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.
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 are 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.
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.
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.
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.
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.
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.
Basically if p(W) = 0.5, i.e. both players have same probability of winning, then 2set vs 3set are the same. otherwise, it's more likely to end in a 2 set. (WW or LL is more likely than WL/LW)
You don’t need third match, because third match guarantees victory for the winner with 100% probability, when played. In actually you only need to calculate probabilities up to two matches, also pay attention to the question, if the match end in two matches or 3 matches, not based on players side
If we call the players A and B, there are four possible combinations of A wins and B wins: AA, BB, AB, and BA. The first two end the match after two sets, while the second two force a third set. Assuming we don't know if A is a much stronger player than B or vice versa, one would assume that we have an equal chance of seeing a two-set match or a three-set one. Mathematically, this question isn't so much different from the socks-in-a-drawer riddle. Assuming the number of socks of each color is equal (but also infinite, such that pulling out a sock of one color does not deplete the supply of socks of that same color!), you are equally likely to get a matching pair with either 2 socks or 3-- your first two socks have a 50/50 chance of matching, and if they don't, your third sock must match one of them: there are not three different sock colors, just as there cannot be three different winners in a three-set tennis match between only two players.
Without knowing how the players rank against each other, 2 sets would be more likely than 3 to produce a match win. There are 3 possible outcomes after 2 sets: 2 are one of the players winning both sets and 1 where each player wins a set. It’s twice as likely that one player will win both sets (2/3) than each wining one (1/3).
Easier thinking is if both player are equally good, then its coin toss, so 2or3 game is equal, but if one player was so good, they would always win in 2. So if one player is better than the other then there is more chances it ends in 2 games.
The thing that stood out to me is that bets are typically not even odds. Given that the bookie would have done this calculation or looked at historical data and you would have an approximately even chance of winning by betting for either 2 or 3 sets. That said I still liked the video and the approach!
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.
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
Without doing all the maths... If two players are of equal ability and have a 50/50 chance of winning a single set, then after set 1 there's a 50/50 chance of it going to 3 sets (whoever wins set 1). But if one player is stronger and has a greater than 50% chance of winning an individual set, then they have a greater than 50% chance of winning the first 2 sets. So for equally matched players, the chances of 2 or 3 sets is 50/50. But with unequal players, the chance of a 2-set match is greater. So always bet on 2 sets.
Kinda crazy. Is it because most of the 2 win matches are "big fish from small ponds" that just get absolutely bodied by the "real big fish" at the nationals? Like if my mommy got matched against the siren Williams or whatever
Another easy solution: Imagine it was an equally fair match, then p=0.5. whoever wins set 1 has a 0.5 probability of winning set 2, so there is a 50% chance of 2 sets, 50% of 3 sets. Intuitively we can tell that if the probability of player 1 winning was above 0.5, then they are more likely to win in 2 sets. To prove this, the prob of 2 sets is WW + LL: (0.5+e)^2 +(0.5-e)^2 which is 0.5 +e^2, therefore greater than a half. Here e is the additional probability above 0.5
Impressed with all of Presh's math, but 1 second worth of common sense says that if one player is superior to the other, then he/she is most likely to win both sets. Skill doesn't dissipate in this example. If the players are exactly even, then the outcome is random, and the odds of two sets is exactly the same odds as the match going to 3 sets.
Because win has probability p and loss probability 1-p. Hence p^2 vs (1-p)^2 Or you could say that loss of first player is the win of the second with the probability - say - k. And get p^2 vs k^2. But again, k=1-p
The thing I jumped to first (considering it’s an interview question, so usually the answer is less important than the outside the box reasoning), I just thought that it’s more likely that one of the players will be better than the other, so the games will be weighted for the better person to win each set, thus 2 sets is more likely than 3. The best chance for 3 is if they were both very equal, which in a random game is not probable. Nice to know my gut pans out in the maths. 😄
Can we gather data from the field, and measure how many matches go 2-0, and how many go 2-1? And maybe then compare how well predictions match reality.
How significant is having first serve? Since the players are arbitrarily named, let's say that player 1 served first. So player 1's win percentage in the first game is not only based on skill, but also influenced by serving first. If serving first conveys a significant advantage, relative to the skills of the players, then 3 may be the proper bet. It all depends on how significant that advantage is.
I can't wrap my head around this one for some reason. A game 3 is only played if the winner of Game 2 is different than the winner of Game 1. If the probability of either player winning is equal (50%) then the probability of a Game 3 being played is 50%. With equal odds for either player winning each match, the probability of the match ending after Game 2 should be equal to the probability of the match going to 3 games. The reason real outcomes don't match this is because matchups are not equal and the player who wins Game 1 is likely the better player that day and has much higher odds than 50% to win any game in the matchup. I don't understand how the odds of a 2 game match could be statistically higher than a 3 game match in the case of even odds for each game (I ran a quick simulation 1000 times to confirm my thoughts before I posted).
What is the probability of a player winning 2 sets in, say, 100 sets played? Answer: probability very High. Win 2 sets in 99 sets played? Answer: high, but slightly lower... Win 2 sets in 98, 97, 96, etc sets played? Answer: probability is reducing... Win 2 sets in 3 sets played? Answer: probability P Win 2 sets in 2 sets played? Answer: a little less than probability P So: A 2 set win is more likely in 3 sets than 2 sets!
