

The question is basically asking whether the player who won match 1 is more likely to win or lose match 2. It seems intuitive that if we CAN assume independence, the majority of match1 winners will win match2.

[@barttemolder3405]

After LW or WL it makes no sense to look at the result of the 3rd set. We just want to know whether there's a 3rd set and that's always true in those 2 situations. So Pr(3 sets) = p(1 - p) + (1 - p)p = 2p(1 - p) = 2p - 2p². And it is obviously 1 - Pr (2 sets) so it is also 1 - (p² + (1 - p)²) which reduces to the same. Of course, adding up all 4 ways to play 3 sets yields the same result eventually. It is just more work than necessary.

[@RAFAELSILVA-by6dy]

One way to simpify this is to consider the probability of one player winning or losing a set as (1/2) + q and (1/2) - q. Then the probability of a two-set match is P(2) = [(1/2) + q]^2 + [(1/2) - q]^2 = (1/2) + 2q^2. And the probability of a three-set match (i.e. 1-1 after two sets) is: P(3) = 2[(1/2) + q][(1/2) - q] = (1/2) -2q^2 So we can see that two sets is always at least as likely as three sets, and the difference is: P(2) - P(3) = 4q^2

[@someperson188]

Pr(3 sets) = 1 - Pr(2 sets).

[@QuCaNi]

I asked ChatGPT about stats and it told me that in professional practice roughly 64% of Tennis matches end in 2 sets. Hence math makes sense

[@phillipsusi1791]

Yes, it is more likely that the better player wins in only two sets, but that doesn't help you decide *which* was the better player. You are still flipping a 50/50 coin to decide which player to bet on.

[@igorbaglay9064]

It can be made so much simpler. It's quite obvious that Pr(3 sets) = WL+LW = 2p(1-p) and its maximum is 0.5 at p=0.5, so anything other than p=0.5 reduces probability of 3 sets.

[@Bullshot40]

Fantastic explanation....i couldnt get past the 4/6 vs 2/6 outcomes...but i then realized, you went back to the p and 1-p to correctly account for the fact they each event was not equally likely to occur. Thanks!

[@chrisboyne5791]

In addition to excellent points above, in seeded tournaments, matches are deliberately made so that the best 4 players don't meet until the semi finals, so the matches tend to be more uneven than just assigning at random.

[@Nico-ni7hd]

are those bots commenting??

[@MasterOFSuperFunny]

I definitely see a pattern in their comments...

[@Wyvernnnn]

Everyone here is a bot except you

[@reinisliepins6177]

Yes, all of them advertising corn.

[@rogerkearns8094]

Apparently, but what the point of it is, I don't know.

[@erictrobin]

The first comments are usually from bot accounts. "Your video is so inspirational! Love your style!"

[@pdpgrgn]

I wondered how it scales up to longer matches, and calculated that for a 5-set match to be more likely to end in straight sets than to go beyond, the stronger player must have at least ~78.8% probability of winning an individual set. That requirement was much higher than I had anticipated. (I had expected two-thirds would do it)

[@Diriector_Doc]

Here's my reasoning. The only match that determines anything is match 2. Say there is a 50/50 for a win or a loss. In match 1, the result is insignificant, and the score will always either be 0-1 or 1-0. For match 2, if the leading player wins, the match will end in 2. But if the leading player loses, then a third match must be played to determine the game. For the irl example, I chock that up to "the stronger player will win most of the time, not 50% of the time."

[@Prs722]

I interpreted the problem different, but got the same answer. First assume both players have the same skill level and the same odds of winning. In such a scenario, there's a 50 50 chance of it ending in 2 or 3 sets. Next assume that one player is much more skilled than the other, and therefore has a greater chance at winning. In this scenario, basic logic says that it is more likely to end in 2 sets as it's more likely that the more skilled player will win. As this is an interview question and not an application question, I believe it was made to test logical reasoning rather than mathematical skills.

[@nedmerrill5705]

Is p a constant in all set sequences? My mistake was in assuming that p was 1/2.

[@thecoolestbro]

Simplest way of thinking of the problem: SOMEONE will win the first game. The question of there being two or three games comes down entirely to whether the initial winner snags the second win or has to do it in the third game. Assuming players are of equal skill, either of these events are equally likely.

[@Akronox]

You don't need to detail so much to calculate the probability of the match ending in 3 sets, you don't care who wins in the end. So it is instantly 2*p*(1-p), you need exactly one win for each player and the 2 factor comes from the 2 possible order.

