 Hello folks and welcome to the mathematics of the bridge game today. We're looking at the probability of surviving The bridge game in the fantastic Netflix series squid game There are 16 players remaining in the game and they have to cross a bridge made up of 18 Nodes as we'll call them at each note There are two panes of glass to choose from to stand one is made of tempered glass and will hold your weight The other is made of real glass and will break if someone stands on it So to do the maths we are going to make a few assumptions The first is that all the players play quickly. We're going to assume that everyone plays fairly and no one can die by accident We're going to assume and this is an important one We're going to assume that all remaining players in the game have perfect memory of how much of the bridge has been solved so far We're also going to assume that the tempered glass and the real glass patterns are chosen perfectly randomly We're also assuming that no player has any special talent or ability over any other player all are equally skilled And we're also going to assume that choosing real glass will always mean certain death It's not like it can maybe start cracking and you can quickly leap off it if you see that you you've chosen it wrong All right So what are the questions we're trying to answer firstly and foremost? What are the probabilities for each individual player to survive the game from player one through to player 16? Secondly, can we generalize the solution for any number of players? We'll call it m players in an n-node game And finally, can we generalize for the probability of surviving a particular step to be different than a half? All right, so let's get started poor old player one This one's the easiest and most trivial to figure out This unfortunate soul has to traverse 18 separate decisions in a row and guess correctly 18 times in a row So his chances of surviving are in the region of 1 in 300,000 ish Exceptionally unlikely that this poor chap is going to live his main job is to just try to do his best and solve as much of the puzzle as he can Already though we can start to generalize instead of writing 18 We're going to write n for the number of nodes and instead of writing 0.5. We're going to write little p. Okay, so let's play a one Let's have a look at player 2 played by this lady in the in the show itself and Here this is a an important Separation we're going to make between the two scenarios for player 2. There are two different paths Via which player 2 can live the first is that player 1 miraculously lived and showed her the way And she has no decisions to make it all other than to memorize the pattern and follow what player 1 did We're going to call that case a The second and far more likely scenario in the case of player 2 is that player 1 died somewhere On node k and we're going to solve each of these two cases separately and then glue the probabilities together because they are Obviously distinct player 1 has either lived or died and there's no other potential possibility for player 1 So in case a we simply carry forward the probability that player 1 lived which is p to the n Or as we saw before 0.5 to the power of 18 So we'll put that to one side we can think of case a is the free roll It's kind of the the free gift that player 1 gives to player 2 In the context of them living they also give a far greater gift in the context of them dying Which is what we will calculate in case b So the probability of being in this exact position as player 2 is equal to the probability of player 1 dying at exactly node k Okay, and player 2 then has 18 minus k decisions left to mate. Okay So what is the probability that player 2 goes on to live after player 1 died at node k? Well, we can work that out by simply multiplying Here if you can see where I'm pointing my mouse p to the n minus k is the rest of her journey And this is the probability that player 1 in square brackets got her here in the first place. There are This is all thought well and good for calculating What is the probability for player 2 living given that player 1 died on node k? But there are 18 different nodes that end different distinct nodes for player 1 to have died on so the number of different possibilities statistically for player 2 to live and player 1 to die is simply 18 times whatever the particular result for one of the nodes is that does change as we go up the List of players and it becomes a little bit more complicated Now cases a and b are mutually exclusive exclusive as I said before player 1 has either lived or died not both And not neither so we can apply a simple addition with the probabilities and we could calculate this formula when you plug in the value of n equals 18 and P equals 0.5 you get this value here point double o seven two five percent Which is 19 times more likely than poor old player one However is still astronomically unlikely She has less than a one in ten thousand chance of survival Player three is where things do get very very interesting and it's from player three onwards that I've been able to derive a general formula for Any player so again, but we're going to use the same structure We're going to start with two possibilities. The first is that player two Miraculously live so we are going to take the probability from player two and carry it forward into case a of course player three This probably isn't the case probably both players that went before him are dead case B same again as what we saw before Okay, so similar structure to what we talked about before with player two Let's have a look and I've drawn a makeshift diagram here to sort of illustrate this point the conundrum for player three standing at node K The probability that player three lives in this scenario is Shown here below you've got here P to the end minus care if we work away from this side This is the work that player three still has to do, but how did we get here? Well player one Died on node one there. He is at one minus P and then player two did K minus two correct guesses That's four correct guesses followed by an incorrect guess on the node. We're standing on okay Now here's the thing there are K minus one different distinct nodes for player one to have died on yes Player two certainly died here on node K When we are restricting ourselves to node K But if we begin to generalize away from node one for player one We see that there are actually K minus one different places He could have died anywhere from one to five in the example So in order to get rid of the P one died at node one and generalize for all of the different nodes in which player one could Have died on this journey We simply have to add them all up So there are five of these in this example in other words We multiply our formula by K minus one to count up all of the Combinations of events that lead to player three standing and starting the game at node K and then going on to win okay, so an important