This is awesome! There's a nice analogy, there's Math theory, there's code and a good visual representation. Your explanation is insightful and clear. Anything one could hope for watching the solution to a puzzle. Really inspiring stuff, man!

Interstingly, the expected value for the score difference is 0, and it doesn't contradict this result! Alice's less frequent victories simply have a larger score difference on average than Bob's more consistent victories, and it all cancels out!

Statisticians after doing 4 continuous hours of complex math: yeah this may or may not happen

My first reaction: it only takes 3 flips for Alice to get 2 points, whereas it requires 4 for Bob

new Monty Hall problem just dropped

As soon as you suggested the dog running onto the field when Bob scores, preventing both teams from scoring, I very quickly realized why that's applicable in the coin flip scenario. Excellent analogy.

A cool use case of this occurred when the “hot hand fallacy” (the belief that basketball players who score a shot are not more likely to score the next shot) was reviewed. Originally statisticians found that it is indeed a fallacy and people aren’t more likely to score based on whether their previous shot was a made one, but it was found that a major statistical flaw was present in their workings. For any finite sequence of iid coin flips the expected proportion of HT will always be higher than HH (converging to equivalent as sequence length approaches infinity, but getting further apart from the ‘naive’ expectation the longer the chain of H’s). This means when analysing a series of misses and makes you expect there to be more makes followed by misses than makes followed by makes. Highly recommend reading miller and sanjurjo’s paper for those intrigued.

Creating a program for this, and running the same simulation 10,000,000 times, the following values appeared: Alice won 4,576,558 times Bob won 4,859,705 times They drew 563,737 times So it's pretty spot on compared to the probabilities at 3:22

Mihai: I'm gonna describe a situation that we're more familiar with, a Soccer game. Me: Oops, I'm in the wrong classroom.

Every time a head comes up there is an equal opportunity to score on the next throw, so if you enumerate all possible results they score the same number of points. Alice can get a higher score than Bob in a single game. With six throws Bob's best win is 3-0 (HTHTHT) but Alice can win 5-0 (HHHHHH). Alice has more decisive victories but they score the same number of points across all possible games so Bob must win more often.

Good anlogy regarding wasting time in the game only after Bob scores. 🎉 I caught on when you started talking about it.

the intuition with the dog makes total sense. thanks, very interesting problem.

Beautifull! I ran a montecarlo sim to check the results and thought I did something wrong when I saw that Bob had the advantage.your explanation made everything clear. Thanks for the content man!

Somehow when this was going around Twitter, it never occurred to me think of it as a Markov chain. But then I opened this video and was like "obviously you can solve this with a Markov chain" and then felt validated that that's what you did.

simple answer: while HT is just as likely as HH for a two-coin pairing, HH combinations can overlap while HT combinations cannot. After scoring a point, there is a 50% chance that Alice will score again. Bob gets no such equivalent chance. Therefore Alice has a higher chance to win. Another way of noting it is that while each pair's individual odds to score are equal to both side, Alice's theoretical maximum is 99 points to Bob's 50. Therefore she is getting more chances to score. Edit: Well shit my answer is wrong.

I like thinking about the "end game". Tie game, two turns left, Heads was last flipped... what could happen next? If it's tails, Bob takes the lead, and indeed the game, as he has victory locked up; there's not enough coin flips for Alice to mount a comeback. If it's heads, Alice takes the lead, and *could* even grow her lead (but a win is just a win, no matter by how much), while Bob could also come back and score a tie. That situation seems straightforward to work out mentally, clearly demonstrating an advantage for Bob. From there, it doesn't seem hard to extrapolate that even if it's 50-50 up until that endgame, there will always be some advantage for Bob. Good puzzle!

Easy explanation: The difference (Alice - Bob) is a random walk with mean zero. Every time after it goes up (HH) it immediately takes another step, while after it goes down (HT) it could hit a dry sequence of tails before taking another step. So the walk has the tendency to stay low and that's the advantage for Bob.