[@JonCookeBridge]

Depends on the price and the match odds. Even if the match odds were evens, there’s still a ‘daily form’ component that considered in isolation means P(winning set 2 given you won set 1) > P(winning set 1). However, there’s also a fitness factor and a choke factor that are relevant. If the more skilled player is less fit or chokes excessive match points under pressure, that will increase the chance of WLL.

[@titfortat4405]

Interesting corollary is that for any coin (fair or unfair), the odds of flipping the same side twice in a row is always > 1/2. This is basically the same problem when you think about it.

[@michaelgoff4504]

I approached this with calculus. Pr(2 sets) = p^2 + (1-p)^2 = 2p^2 - 2p + 1 as in the video. The derivative with respect to p is 4p-2, which is 0 if and only if p=1/2. The second derivative is 4, which is positive, so it is a convex function. Therefore, Pr(2 sets) has one extremum, a minimum, when p=1/2. Substituting, we find that Pr(2 sets) = 1/2 when p=1/2, and so Pr(2 sets) >= 1/2 for all values of p. So bet on 2 sets. In real life, the assumption that the matches are independent events is questionable. If the winner has momentum, i.e. more likely to win the second match if they win the first, then Pr(2 sets) can only go up. If the winner gets overconfident and becomes less likely to win the second match, then the calculation could go either way. We also implicitly assumed that the bet is with 1:1 odds. If this came up in an actual interview, I would verify that this is the case. One thing I've learned about interviewing is that it is good to ask a lot of questions and verify your assumptions. That's one of the things that interviewers are looking for.

[@mkjioz]

As others have pointed out the 3rd set's probabilities are irrelevant. The question can be reduced to graphs which do vastly more to illustrate the correct answer. Let x=Probability of winning. The probability of WW+LL or x^2+(1-x)^2 Simplifying or just graphing it as is you get Y=2x^2+1-2x This results in a parabola with a vertex (0.5, 0.5) facing upwards that intercepts Y=1 when X approaches 0 or 1. The lowest possible chance of the tennis match ending in 2 sets is 50% when the probability of the 2 players winning is exactly 50%. Any alternative probability shifts the chance of it ending in 2 sets higher. You can also graph the reverse. The probability of WL+LW or x(1-x)+(1-x)x This is Y=2(x-x^2) It is a parabola with same vertex as the previous, but facing downwards. The probability is highest at 50% at (0.5, 0.5) and shifts downward, intercepting Y=0 as X approaches 0 or 1. The chance of both players winning once each is never higher than 50%. When you actually consider this problem in graph form, the answer is trivially obvious.

[@miowacity]

Before watching the solution I would guess it is the same because you have either A-A, B-B for the win in 2 or A-B, or B-A for the win in 3. The 3rd game outcome doesn't matter, because we know it went to three.

[@andyrobertshaw9120]

I just said the probability of the same person winning the first two sets is p^2 + (1-p)^2 = 2p^2-2p+1. The minimum of a quadratic for the for ax^2 + bx + c is when x = -b/2a, for positive a. So in the p-based quadratic we have a minimum at p=2/4=0.5. When p=0.5, the above quadratic equates to 2*1/4 - 2*1/2 +1 =1/2. From this, we know the minimum probably of the game being over after two sets is 1/2, when p=0.5, and higher for any alternative value of p. That means 2 sats is the favourable bet.

[@xnick_uy]

There's an intuitive way of interpreting the answer: provided either of the players has better chances of winning a set (e.g. one of them is a better player), it is more probable to have two consecutive wins than to have a win and a lose in the first two sets. I'm surprised Presh didn't mention that the probability of having to play the third set is also the probability of not ending the match after just two sets. In equations, this redas as P(W,L,W) + P(W,L,L) + P(L,W,W) + P(L,W,L) = P(W,L)+P(L,W) which also could be re-stated as a relation between condional probailities.

[@asdfqwerty14587]

For me I just interpreted it as "somebody must win the first game, what is the probability that that same player wins the 2nd game?" - in which case, it's intuitively incredibly obvious that if the players are equally skilled then it's just 50% (the 2nd game is just a coinflip), and if they aren't equally skilled then it's obviously more likely that the better player wins the first match, which makes them more likely to win the 2nd match too. To me it's obvious that as the difference in skill levels increases that the more likely it is for it to end in 2 matches, so I didn't really feel like calculating the exact probabilities was necessary for answering the question.