result from this is that as Player three's game begin if player three's game begins later He has more there are more combinations in which his game begins later and he goes on to live Again not to be confused with the fact that in reality He's starting from node three or four or five almost all the time and he's almost always dying Don't confuse that with the Counting of scenarios in which player three lives. All right, there are however n different values of K So we need to move player three along He could be starting on any one of these 18 nodes and we're going to need to add up all the probability contributions from all 18 So instead of while we're at it We're also going to replace K minus one with K minus one choose one It might seem trivial and stupid to do that But we're going to use it later when we come to generalize for player four five six and seven and so on So what we're going to do is we're going to add up all of the values of K Where player three is starting his game from using our formula now To double down what I said before when player two died on node 18 There are 17 different ways that player one died. So this becomes a 17 by contrast when played when player two Dies on node two. There's actually only one combination. There are not 17 ways the player three Finds himself winning in that scenario. There's only one way he finds himself winning Which is that he has to do all the work as opposed to having any help from his Predecessors So that's an important result is that player three can never begin the game on node one because player one will certainly Play one or two will certainly Have lived on node one Okay, so let's have a look at let's close out player three and Generalize for player J So to bring all of that together. We've got the scenario on which player two lived That's the free role where we don't even have to play the game We're adding that to the product to the scenario on which player two died as we derived on the previous Slide there it feels like we were at a stage where we should be able to begin to generalize and indeed we can So let's check this let's check this formula out So the probability instead of p3 living if we call them player J Where J is a value between three and M We get the contribution from player two and we carry that forward for all future players player six player 16 Both have all have this contribution from player two We then have the contribution from out ourselves the idea that if player two if player J minus one is dead J minus one and all of the people before him have gotten me to node K What is my likelihood of living from node K onwards? So this formula here is exactly the same as this stuff in the brackets with Q equal to zero if you imagine Q is equal to zero For a moment. You'll see that and J is equal to three You'll see that this is just K minus one and this thing down the bottom is a one And so yeah, just to label that this is the scenario where player two lived And this is the cumulative survival of player three Onward plus the effort of J themselves So here in the diagram for example player six standing on something like node 10 after Five deaths have happened the fifth death death death happened on node 10 Four deaths happen before that and what you'll find is that this combinatorial thing is telling you the number of combinations the patterns in which The four deaths can happen on K minus one nodes So this is a formulaic way to derive the number of combinations as opposed to having to count them all Manually, and I think it's this that gives rise to the intuitive kind of complexity of solving this problem No, I'm ahead of my I'm ahead of my diagram. I'm ahead of my diagrams But hopefully you guys picked up what I meant by that So here's the results and we have this fascinating Behavior this is amazing graph I was really blown away when this dropped out of excel when I when I plug these formulas in Here's our player one and our player two stats that we saw before and as you can see the situation for player three is substantially better We rise to point oh six of a percent instead of point double oh seven so nearly tenfold Sadly, it's still one and a thousand ish. Yeah, whatever that is What it's with worse than one and a thousand But things begin to sort of geometrically ramp up player four is Nearly getting up to one percent chance of survival Player five has a greater than one percent chance of survive And then things really begin to rock at player six has four percent or it's five percent player seven twelve percent player eight twenty four percent Player nine forty percent we get this what I call what I would describe as a sigmoidal pattern Now here's the fascinating thing Once 15 players have tried to cross the bridge it's almost mathematically impossible for player 16 to die because of the overwhelming like Overwhelmingly high likelihood that the game has been solved by then So the contribution from all of the deaths of the previous players will mean that player 16's game Probably won't even have to be run and actually in the episode Player 16 Actually doesn't end up making any decisions in terms of leaping from one No to the other I did extend this table down to player 19 And I have confirmed and verified that these equations generate a value of 100 percent for player 19 With 18 players having gone before him it's mathematically impossible for the bridge not to have been solved Before he embarks on his journey a player 18. It is still technically possible that 17 people died Before you and you're standing on node 17 having to make a 50-50 guess on node 18 all right finally entered into the Excel spreadsheet some different values of P and Unsurprisingly what this does is it pushes the curve from to the left and to the right So as you increase the likelihood of tempered glass up to seventy five percent Player five now has a fifty percent chance of living as opposed to what we saw before which was a 1% chance of living Okay, as you make the tempered glass rarer things start to get really really bad Player eight goes down to a 5% chance of living As an example there and if you send it to a ridiculously low level you have 25% even player 16 And 15 have a very tangible chance of death noting though still that player 16 is right up in the 80s In terms of the chance of that player surviving So anyway, I'll I'll end back on the table there. I Final word. I am a mathematician by training, but this work has not been peer-reviewed. So If anyone has spotted any gremlins in my narrative or worse yet any carry-the-one problems or other logical Problems from a probability theory perspective. I would appreciate constructive negative feedback and I'd be happy to do an Arata I might even if I have to I'd film the thing again, but I Hoping that won't be necessary and I think I think I've nailed it here So if you're still with me, thank you very much. I hope you've enjoyed the video and I'll see you again soon. Cheers