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Stadium 21s

Practically impossible.
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To play 21s you get a group of 6-15 people in a room and have them close their eyes. The object is to count to 21, one number at a time, without anyone speaking over anyone else. You have to sense when no one is going to speak and say the appropriate number. If two or more people go for it at the same time you start over.

Now take that game to the crowd at a sports stadium, perhaps as a prelude to kick off. Participants have to scream the number so everyone can hear it. Silence but for the numbers emanating from the distance and the collective sighs when 2 (or likely more) people mess it up.

The only problem is the type of people that go to sports matches are commonly the type of people who would take enjoyment in spoiling the game. Give it time to become a commonplace tradition (a few decades) and the spoilers will be less inclined to spoil.

However, the temptation to be a part of the winning crowd would be too much to keep you quiet and failure would be commonplace.

But imagine the day they succeed! Imagine filling your lungs and screaming at the top of your voice "TWENTY ONE!" and the crowd going flippin' wild!

theleopard, Jan 17 2008

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       This sounds like an exercise for a tutorial in statistical analysis or traffic arrival rate in telecoms...
Jinbish, Jan 17 2008
  

       Okay folk, for this game of 21s, you all keep quiet, and I count to 21. We all win.
vincevincevince, Jan 17 2008
  

       Ah yes, that would certainly work, but where would be the sense of achievement?   

       Most suitable sporting events might include the Olympics, Wimbledon's centre court, and maybe a cricket ground, but cricket supporters are about as loutish as football or rugby's.
theleopard, Jan 17 2008
  

       //Wimbledon's centre court// ... I'm thinking you get ejected before you get to 4 ...
vincevincevince, Jan 17 2008
  

       What, were you watching the San Diego / Indianapolis game last weekend?
normzone, Jan 17 2008
  

       Should this not have been posted by [21Quest]?
MaxwellBuchanan, Jan 17 2008
  

       What happened in the San Diego / Indianapolis game last weekend?   

       (Ah... I miss [21]...)
theleopard, Jan 18 2008
  

       I ought to work out the statistics on this. Basically, a lot depends on the time it takes to utter each number, and on how you judge whether two numbers are "overlapping". For example, if someone says "ten" and then someone says "eleven" immediately afterwards, it is very difficult to tell exactly where the "nnnn" of "ten" ended or the "eh" of "eleven" began.   

       OK. So, for the sake of argument, suppose it takes 0.5s to utter each number (a fair average, allowing for a discernable gap). What next?   

       Well, let's suppose that we divide time into half-second windows (this quantization shouldn't affect the answer too much). In each window, each person has the option to remain silent, or to shout out the next number in the sequence. If two people shout in the same window, the game is void, yes?   

       OK. Suppose we have N people in the crowd. Suppose also that they are all equally likely to shout a number. Suppose also that each person adjusts their behaviour so that the probability of their shouting a number in any given window is P. What is the optimal value of P, given N?   

       Well, if P is very, very small (P<<1/N) then the probability of two shouts coinciding is also small, and hence the game will be "won" every time. The only problem is that, for very small P's, the time taken to complete the game will be very long (imagine a thousand people, where each person has only a one in ten million probability of shouting in any one interval; there will be an average of one shout every three hours, and it will take 60 hours to finish the game).   

       If P is larger, then of course games will proceed faster, but most games will be "lost" because of coinciding shouts.   

       So, suppose we want to find P (given N) such that the average time to complete a "winning" game is minimized.   

       Let me go away and think about that.
MaxwellBuchanan, Jan 19 2008
  

       OK. I've thought. The following statistics are quick and dirty but approximately correct. We have to assume that P < 1/N for them to work.   

       So, consider any one utterance. What is the probability that it is "killed" (and the game voided) by another utterance in the same quantum of time? Well, as long as P <1/N, that probability (which we'll call K) is roughly NP.   

       So, what is the probability that a game gets through all 21 numbers without failing? It is obviously (1-K)^21.   

       How long will it take to get through each game? Here, we have to make a simplification, by assuming that, if a game "fails", it gets *on average* half way through before it does so. (This just makes the maths easier). So, on average, each game goes from 1 to 10 or 11.   