This is literally the Humble-Nishiyama randomness game that Vsauce2 covered before. This is such a nice, technical perspective!

i hope Dr Litt sees this, honestly the only intuition to proof that ive seen proposed

Amazing video! I'm not the biggest fan of probability but something about the way you shared intuition for this seemingly simple problem is absolutely beautiful.

I intuitively guessed the correct answer, but getting my head around why took a bit of effort. It helped me to look at just a 3 coin toss. I found that the expected number of points for Bob and Alice were the same, but two of Alice's points come from a single win when the result is HHH. The only other winning outcome for her is THH. Bob has three winning outcomes: HTH,HTT and THT.

What's equal is the probability of both having score the same goals after large enough N matches. But scoring equally goals does not mean matches are won equally since 1-0 is a win, but so does 5-0. Alice is then more likely to win by a greater difference of goals, because a sequence of HH is intuitively easier, at the cost of loosing more matches.

That analogy is so smart… you are an amazing teacher

Stunning video: fascinating topic and explanation allows for even me to understand where you're coming from on each step.

Knew it was Bob from the initial question. It's the same problem you get from modelling the value of the ability to pause the clock with timeouts toward the end of a sporting event. But I'm quite surprised by the magnitude of the effect. My guess would have been closer to 0.3% than the 3% it turned out to be. That's over half a point worth of value.

The explanation that felt the most intuitive to me emerges from this short series of observations: - There are only points on the line when the previous result is H - If the previous result is H, both players have a 50% chance of scoring (therefore, for an infinite game, the odds are even) - A win for alice is always followed by either another win or a win for bob (i.e. the sequence produced by adding H or T to a sequence ending with HH will always end with HH or HT) - Therefore, if the last pair of heads is anything other than HH, we know that the last point of the game went to Bob, which is a 3/4 chance. Ergo, in a game in which x points are scored, Point 1 through Point x-1 have a 50/50 chance of going either way, but Point x has a 75% chance of going to bob, which gives him the slightest edge. Now that i spell it out i guess it still isnt that intuitive, but it basically comes back around to "the game is more likely to end after a Bob win than an Alice win, giving him a slight edge"

Thank you, professor, for this stimulant problem. I thought of a simplest way to prove that Bob wins in the most of case. This is my reasoning: I represent all issues as a binary tree going from the level 0 to the level 100. When in a node (H or T), I toss the coin and have H or T for the next level, then mark the node with +1 if Alice wins, and -1 if Bob wins and 0 otherwise. When all my tree is full, I have certainly 0 as sum of all the nodes (because on each node I have +1, I have a node with -1 just at the same stage). Now I go to the leaves at the level 100 (I have 2 power 100 leaves). For each leaf (i), I sum the points from the root to the leaf, call it S(i), the Alice wins if S(i) is positive, Bob wins if it is negative, and no one wins at 0. The I must count the number of win leaves for each player. A player wins the game if it wins in more leaves. Or the sum of all S(i) must be 0 (because a each level the some of points is 0), so the sum of S(i such Alice wins) is the opposite of S(I such Bob wins). But when Alice wins, she wins with a major score (because of the possibility of succession of H, example Alice can win with score 100, and the best score for Bob is 50, and more precisely for ever branch where Bob wins we can find a branch where Alice wins with best score, so the mean for Bob is less then Alice), then in the count of leaves Bob must wins. I hope that my reasoning was clear. And my English understandable. Thank you very much.

Thanks for this! Was looking for a proof that I was satisfied with. When I first saw the problem, my intuition was that for any finite N flips, E[score difference at time: N-(time of last win - 1)] = 0. At the next flip either Alice or Bob wins with equal probability, whoever wins would now have a higher chance of winning. If Bob wins, Alice has to “wait” for the next heads so Bob is effectively time wasting meaning Alice has less time than Bob to catch up compared to how much time Bob has to catch up if Alice wins.