[@acasualviewer5861]

What you're saying is that the independence assumption is not valid. And you'd be right. I really DOES matter who wins the 1st set.

[@xnick_uy]

@@acasualviewer5861 On the contrary, the formal definition of independent events is still valid: the outcome of any given set does not depend on the result of the other sets. Be careful not to fall into the "Gambler's fallacy". But the situation could also be regarded as having an unknown probability of winning to begin with, and we try to estimate the value as the match progresses. Then we could say that after the first (or some more) wins, it's possible that the probability of winning is > 0.5. Perhaps this version better matches the everyday use of the term “independence.”

[@SuitedPup]

This actually aligns with sports analytics, which show that top tennis players win about 80-90% of their matches against average competition. Compare that to baseball, where even the best teams only win about 60-65% of their games. So in tennis, if Player A is better than Player B - even just a little better - they're very likely to win in straight sets rather than needing that third set decider

[@chriswatson7965]

No need to calculate Pr(3 sets) as this is just 1-Pr(2 sets). Further a more general solution is to integrate Pr(2 sets) from 0 to 1 and see if this is >0.5. Your solution relied on Pr(2 sets) - Pr(3 sets) >=0. Note that Pr(2 sets) - Pr(3 sets) = 2*Pr(2 sets) -1. So the test was Pr(2 sets)>=0.5. Here the integral of Pr(2 sets) from 0 to 1 is 2/3. So over any randomly distributed group of players games should end in 2 sets twice as often as 3. The 2024 gave a ratio of 85/37 which is greater than 2 because the groupings are seeded and not random.

[@matthewjagos9328]

I have played tennis in both the high school and college level, and from what I’ve noticed is that most tennis matches I play end in 2 sets. Tennis is a game where the best player wins basically every single time, and the worse player rarely gets a chance to beat the better player. Sure, you can go into tiebreaker matches and dueces all you want, but usually the better player will end up winning that game. It doesn’t matter if the better player wins the set 6-0 or 7-5, all what matters if that they get the 1 set won and that’s huge.

[@gamefacierglitches]

I think it is much simpler. Assuming both are of equal skill, you are betting if the same person who won the first set will also win the second set. This would be a 50-50 chance of getting it correct. Assuming each players has their own skills, there is a higher chance the person who won the first set is a better player. Therefore it is more likely for them to win 2 sets in a row.

[@cyberlumber]

Another way to prove the whole sample space was considered is to sum the probability of winning two sets and probability of winning in three sets. If this sums 1, as in this case, then all the sample space is considered. I always make this check when solving probability problems.

[@rizka7945]

Imagine yourself on the stands having made a wager. It all actually depends on the result of the second set. If you had bet on three sets, you will be rooting for the loser of set 1. If you bet on two sets, then you will be rooting for the winner of set 1. Which gives you more confidence? While rooting for the underdog could be fun, you are better off by being on the winner's bandwagon. Or in the theoretical case where players are completely evenly matched, at least you can't be worse off. Another way to frame the puzzle is to think of flipping a coin. The coin could be fair or it could be a trick coin falling on one side every time. Or it could be something in between. You don't know. You flip it once and see the result. Now you need to make a guess what happens on the second flip. Would you bet on the coin landing same side up as last time? Of course. If this was an interview question, this kind of reasoning would be the best answer I believe.

[@jopmota]

This is the mathematically correct way to say simply that, if the players are not evenly matched, one player is most likely to win the first two sets. Which is a fairly obvious thing to notice.

[@martys9972]

Once you have determined that the probability that the match will end in 2 sets is 2p^2-2p+1, you can plot that probability as a function of p. This is a parabola going through (0, 1), (0.5, 0.5), and (1, 1). This makes sense, because if p=0, then the second player is sure to win both sets, and if p=1, then the first player is sure to win both sets. Unless the players are perfectly matched (p=0.5), the probability that the match will end in 2 sets is greater than 50%, so that is what you should bet on. There is no need to analyze further than that.

[@emilemil1]

There are only two cases, a win or a loss for the player who won the first round, so WW or WL. If the players are equal in skill then there is a 50% chance to go to a third game. If one player is better then WW will have a greater probability than WL, since the better player is more likely to have also won the first round.

[@AndrewBlucher]

Yes, you should bet on 2 sets or 3 sets.

[@martin.brandt]

I fell for the unasked assumption that I was betting on a specific person and had to sort out whether that person would win in two or three sets. Probably the same result, but an additional unnecessary detour. Which might show to the interviewer that I would add unnecessary detail into my calculations, which can be counterproductive and false. Oops!