       With this assumption, the average game requires 10 utterances. On average, there will be NP utterances in each "quantum" of time, so the average game will last for 10/NP quanta of time.   

       OK. With me so far?   

       So, the probability of a game succeeding is (1-K)^21, and the average time taken for a game to run (either losing or winning) is 10/NP.   

       To have a 50% chance of winning a game, we need to run 0.5/[(1-K)^21] games, which will take a total time (T) of:   

       T=(10/NP) x (0.5/[(1-K)^21] quanta of time.   

       We want to find P, given N, such that T is minimized. Remember that:   

       K=NP.   

       So:   

       T=(10/NP) x (0.5/[1-NP]^21)   

       OK. Time to go away and think again.
MaxwellBuchanan, Jan 19 2008
  

       OK, back again. A little calculus would give us a function that would find the P necessary to minimize T for any given N.   

       However, a little calculus is a dangerous thing. Instead, we can start by trying to find the optimal value of NP, since this term appears as an entity in the foregoing equation.   

       Calculation gives the following (each line gives the value of NP, and the corresponding value of T):   

       0.01 -->617
0.02 -->382
0.03 -->315
0.04 -->294
0.05 -->293
0.06 -->305
0.07 -->328
0.08 -->360
0.09 -->403
0.10 -->456
etc.
  

       In fact, the minimum value of T comes at NP=0.045 (where T=292).   

       So now, we can apply this to a real situation. Average attendance at major league games is, say, 10,000 people (ie, N=10,000). At such a game, if we want NP=0.045, each person should have a probability P of shouting in every "quantum" of:   

       P=0.045/N
=0.045/10000 =4.5 x 10^-6.
  

       or once in about 0.2 million "quanta". Since we defined a "quantum" as 0.5 seconds, it follows that each person should shout, on average, once every 0.1 million seconds, or once every 27 hours.   

       In this way, there will be a "win" at the earliest possible opportunity, and it will typically take 292 quanta of time, or 146 seconds, or about two and half minutes. Eeeeeeeeasy.   

       How would you do this in practice? Even easier!! Upon entering the stadium, each person picks a random number between 1 and 100,000. They then count seconds, in their head, starting with that number. If a person gets to 100,000 they shout the next number in the sequence.   

       Of course, this makes a few assumptions about human behaviour. It assumes that everyone picks a truly random number and counts at roughly the right speed, doesn't cheat, and isn't influenced by other utterances or by unseemly (but random) silent pauses.   

       Statisticians conferences would be the place to try this out.
MaxwellBuchanan, Jan 19 2008
  

       If you ignore all the other rubbish, then //Well, as long as P <1/N, that probability (which we'll call K) is roughly NP.//   

       Please report to the Clay Institute to collect your <places pinky in mouth> ONE MILLION dollars.
4whom, Jan 20 2008
  

       It's very kind, but I'm too busy applying the viscothixotropic theory of custard to an axiom-free solution of Goldbach's conjecture.   

       The main point is that there is a simple strategy whereby the game of "Stadium21" is easily winnable by a large crowd of indendent individuals.
MaxwellBuchanan, Jan 20 2008
  

       Uncle Petros would be so proud! I disagree with the view that it is a //simple strategy//.
4whom, Jan 20 2008
  

       Quite so, but what would the Parrot say?
MaxwellBuchanan, Jan 20 2008
  

       The biggest difference between game theory and decision theory (which you are proposing) is that game theory relies on the *interaction* between participants. Not enough attention has been paid to the fact that information relevant to the entire game is accessed at differing time intervals for each participant. The speed of sound, and the stadium size, are effecting different participants in different ways. This situation does not exist in the // 6-15 people in a room and have them close their eyes// scenario. You would either, have to eliminate the "speed of sound"/"distance" part (as you have done), or reformulate the strategy including the varying rate of information transfer for individual players. The solution you propose is less game theory and more decision based algorithm. I would imagine Guedj's parrot would say something along those lines.
4whom, Jan 20 2008
  

       Although the model doesn't take into account several subtleties (such as the speed of sound, the size of stadium, and the position of the judge who is listening for the shouts, which could be factored in but would make the maths more complex), I believe that is a game theory.   