The calculation is so unnecessary 1. the match will either end in 2 sets or 3 sets, just see if p^2+(1-p)^2 is larger than 0.5 2p^2-2p+1 = 2(p-1/2)^2+1/2 >= 1/2, so bet on 2 sets or you can use inequality sqrt(a^2+b^2/2) >= a+b/2
This is a ridiculous interview question unless a detailed explanation of working out the maths is required. 50/50 to simply get lucky and guess the answer.
interview question is to show thinking process. hedge funds probably wouldn't require this level of maths, but this level of problem solving to at least structure the probability space is reasonable.
You can logically think what would be the right answer. After one match, the person who won is more likely to have more skill and win again. Unless they both have exactly the same chance of winning, the outcome favours the match ending after 2 games.
@@simoncirkel6348no, because players have an average that they sometimes play better than, and sometimes play worse than. Daniel Kahneman does a brilliant job of explaining this in his book: thinking fast thinking slow.
Nope, once again, MindYourDecisions doesn't know tennis and thus uses a completely wrong method. Given a random tennis match, always bet on 2 sets, since the vast majority of 3-set matches end in 2 sets. That's literally all there is to it. The VAST majority of tennis matches end in 2, so given no other info about the match, always bet on 2.
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.
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 are 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.
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.
My conclusion also, came to that almost instantly. Players aren't evenly matched.
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.
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.
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.
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.
9:19 WW? Woodrow Wilson? Willy Wonka? Walter White?
"Ha, you got me!"
are those bots commenting??
I definitely see a pattern in their comments...
Everyone here is a bot except you
Yes, all of them advertising corn.
Apparently, but what the point of it is, I don't know.
The first comments are usually from bot accounts.
"Your video is so inspirational! Love your style!"
How is there more options for a 3 set victory and only 2 options for a 2 set victory, but the 2 set win is more probable?
Because the options for the 3 set victories are each (and as a sum) less likely than the options for 2 set victories
Basically if p(W) = 0.5, i.e. both players have same probability of winning, then 2set vs 3set are the same. otherwise, it's more likely to end in a 2 set. (WW or LL is more likely than WL/LW)
You don’t need third match, because third match guarantees victory for the winner with 100% probability, when played. In actually you only need to calculate probabilities up to two matches, also pay attention to the question, if the match end in two matches or 3 matches, not based on players side
But when summing probabilities of 3 matches is the same of probabilities of 2 matches with one win 1 in loss, WL and LW
If we call the players A and B, there are four possible combinations of A wins and B wins: AA, BB, AB, and BA. The first two end the match after two sets, while the second two force a third set. Assuming we don't know if A is a much stronger player than B or vice versa, one would assume that we have an equal chance of seeing a two-set match or a three-set one.
Mathematically, this question isn't so much different from the socks-in-a-drawer riddle. Assuming the number of socks of each color is equal (but also infinite, such that pulling out a sock of one color does not deplete the supply of socks of that same color!), you are equally likely to get a matching pair with either 2 socks or 3-- your first two socks have a 50/50 chance of matching, and if they don't, your third sock must match one of them: there are not three different sock colors, just as there cannot be three different winners in a three-set tennis match between only two players.
Without knowing how the players rank against each other, 2 sets would be more likely than 3 to produce a match win. There are 3 possible outcomes after 2 sets: 2 are one of the players winning both sets and 1 where each player wins a set. It’s twice as likely that one player will win both sets (2/3) than each wining one (1/3).
You can't sum up probabilities. Winning two sets of three in a row doesn't happen 2/3 of the time.
Easier thinking is if both player are equally good, then its coin toss, so 2or3 game is equal, but if one player was so good, they would always win in 2. So if one player is better than the other then there is more chances it ends in 2 games.
The thing that stood out to me is that bets are typically not even odds. Given that the bookie would have done this calculation or looked at historical data and you would have an approximately even chance of winning by betting for either 2 or 3 sets. That said I still liked the video and the approach!
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.
Not only do I not know the answer, I don't even know what the question is.
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
Without doing all the maths... If two players are of equal ability and have a 50/50 chance of winning a single set, then after set 1 there's a 50/50 chance of it going to 3 sets (whoever wins set 1). But if one player is stronger and has a greater than 50% chance of winning an individual set, then they have a greater than 50% chance of winning the first 2 sets. So for equally matched players, the chances of 2 or 3 sets is 50/50. But with unequal players, the chance of a 2-set match is greater. So always bet on 2 sets.
Kinda crazy. Is it because most of the 2 win matches are "big fish from small ponds" that just get absolutely bodied by the "real big fish" at the nationals?
Like if my mommy got matched against the siren Williams or whatever
Another easy solution:
Imagine it was an equally fair match, then p=0.5. whoever wins set 1 has a 0.5 probability of winning set 2, so there is a 50% chance of 2 sets, 50% of 3 sets.
Intuitively we can tell that if the probability of player 1 winning was above 0.5, then they are more likely to win in 2 sets.