I couldn't wrap my head around this until I saw the distributions themselves: Points Distribution: HHHHH = Alice + 4 HHHHT = Alice +3, Bob +1 HHHTH = Alice +2, Bob +1 HHHTT = Alice +2, Bob +1 HHTHH = Alice +2, Bob +1 HHTHT = Alice +1, Bob +2 HHTTH = Alice +1, Bob +1 HHTTT = Alice +1, Bob +1 HTTHH = Alice +1, Bob +1 HTTHT = Bob +2 HTTTH = Bob +1 HTTTT = Bob +1 HTHHH = Alice +2, Bob +1 HTHHT = Alice +1, Bob +2 HTHTH = Bob +2 HTHTT = Bob +2 I realized that Alice and Bob's total scores if every combination happened was equal for sets of 2, 3, 4, 5, etc. BUT!!! Advantage distribution: Alice +4 = 1 Alice +2 = 1 Alice +1 = 4 Equal = 3 Bob +1 = 4 Bob +2 = 3 Alice loses the number game because she is reliant on getting that single +4 string while Bob is more flexible and thus more likely to have a win.

Before I watch the vid: - reminds me on Penney's paradox - by linearity of expectation, both are expected to get 99/4 = 24.75 Let's go case by case: - Start (State 1) T or H? - If T, same as start (State 1) - If H, both just need 1 more (State 2) - T or H? - If T, Bob gets points, go back to state 1 - If H, Alice gets points, back to state 2. So essentially we're asking to look at all the heads and see if there's more tails or heads afterwards. I think that would slightly favor Bob since if we look at a heads, that removes 1 possible heads (i.e. let's assume there's 50 heads and 50 tails, looking after a heads, means the next is 49.49% to be heads vs 50.51% to be tails). As a quick example, let's look at the case were there's only 3 flips. HHH - Alice wins 2-0 HHT - tied 1-1 HTH - Bob wins 1-0 THH - Alice wins 1-0 HTT - Bob wins 1-0 THT - Bob wins 1-0 TTH - Tied 0-0 TTT - Tied 0-0 So Bob won 3x, Alice won 2x and 3 ties. As we expected both are expected to get 2/4 = 0.5 per game, but the key insight is that when Alice wins, she wins by a bigger margin on average, so Bobs will win more often by a smaller margin

How would a 4 player variant of the game go? Naïvely, I'm thinking the analysis presented here doesn't interfere with itself. So the HH/HT and TT/TH can be analyzed separately to get expected points, giving us a double tie between the alternating options winning, then the constant options 1/2sqrt(nt) behind. Is that valid, or is there some part of the reasoning I've caught on?

So it’s the finite length of the game that gives advantage to Bob. Makes sense.

Possibly the easier way to intuit it is to ask how much Alice or Bob can win by. The average score difference is 0, but when you get 100 heads in a row Alice wins by 99. Optimal for bob is only 50. We can then ask the question, given the games that Bob won, what was the average score difference? Likewise with Alice, we can ask what the score difference when Alice wins? Lets annotate then S|B as average score given Bob won, S|A as average score given Alice won, and score given a tie, S|T, which is zero. P(A) is the probability that Alice wins, and so on. The average score is P(B)*S|B + P(A)*S|A + P(T)*S|A. The score given a tie is zero, so that disappears. We can see by analysing the game that the average score difference is zero. This means that P(B)*S|B = -P(A)*S|A. The fact that Alice can win by more implies that she will win less frequently. Not strictly rigorous, as we cannot always get from looking at the peak win to the average win, but a useful estimation trick in my experience.

Dog on the field, very nice analogy, immediately understood the whole thing intuitively. Nice vid.

I intuitively picked Bob, by a hand-wavy sort of version of your proof. I asked, "Who is more likely to have scored last at any given point?" The answer to this question is Bob, because Bob scored last in every sequence that ends in a tails, but he also scored last in sequences that end in exactly one heads. This reasoning holds at every point in the game going right back to the beginning, so it seemed sensible that Bob maintains a small advantage.

it's very fitting that the dog runs out on the field every time "tails" comes up! XD In my head the dog is called Tails

I wrote a simple program to simulate a million games and print out the percentage that Bob and Alice won, and your math was correct! I just generate a single 128-bit random number, such that every bit in the number will be a random 0 or 1, and then loop over the integer 100 times, bit-shifting it by i each iteration and "and"ing that with 0b11 to get the first two bits. If the value in the new integer storing the first two bits is 0, that means that Alice won, and if it is 1, that means Bob won. Pretty efficient and the math checks out!