       The cornerstones of game theory are that (a) participants try to achieve the optimal outcome for themselves (regardless of whether this is for the common good) (b) participants play to a set of rules imposed by the game (c) participants do not know in advance the intentions of other players and (d) participants can benefit by taking into account the *likely* intentions of other players (usually by assuming that the other players have the same goals as themselves).   

       My model fits all of these criteria. Each player's goal is to see the game completed; the rules are that no two utterances must overlap; each player has no idea when another player will utter; and each player must estimate the likelihood that another will utter.   

       The optimal solution is when all players utter rarely enough to give the greatest chance of completing a game, in the shortest possible time. In what way is this not game theory? Unless each player is familiar with the habits of the other players, what information could they use in order to implement decision theory?
MaxwellBuchanan, Jan 20 2008
  

       I've just done some more thinking on this, and I'm pretty sure that my strategy (which I argue is game theory) is the optimal solution (aside from the corrections for speed of sound etc). Here's why.   

       The players are all equivalent or, more importantly, have no prior means for distinguishing amongst themselves, and therefore must consider themselves to be equivalent. Under these conditions, and barring any prior agreement between the players, the *optimal* strategy will result in a random distribution of utterances across time. If players try to develop a rule to avoid randomness (for example "if no one has uttered for three seconds, then utter"), then this will lead to clustering of utterances.   

       Therefore, randomly-timed utterances are the optimal solution (without collusion), and my algorithm allows a way for all players to calculate the average frequency with which they should utter in order to achieve the best outcome.
MaxwellBuchanan, Jan 20 2008
  

       [boys'parks] all that you say is true, but depends upon strategies either agreed in advance, or evolved further into the game. However, I maintain that, for a crowd of previously unconnected individuals, the a priori strategy I proposed is the best one.   

       In reality, of course, two factors will intervene:   

       (1) most people will not be aware of this optimal strategy, and hence will not follow it. Nevertheless, this is equally true in all game theory: the optimal result depends upon all players being intelligent and independently converging upon the same, optimal, strategy.   

       (2) as you point out, better strategies are available if the crowd discusses or organizes amongst itself, up to the point of electing 21 individuals and having them count from 1 to 21 in order. However, bring 10,000 people together "cold" and ask them to start playing without preamble, and my strategy will be optimal.   

       In any case, my point was that the game is by no means "practically impossible" as the subtitle says: a crowd of people can win the game in under three minutes, just by guessing how many other people are around and then each adjusting their tendency to utter accordingly.
MaxwellBuchanan, Jan 20 2008
  

       True enough. However, some parameters still apply even after iteration. It's unlikely that a complete consensus will be reached within three minutes (a handful of tries at most), so it's likelier that (as you proposed) local groups will form, each involving perhaps a hundred people. (I'm guessing that one central individual will be readily visible and audible by a hundred neighbours, thereby defining a locally coordinated cluster).   

       In this case, then, we have perhaps a hundred local clusters in the stadium, and each cluster can behave like a single indivudual, independent of its neighbours.   

       In this situation, we still want to aim for 0.045 "utterances" per unit time (or 0.09 utterances per second, if the quantum of time is 0.5s and if we ignore the speed of sound). In this case, all we have done is effectively reduce "N" from 10,000 to 100. Each group should therefore make an utterance (through its spokesperson) once every 1100 seconds or so. Unless there is long-range coordination between groups, the best strategy will be for each group to utter at a random time (but still averaging only one utterance per group per 1100 seconds). Once again, the average time to complete a game successfully will be the same - about 2-3 minutes on average.   

       Clearly, a "Mexican wave" strategy could evolve, but this requires coordination over longer distances (for example, to avoid two simultaneous starts, or a wave running in two directions from one starting point). Once this degree of coordination evolves, the game ceases to become interesting.
MaxwellBuchanan, Jan 20 2008
  

       <waves palm over top of head, with bemused look on face, whilst blowing out of mouth>   

       Er... 1?
theleopard, Jan 22 2008
  

       If this were a statisticians' conference, of course, you've had been more logical and started with 0, [theleo]
MaxwellBuchanan, Jan 22 2008
  
      
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