To prove this, the prob of 2 sets is WW + LL: (0.5+e)^2 +(0.5-e)^2 which is 0.5 +e^2, therefore greater than a half. Here e is the additional probability above 0.5
So what's the resulting p from the database you used?
Impressed with all of Presh's math, but 1 second worth of common sense says that if one player is superior to the other, then he/she is most likely to win both sets. Skill doesn't dissipate in this example. If the players are exactly even, then the outcome is random, and the odds of two sets is exactly the same odds as the match going to 3 sets.
Can someone please explain why the equation for losing both sets is different than the one for winning both sets? I don't get it... xD
Because win has probability p and loss probability 1-p. Hence p^2 vs (1-p)^2
Or you could say that loss of first player is the win of the second with the probability - say - k. And get p^2 vs k^2. But again, k=1-p
You have to remember that the best player will win 2 games. Odds are, if there was only one game then the best player will win.
The thing I jumped to first (considering it’s an interview question, so usually the answer is less important than the outside the box reasoning), I just thought that it’s more likely that one of the players will be better than the other, so the games will be weighted for the better person to win each set, thus 2 sets is more likely than 3. The best chance for 3 is if they were both very equal, which in a random game is not probable.
Nice to know my gut pans out in the maths. 😄
We have two graphs about this probabilities like that
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Can we gather data from the field, and measure how many matches go 2-0, and how many go 2-1? And maybe then compare how well predictions match reality.
Pr(3 sets) = WL | LW.
This helps me a lot. But i still get confused over when i should count possible outcomes vs use the probability of an outcome.
As someone preparing for hedge funds interview.... I'd have said 2 sets without making any calculations hoping they dont ask for explanations
I haven't watched the full video yet I hope 2 is the answer
How significant is having first serve? Since the players are arbitrarily named, let's say that player 1 served first. So player 1's win percentage in the first game is not only based on skill, but also influenced by serving first. If serving first conveys a significant advantage, relative to the skills of the players, then 3 may be the proper bet. It all depends on how significant that advantage is.
I don't do betting.
anyways nice video!
Is this a joke
Switch doors 🤷♂️
binomial probability
Maybe i should work for a hedge fund, that was easy
Thank you for your efforts! Your videos are always so interesting and professional. Keep it up!🏵🏈⚾️
idk i guess 2
I can't wrap my head around this one for some reason. A game 3 is only played if the winner of Game 2 is different than the winner of Game 1. If the probability of either player winning is equal (50%) then the probability of a Game 3 being played is 50%. With equal odds for either player winning each match, the probability of the match ending after Game 2 should be equal to the probability of the match going to 3 games. The reason real outcomes don't match this is because matchups are not equal and the player who wins Game 1 is likely the better player that day and has much higher odds than 50% to win any game in the matchup. I don't understand how the odds of a 2 game match could be statistically higher than a 3 game match in the case of even odds for each game (I ran a quick simulation 1000 times to confirm my thoughts before I posted).
What is the probability of a player winning 2 sets in, say, 100 sets played?
Answer: probability very High.
Win 2 sets in 99 sets played?
Answer: high, but slightly lower...
Win 2 sets in 98, 97, 96, etc sets played?
Answer: probability is reducing...
Win 2 sets in 3 sets played?
Answer: probability P
Win 2 sets in 2 sets played?
Answer: a little less than probability P
So: A 2 set win is more likely in 3 sets than 2 sets!
The calculation is so unnecessary
1. the match will either end in 2 sets or 3 sets, just see if p^2+(1-p)^2 is larger than 0.5
2p^2-2p+1 = 2(p-1/2)^2+1/2 >= 1/2, so bet on 2 sets
or you can use inequality sqrt(a^2+b^2/2) >= a+b/2
This is a ridiculous interview question unless a detailed explanation of working out the maths is required. 50/50 to simply get lucky and guess the answer.
interview question is to show thinking process. hedge funds probably wouldn't require this level of maths, but this level of problem solving to at least structure the probability space is reasonable.
You can logically think what would be the right answer. After one match, the person who won is more likely to have more skill and win again. Unless they both have exactly the same chance of winning, the outcome favours the match ending after 2 games.
@@simoncirkel6348no, because players have an average that they sometimes play better than, and sometimes play worse than. Daniel Kahneman does a brilliant job of explaining this in his book: thinking fast thinking slow.
@@simoncirkel6348 agree, I would do common sense method first and sketch a graph before doing the maths.
Nope, once again, MindYourDecisions doesn't know tennis and thus uses a completely wrong method. Given a random tennis match, always bet on 2 sets, since the vast majority of 3-set matches end in 2 sets. That's literally all there is to it. The VAST majority of tennis matches end in 2, so given no other info about the match, always bet on 2.
Your videos are always so insightful and intelligent. Thank you for your contribution to the education and development of the community!🐞🏄🦎
Every video is a real treat! Thank you for your creativity and professionalism.🏆❗️🏄
Your ideas are always so creative and interesting. Thank you for your hard work!🏄⛹️🌃
Every time I admire your ability to create quality content. Keep making us happy with your videos!🙊🧢🟫