There’s a very interesting connection between this principle and why the stats behind the “hot hand fallacy” were wrong.

For anyone that wants to try a similar problem that is a bit more guided can try:"15-S2-Q12 step problem". This is question 12 of step 2 at 2015. Step is Cambridge maths entrance exam.

Who made the music is playing in the background? Great video btw

I would explain this by invoking dependent vs independent probability. Because the game is written up as a single chain of 100 rolls where the pairs are dependent, the probability can deviate from the obvious answer assuming independent samples that they should be even. In other words if you played the game such that you rolled 50 independent pairs, then the chances of winning would be 50/50. But since the rolls are dependent, the H assumed for the first roll in either scoring scenario influences the rest of possible scores

Very interesting. What if we consider a modified version of the game “Alice HH vs Bob TH”, so Bob now score for every “TH” instead of “HT”. Intuitively it seems that the answer should be the same: Bob is more likely to win (I checked, it is indeed the case). However, I do not know how to apply your argument with the dog now. I guess here we should just say: if C=H, both are as likely to earn a point, but if C=T, only Bob can score on the next time step. So instead of the dog, I would say that “special rule” is that a brick wall randomly appears in front of Bob’s goals and it stays there until Bob scores. Once Bob scores, wall is removed and both players are equally likely to score again.

I havent seen this video yet but I found them to have tied. imagine grouping every THHH...HHT into chunks (ignoring consecutive tails completely) the expected value of every chunk for Bob is 1 the expected value of every chunk for Alice is 1/4 + 2/8 + 3/16 + 4/32... = 1 Since the expected values are both 1, they should tie in the long run, no? unless this question doesn't ask for a long run

Since they have the same expected score, but Bob is expected to win, it is reasonable to assume that Bob is more likely to lead by a small margin if he wins, while Alice is more likely to lead by a larger margin if she wins. This also meshes well with Alice's ability to score points several throws in a row, while Bob can't: It is easier for Alice to get extremely good scores, while Bob's scores are usually more moderate. Which again is what we see in the probability distribution plot at the start.

Of course you publish this proof and video AFTER my stats class is over so I can't show this to them.

The part about adding the TT points to both is crazy to me. I think I understand that that makes their distribution of points identical, but my brain refuses to accept that they can have identical distributions of points with Bob still having an advantage to win. Really great video! Super well done! Bob scoring and then stalling makes sense as a "strategy". Alice being able to rally a bunch of points also feels like an effective "strategy" but for winning more decisively not necessarily more consistently. It feels like Bob is playing defensively while Alice is playing offensively and the old adage "offense wins games, defense wins championships" comes to mind. Thanks again!

Lovely analogy that instantly made it intuitive, nice!

If a heads just occurred, they're both equally likely to score. So every time alice scores, bob could also score next round. But if a tails just occurred, they're both two away from potentially scoring. So when bob scores, alice is locked out for a turn. Alice's sequence may be able to self-intersect, but on a fair coin a run of heads is going to show up less often than alternation. I think any advantage she might have had from that is engulfed by her requiring (at least) a triple heads to ever benefit from it, which is going to be less likely than a sequence of three flips containing at least one tails. To contrast, I think if alice scored on "HTHT" and bob scored on "HHTT", the situation would match people's general intuition.

I'm an amateur soccer player My takeaway from this video is that we should let a dog into the field every time we score a goal and that will increase our chances of winning Thanks!

i just did two thought experiments - if there's two flips, they're both equal, if there's three flips, then Alice scores 3 and Bob scores four, so I'd guess Bob wins because the length of the sequence shouldnt make a difference after 3 flips. that doesn't give much of an intuition for *why*, though

You can see the phenomenon early on with just 3 flips. Of the 8 scenarios, 2 have Alice winning (HHH, THH), 3 have Bob winning (HTH, HTT, THT) and 3 have a tie (HHT, TTH, TTT)

Really intuitive explanation! Great vid

H Draw T Draw HH A HT B TH D TT D THH A THT B TTH D TTT D HHH A HHT D HTH B HTT B When adding a T (to the start of the chain), the result would be the same has the previous chain. When adding a H (to the start of the chain), Alice is wasting 1 H on a chain she’s already winning while Bob has not wasted any yet. Adding a T gives Bob the previously advantageous scenario while adding a H would grant both sides an equal amount of extra points. However, Bob is more likely to gain points in the scenarios with more T (more likely to go from Draw to Bob), while Alice is more likely to gain points in the scenarios with more H (Alice gains less overall wins). Therefore, Bob would win more than Alice.