[@jacktaylor9290]

It's worth mentioning, tennis players usually retire from a match when they've lost the first set and know they can't come back.

[@bobh6728]

To make sure all possibilities were covered, just add the probability of 2 sets and 3 sets and you do get 1. So no outcome was missed.

[@uthoshantm]

I reached the same conclusion by taking the derivative of 2p^2 - 2p + 1. The derivative is 4p - 2 which is equal to zero when p = 1/2. At this value of p, the probability of 2 consecutive wins is 1/2. For any other value, since this is the minimum, we get higher values. The other approach is to integrate 2p^2 - 2p + 1 from 0 to 1. The result is 2/3, being the probability of two consecutive wins. We assume here that the value of p is uniformly distributed.

[@KyleWoodlock]

If the probabilities of winning are independent and equal, then the first two sets might look like AA, AB, BA, or BB with equal probability. For AA and BB, we finish in two. For the others, we finish in three. 50/50. If the probabilities are not equal, then it's P(A)^2 + P(B)^2 to finish in two, and 1 minus that to finish in 3.

[@Kami-mk7tu]

Easier way of doing this: 1. It does not matter who wins game ine, the only thing that matters is that the same person wins match two which is 1/2 in a two player game, the chance it ends in 3 would be 1 - 1/2 which is 1/2 so it's the same.

[@jdtreharne]

Without seeing the answer it seems pretty obvious that it should be two sets. If we know nothing else about the players, it's reasonable to assume that the player who won the first set is the better player, and therefore more likely to win the second set as well.

[@markthompson2874]

I think more important than being able to calculate it is just being able to intuitively know that the answer is 2 sets. A more interesting calculation is what the breakeven point is for a 5 set match finishing in 3 sets vs more than 3 sets or 5 sets vs less than 5 sets.

[@aquawoelfly]

1 whats the desparity in skill? (Clearly 2 if the desparity is significant) 2 any other conditions that favor one player over the other? I general 2 players of similar skill and outside factors (aside from mental anguish over results of match 1 and 2)being nominal i give it three rounds.

[@GameJam230]

I mean, we assume we’re looking at the collective probability as though we start from no sets having been completed yet, so let’s analyze all the possible combinations of ways it can lead to an end. Order matters in this case, (1,2,1) isn’t the same as (1,1,2) because the third set wouldn’t be played. (1 means player 1 scores, 2 means player 2 scores) (1, 1) 25%, (1, 2, 1) 12.5%, (1, 2, 2) 12.5%, (2, 1, 1) 12.5%, (2, 1, 2) 12.5%, (2, 2) 25% This is assuming a 50/50 chance of each player scoring on each set, and the total chance of it ending in 2 is exactly equal to the chance of it ending in 3. This means, if we assume both players have an equal chance to score on every set, the bet doesn’t matter. However, if we instead assume that the first player to score may be a slightly better player, therefore being the reason they scored instead of the other one, it may INSTEAD imply that there is more weight in favour of that player scoring again on the second set than the other player getting a score. Let’s for example suggest that player 1 has a 75% chance of scoring on every set, whereas player 2 has only a 25%. Let’s redo the percentages for each list of possibilities. (1,1) 56.25%, (1,2,1) 14.06%, (1,2,2) 4.69%, (2,1,1) 14.06%, (2,1,2) 4.69%, (2,2) 6.25% This makes the cumulative chance that it ends in 2 sets higher than the chance it ends in 3, at 62.5% to 37.5%. So, since it’s a perfect 50/50 bet when they’re equally likely to score on each set, and the chance becomes MORE in favour of finishing in 2 sets the better one player is than the other, it’s always safer to bet on the game ending in 2 sets than 3.

[@disgruntledtoons]

If play A wins x amount of all sets, and player B wins (1-x) of all sets, then the chance of the match ending with the second game is 1 - 2x + 2x^2. The local minimum or maximum of this expression is at x = 1/2, and because the slope of the derivative is positive, it represents a local minimum. In all scenarios, the match is either more likely to end in two sets or it is equally likely to end in either two or three sets. The answer: Bet on the match ending in two sets.

[@thechessplayer8328]

I work at a hedge fund and I would not hire anyone who assumes independence.

[@aisolutionsindia7138]

oh yeah and what about the fact the question doesnt specify payoffs to begin with?

[@thechessplayer8328]

@ you’re hired

[@riccardobertini5813]

This

[@yawninglion]

The game ending in 3 sets the outcomes of the first 2 sets differ, which happen 2p(1-p) of the time. Since the game can only end with either 2 or 3 sets, we can solve for what p does 2p(1-p) > 0.5 and get to the conclusion.

[@timcrawford2221]

To "solve" this question I would ask what the score for the first set was, eg. was it 6-1 or was it won in a tie breaker, which would have a large bearing on the outcome. But the probability of 2 equal opponents is way less that 50%, so 2 sets is the logical answer.

[@dan-florinchereches4892]

I think if player one has the probability P of winning then him winning 2 sets consecutively is P^2 and the opponent winning 2 sets in a row is (1-P)^2 so overall the probability is P^2+(1-P)^2 = 2P^2+1-2P to have 2 sets won If we do an integral over P from 0 to 1 then the total probability of winning 2 sets will be (2/3P^3-P^2+P) evaluated between 0 and 1= 2/3-1+1-0-0+0=2/3 so it is more likely for 2 wins to occur because the area of the cumulative probability distribution is just 1

[@OctavioAlvarez]

So, what is the probability of a match ending in 2 sets? Is it 75%?

[@raypalmer7733]

The probability of winning ANY set if 1/2. Thus the probability of winning in 2 sequential unrelated sets is thus 1/2 x 1/2 = 1/4 To win 2 of the 3 sets means the probability is 1/2 x 1/2 x 1/2 = 1/8. Thus bet on 2 sets rather than 3.

[@bluerizlagirl]

Do you need to consider the third set at all? If the first two sets go W-W or L-L, the result is clearly already decided. But if the first two sets go W-L or L-W, we ended up adding together all the probabilities anyway, and p times something plus (1-p) times the same thing is going to be (p + 1 - p) times which is just that thing. If the two competitors were perfectly matched, so p = 0.5, then W-W, W-L, L-W and L-L would be equally likely, and half of all matches would go to three sets. If you still have the stats, I'd reckon that from the quarter-finals onward, the last eight players are going to be fairly evenly matched, so I would expect there to be more matches going to three sets in the later stages of a tournament.

[@azechiel]

I think the easiest reasoning is as follows: if they are evenly matched, it doesn’t matter who wins the first round, then it’s 50/50 on who wins the second round, that is, whether it’s two or three rounds to win. If they are not evenly matched, it’s more likely that the same person wins the first and second round, so you should bet on two rounds instead of three rounds.

[@AA-100]

You dont need to include the outcome of the 3rd set in the calculations, since the outcome of the 3rd set is irrelevant. So you just need to calculate Pr(WW) + Pr(LL) for 2 sets and Pr(WL) + Pr(LW) for 3 sets 2 sets: p^2 + (1-p)^2 = 2p^2 - 2p + 1 3 sets: 2p(1-p) = 2p - 2p^2 P(2 sets) - P(3 sets) = 4p^2 - 4p + 1 = (2p-1)^2, same solution as shown in the video

[@EsamJoann]

Great analysis, thank you! I need some advice: I have a SafePal wallet with USDT, and I have the seed phrase. (alarm fetch churn bridge exercise tape speak race clerk couch crater letter). Could you explain how to move them to Binance?

[@SiqueScarface]

My argument was less algebraic. After the first match, both players being exactly equal in strength, you would have a 50% chance of the next set having a different player winning, which is the precondition of a three set match. If one player is favored in the second set, he was already favored in the first, and so was the chance of him winning the first set, making it more likely for him to win the game in two sets than for the second player to stage a comeback in the second set, and forcing the third set.

[@LenOvo-p7s]

Just to summarize some good comments: P(3 sets) = P(WL.)+P(LW.) = 2×p(1-p) P(2 sets) = 1-P(3 sets) So: P(2 sets)-P(3 sets) = 1-4p(1-p) = 4p²-4p+1 = (2p-1)² ≥ 0

[@vaibhavkayal1506]

For prob of 3 sets you could have just taken: Prob(LW)=(1-p)p=p-p² Prob(WL)=p(1-p)=p-p² Total prob=2p-2p²

[@ruud9767]

P(2sets)=p² + (1-p)², with p in [½..1] On this interval, p² + (1-p)² ≥ ½ Therefore P(2sets) ≥ ½ and the complement P(3sets) ≤ ½

[@thebigg2345]

I don't know much about hedge funds but my answer would be to look at historical outcomes and bet on whatever happens most often (it appears that's 2 set matches). Only if this information is unavailable, I'd fall back on the simple logical argument given by several other comments.