Fun follow up question: If we had an infinite sequence of score differences, and randomly checked some element of the sequence if Bob or Alice is winning, would the probabilities of them winning in that step be equal?

Markov chains is a nice cameo. Is there a generalization that would allow to solve a similar problem with combinations of length three instead of two?

Once Bon wins a point, the probability of either of them winning a point the next flip goes to 0, when Alice wins, it goes back to 50\50

This is really well laid out. Thanks for a great vid!

Assume any divisible by 4 flip count. Consider all 16 length 4 subsequences. Now all subsequences are equally likely (this is key). We take a look and see half start with H, half end with H. So half the sequences end in H and from there have a 50% of point for Bob or Point for Alice. QED.

I think the thing that breaks intuition here is that the results depend on the game having a fixed length. If the game stopped after one player reached a certain number of points, it would be different! (If I'm not mistaken, that case really IS 50/50.)

Great video! This averaging feels like convolution, which feels like diffusion. The limiting process of a discrete random walk is Brownian motion. What would be the limiting process (continuous time, continuous space) for this more complicated Markov chain?

For every H that comes up, it increases total H count, so Alice should be slightly favored I think. Another ay to say it is that Alice can string together wins but Bob cant. Oops im wrong because I was thinking too long term. This is why I also simulate before answering if the answer matters

I don't understand the intuitive proof, why is the dog coming on the field after bob scores an advantage? both bob and alice are equally likely to score, so it still seems random who would benefit. I guess the only use case is when bob is ahead, and it's the almost the end of the game, ohh i get it now.

I'll note that you get the same probabilities of Bob winning if he scores a point on TH instead of HT. I guess the corresponding soccer intuition would be that every so often a dog runs on the field, and only Bob can score immediately after the dog runs onto the field?

Interesting, the limited amount of rounds is what driving bobs advantage. What are the odds if it was the first to 100 points?

The way I concluded Bob would come out on top is absolutely not rigorous, but I wondered if it could be improved upon. I considered any chain of flips beginning with a head. Using Alice scoring as +1 and Bob as -1 you would have the expected value of the chain as 1/2*(-1) + 1/4*0 + 1/8*1 + 1/16*2 + 1/32*3 +... I didn't care to calculate this limit but I figured it had to be negative due to how quickly the terms decrease, and as such you only have to consider how big of an effect the potential for the 100 flips to end on a sequence of Heads is and if that is strong enough to outweigh the expected value of the individual sequences

I find myself having a hard time interpreting the charts of the victory function that are on screen from minute 19 onwards. Can someone break it down for me? If I try to explain it, I think I'm just going to confuse myself more. I feel like that missing "this is how you interpret this visual information" is the one big flaw in this video.

don't know if this works but i simply tried the simplest case, a case for 3 and noticed that bob has a high probability of winning. i don't really understand induction proofs but i think since we know that the last coin flip has a 50/50 chance of being heads or tails and the same is for the next flip we can say that alices and bobs number of points increase at the same rate and thus by induction we conclude that Alice's probability of winning will never increase past bobs or something

Very cool! Thank you. I have seen value functions in the context of MDPs / RL but nice to think of them in context of more classic problems in probability. Any recommended reference / books on the topic?

Flip it 3 times. Alice wins 2/8, Bob wins 3/8 and Nobody wins 3/8. It's kind of obvious to stick with Bob. It would be equal odds if the question is who would have the max points at the end. The E(' who wins eventually') is significantly impacted by the the choice of "HH, TH" combination. Now on to the video!

Interesting! I thought Alice would have a higher probability of winning because her scoring condition is symmetric whereas Bob is order dependent.

How is it scored if we get HHHHHT 4 points for Alice and 1 for bob?

Best intuition I have for this problem. We know that any given flip has an expected point value of 0.5 Heads results => that the next flip, regardless of what it is, will result in a point. So the next flip is worth more than an average flip. Tails results => that the next flip, regardless of what it is, will result in no points. So the next flip is worth less than an average flip. Alice only scores on Heads. Bob only scores on Tails. So the +1 point advantage Alice gains from a Heads result means less than the +1 advantage Bob gets on a Tails result because there are more points to be had in a game with a Heads result. The longer the game goes on, the less this advantage matters because a longer game also means more points to be had, diluting the effect.

I almost wonder if it’s even more important to outline that that any time there is an H, there will be a T at some point, so bob always has the assurance that he will gain one point, while alice does not get any sort of assurance when an H shows up.

im not good at math so i just calculated by hand every outcome for just 2 steps from point (0,H). kinda obvious there is an advantage and how it occurs.

I honestly thought bob was intuitive and obvious, as every chain of points alice can get will terminate with a point for bob, and the probability of alice getting at least one ahead of bob is slim.

so basically they both have the same chance to score but after Bob scores his coach lets the timer run down while Alice's uses a time-out

What is the math accounting for Alice able to score 2 with 3 flips "HHH" vs Bob needing 4 flips "HTHT"? My intuition told me that gave Alice an advantage, but was not addressed in the video from my novice understanding. Thank you for the video and increased understanding!

Without having watched it, HH feels a but more likely to happen, because you can chain them so easily. For 3 coin tosses, HHH gives 2 points for Alice, while Bob has no way to get 2. with infinite throws this light equal out though Because looking at the chain of events in every state it’s either 50/50 who gets a point (after an H), or noone gets a point (after T)

Just paused at 0:45 - shouldn't it be obvious the answer is Alice? HT only appears in one direction so-to-speak, meaning HTH is only 1 win for Bob. Meanwhile, the equally likely HHH gives Alice 2 points. That alone should immediately answer that Alice is more likely to win. If it were instead a question of same vs switch the question would be different and answer would be equal likelihood

Naive answer before watching the video: I think Bob will win, because he always gets "the last word", so to speak. He's the only player who scores when a tail is landed. When a coin lands H, the next coin has a 50% chance to give a point for Alice or for Bob, but when a coin lands T, neither player can score a point until the next H. Any scoring opportunity for Alice is also a scoring opportunity for Bob, but only Bob can shut out the game by scoring the last point before a long sequence of Ts, thereby giving him a slight advantage over Alice.

A great way of understanding the 100-flip game is to play the 3-flip game: HHH - A HHT - tie HTH - B HTT - B THH - A THT - B* TTH - tie TTT - tie *this one is sorta the one throwing off the tie if you think of a certain parity of HH and HT. It’s the ends of each sequence (or the ends in between sequences of TTs) that have the most impact on the score difference, and Bob wins those slightly more than Alice as seen here.

I went with Bob after some thinking, but mostly inuition -- I pretty immediately suspected they wouldn't be equal. I am very familiar with a 3-value version of this (where you score if a specific set of 3 values shows up), that game has a much more drastic difference between the players for certain combinations. (HHH / THH - the most extreme example, where the first pattern "HHH" can only ever "win" if it shows up as the first 3 coinflips of the game, otherwise, once a single tails is shown, the second player always scores first, as the first player cannot score without the second player getting a point first -- this version of the game was only "who gets the first showing", rather than who had the most, too) In your game, it made sense to me that any time alice has a run, bob automatically gets a point at the end of the run. This means that alice's most common two runs: "HHT" and "HHHT" are automatically going to result in a draw for the former, and a lead of just +1 for the latter, meanwhile bob's run can score without alice getting points. Alice getting "luckier" runs that go longer are just as likely as bob getting repeat scores of his own too, hence why bob wins in this scenario. -- not a formal proof, but it was enough for me to vote relatively confidently in the poll at least.