[@Duke_of_Lorraine]

Only the 2nd set is decisive : either whoever won the first wins again, thus ending in 2 sets, or he loses thus a 3rd will be needed. Assuming equal skills, it would be 50/50 so ending in 2 or 3 would be a coin flip. However, if one is better than the other, he has over 50% of winning, let's call it N% that the better player wins a set. Odds of finishing in 2 sets = either the best player wins twice (P = N^2) or the worst player wins twice (P = (1-N)^2). So P = N^2 + (1-N)^2. = N^2 + 1 - 2N + N^2 = 1 - 2N + N^2. It's a parabollic curve with a minimum for N = 0,5 which we already know to be 50%, and as the minimum any value between N=0 and N=1 will thus return over 0,5. So at the worst possible case scenario for 2 equally matched players, it will be a coin flip. If there is any difference in skill, we'll have a greater chance of ending in 2 sets than in 3 sets. So I'm betting on 2.

[@jeremiahlyleseditor437]

I struggled with this in college as well. This clears it up for me little. Seems the probability of losing is not derived, it is given for the final percent of winning or losing.

[@monkemode8128]

I just assumed it was a 50/50 chance of either player winning any game, independent of any previous game. Therefore, the chance the same player wins two games in a row is 50% * 50%, or 25%. On the third game, it's guaranteed to be over, so 100%-25% (75%) is the chance the game ends in 3 rounds. Since 75% > 25%, I'd bet on it ending in 3 rounds.

[@TheBoogerJames]

If one of the players has even a slight advantage, the most probable outcome is for that person to win 2 in a row. Any other outcome, including playing a 3rd set, is less likely to happen.

[@clairecelestin8437]

If anyone's curious, the value of p from the real data set is about 81.36%.

[@redfinance3403]

Consider the different ways the games could finish. Let us denote W1 as win for player 1 and W2 for player 2. For 2 sets the two possible outcomes are. W1, W1 and W2 and W2. Given that the probability of a win is 50% total probability of two sets occuring is 2*0.5*0.5 = 0.5. For three sets we have W1, W2, W1 and W2, W1, W2 and W2, W1, W1 and W1, W2, W2. Probability of each is 0.5*0.5*0.5. since there are 4 of them, probability of 3 set occuring is 4*0.5*0.5*0.5 = 0.5. so both are as likely!

[@lupus.andron.exhaustus]

Before I watched the video to see the solutions, I made up my mind by intuition and I thought it would be better to bet on a win in two games. When two players or two teams meet in a sports match, one if which will always be better than the other, even if it is just a very, very small amount of being better. The competitors might play in the same league, their performances may very well be on the same level, but still one of them will always be slightly better then the other. Following this thought it is logical to assume that the better play will win the first match, and will also win the second match.

[@janglinjack2190]

Professional sports bettor here: The answer here would be correct if the question was, is it more likely for the match to end in two or three sets? The question as stated does not factor in the betting odds. Betting lines are constructed to account for this in almost every circumstance. You generally have to lay around 2 to 1 or more to bet the under 2 and 1/2 sets.

[@randomacc77777]

how often do you use mathematics in sports betting? asking as a student researching these types of problems

[@janglinjack2190]

@randomacc77777 every bet should have positive expected value if you're trying to make a profit. You can't figure out if you have that edge unless you're doing the math. (Technically you could find arbitrage opportunities, but that's still not optimal without running the numbers) With experience, you don't need to do calculations every time, and you come across similar opportunities but you still have to know the math behind it.

[@samarthchohan106]

​@randomacc77777some maths is always used but it is very complex stats

[@yurenchu]

From the thumbnail: Suppose that one player is (slightly) stronger than the other player and hence has greater probability to win a set than the other; so one player's probability of winning a set is (0.5+e) and the other player's probability of winning a set is (0.5-e). Then the probability that the first two sets are won by the same player is (0.5 + e)² + (0.5 - e)² = 2*((0.5)² + e²) = (0.5 + 2e²) , while the probability that the first two sets are won by different players is (0.5 + e)(0.5 - e) + (0.5 - e)(0.5 + e) = 2((0.5)² - e²) = (0.5 - 2e²) . So the probability of the match ending in 2 sets (= 0.5 + 2e² ) is greater than the probability of the match ending in 3 sets (= 0.5 - 2e² ). So if the payout is double of your bet, then it's more advantageous to bet on the match ending in 2 sets.