Let me see if I can summarize the video: 1. When Bob is ahead in the score, it is tails, which means it takes up 2 whole turns. 2. When Bob is losing, Alice does not waste any time, as Bob is just as likely to score. To answer - When Bob is ahead, they are more likely to stay ahead slightly longer at the end of the game, increasing the chance of victory

Some quick guessing: 1. Obviously both A and B has the same expected score of 99/4 2. A *could* get large scores as high as 99, while B can only get at most 50 3. Tian Ji’s Horse Racing Strategy comes into my mind so I'm guessing B will win more. en.wikipedia.org/wiki/Tian_Ji

Both need H, which happens .5 of the time. Then Alice needs H, which can happen, but if it does, we can assume fewer H in the future, because the total portion of H in the game remains .5. Bob needs T, which is as likely as H and equally likely over the course of the game. Therefore, Bob has slightly more opportunities to win.

Haven't watched the video yet, don't know the answer. I would guess Alice wins narrowly just because a sequence of Bob points takes more space than a sequence of Alice points.

This is a nice way to approximate pi :)

Can’t you reverse Bob’s prediction to “TH” to make this more intuitive? If Bob scores on “TH” then he will score the first point 75% of the time. Every time that a “tails” is the result of either of the first two tosses he will score first. That gives him a small advantage even though the chance of scoring thereafter is the same.

This reminded me of a problem I saw long ago that was similar. Instead of discussing the person who got more overall, it was who would get it first, TH or HH. And TH is more common there, since once a tails comes up, you're guaranteed to get a point because for it to switch over to HH you would need to have it go through TH first. I think these happen for the same reason - this behavior is the dog stalling behavior in reverse, and I would conjecture that given any series of coin flips, whoever's coin flip is more likely to come up first is also more likely to win the game when counted over a long series of flips. edit: that verison is called Penney's Game en.wikipedia.org/wiki/Penney%27s_game

Wait, HOW DID YOU APPROXIMATE LINES INTO FORMULAS?! Is there a simple way to convert line of unknown formula back to approximate formula? Im referring to this: 5:06

Whenever Alice gets a point Bob has an equal chanve to get a point on the next flip as Alice, but when Bob gets a point neither of them have a chance to get another point on the next flip. So over time i think Bob eeks it out

So if Bob instead scored with TH would Alice and Bob be equally likely to win?

My hand-wavey argument was: In the infinite case, on the transition to heads, alice has a 1/2 chance of scoring once, a 1/4 chance of scoring twice, so on and so forth. Bob will score exactly once. Summing up alice's infinite series gives an expected value of 1, and bob of 1 as well. However, since we're truncating the game, the tail end of alice's infinite series is cut off, giving bob a slight advantage. It feels like a good argument, but it's wrong, sadly, I think because if alice wins on a streak, then bob doesn't get a point :((

I have noticed that the case n=3 flips is bad for Team HH, and I suspect an induction might work. But I'm not sure yet. T start reduces to n=2, with 1W 2D 1L outcomes. But HT is a loss, and HH only wins half and draws half.

I knew this result from the same game played with a choice of triplet results, as originally invented by Walter Penney back in 1969 (10 years before I was born, but I read a lot of mathematics books and journals as a kid for fun).

Now the question is, how do I get onto the “interesting math problems” side of Twitter?

I like the music you chose

My solution is as follows: Both players cannot score until a head is tossed. Then, after a head is tossed, a run of heads will be tossed. At the end, there is a HT, so Bob will get 1 point. Alice, however, gets a number of points equal to the number of heads tossed before this point (not including the first one). The expected value of this is: 1/2+1/4+...=1, as at each flip in the run, Alice has a 1/2 chance of adding an extra point on. Hence, every run the expected value for both players is the same. Therefore, both players are equally likely to win. Could someone let me know if this proof works?

Take a similar game: Alice combi is THH, Bob Combo is HHH. Game ends as soon as either combo gets flipped. In this case it's easy to see, thaz Alice almost always wins.

I'll be honest, in the sports analogy, I don't feel like I intuitively get why the dog would be better for Bob. Here's my intuition for why Bob wins: they have same expected value and Alice can hit higher highs, which must mean Bob wins more often to compensate.