@Takket you're close on the math, but a little off. For one, the odds are closer to 2/6 (12/36) for each individual person to match the number in two spins. I appreciate that you accounted for the "first spin win" of someone matching her number on the first spin, but in removing that chance you lost 1/36 somehow. I'm not really sure how -shrugs-
NOTE: when I say 2/6, i mean .3056. See end for details.
Secondly, you fail to keep in mind that there are 9 people playing the game - and as such the odds are different between the four corner, four edge, and one center player(s). For example, the center player has to match the host, which is a 2/6 chance. Furthermore, he must match any one outer player, and the chance of this
not happening (i.e. nobody else matching the HOST) is (5/6)^2^8 (squared, then to the 8th), or .05408, thus the chance that
any one outer player matches the HOST after two spins is .9459. After this, that center person then has to match *one* specific person, who has to match the host's number. That person has a 2/6 chance of getting the same number as the HOST. Thus, the chance that the center player wins (which by definition means two other players win) is
2/6 times .7674 times
2/6 (chance center matches, chance any one outer player matches, chances the specific required outer player also matches). This comes out to a 0.07167 probability, or a
7.167% chance the center player in any given game wins the game. This does not, however, account for the possibility that multiple rows win. However, just based upon some mental analysis, I doubt that comes out to more than .001, or 0.1% chance difference, which I feel is too statistically insignificant for this to even bother calculating it (read: i'm lazy).
Continuing this, we can now calculate the probability the others win. Starting with the corners, it's similar, except instead of matching any 1 other person out of the other 8, they must match any one out of
6 people on the board. To illustrate this, imagine I'm in the top left of the board. I must match one of the following: Someone in the top row, someone in the left column, or someone in the top left-bottom right diagonal. There are two other people in each of these three rows, leaving 6 people. Then it's similar to the center: 2/6 for my chance to match, (1-(5/6)^2^
6) for one of the 6 players to also match the host, and then 2/6 for the final required player to match the host. This comes out to 0.08292, or a
8.292% chance a player in the corner will win.
Lastly, the probability for the edges. This time, the edges must match either their row, or their column. They don't have a diagonal to match, leaving a required match of one of 4 specific people. This comes out to 2/6 times (1-(5/6)^2^
4) times 2/6, or 0.07167, or a
7.167% chance that a player on an edge will win.
Lastly (again, i ran out of transition words :3), the probability that
at least one group will win is within 3% of 2/6 times 2/6 times 2/6, times the number of groups, which is 8. This gives about a
0.2283 chance that at least one row wins, or 22.83%. Again, this does not consider that multiple rows can win one game. Considering that this probability is that 3 people win, and the percent is
relatively close to 1/3, it's acceptable to say that
for every game played, approximately one prize will be given out. That is the number most people probably care about - how many prizes are given out, and what's the chance any game has a winner.
Some more analysis on the statistics I used is at the bottom. Now I'll kinda analyze what this means for the game.
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What does this mean for the game? For one, one player in a given game has about a 2% chance higher than some of the others of winning the game. That's not really fair in and of itself. Second of all, the probability that there's
any row that wins can be calculated as above, and it's not super low in comparison to other games. For example, in a game with the HOST spinning a generator to pick one of 5 or 6 people, the chance of winning is either 20% of 16.66% depending on how many categories there are.
Furthermore, it shows that there's, in fact, around a 7-10% chance you win the game any given time you go in. That's not super far off from other HOST events, which generally have between 10-20% chances of winning for any individual player.
If the number of tries were raised to 4, the chance of getting it within the 4 tries would be around (not exact) halfway between 3/6 and 4/6 - i.e. about a 1 in 2 chance that any one person gets the same number as the HOST. This would result in a much higher chance that at least one group wins - about
70-80% based upon my calculations (i did a few and came up with that based upon my analysis, I can go into further detail with anyone in PM if they really don't trust me).
Por fin, do I think it needs changed? No. Especially since we accept the lower 16-20% chance of winning some other games have. Even taking into account the margins of error introduced by my basic statistics (if I actually caluclated everything out for a smaller MoE you'd be bored to death and I could probably write a dissertation on it), the chance of winning isn't too far off from other games.
The underlying cause of the game's complexity in being able to get an acceptably low win rate for the HOSTs, but still provide a high enough "personal win chance" is, in fact, it's complexity itself. There are many different factors that come together for someone to win the game and if you've read this far I really commend you for paying attention please pm me with a note saying you did so i can hug you but also that those factors are all mixed, yet independent, yet some dependent.. etc.
-Whispered (statistical methods follow)
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For the purpose of this analysis, I assume that there's a 1/6 chance of matching the HOST on any given spin. This gives a 1/6 chance of matching the HOST on the first spin, and a (1/6) * (5/6) chance of matching the HOST on the second spin. The second spin chance is caluclated by the fact that for the person to be on their second spin, they must have spun
not the HOST number on their first spin (a 5/6 chance). This comes out to .3056 chance of matching the HOST on one of the two spins. 2/6 is .3333, a difference of about 8% between those two numbers, which doesn't justify, in my mind, the loss of clarity by putting that each time instead of 2/6. However, it did introduce a statistically significant error (of sometimes 6%), and as such, I used it in my actual calculations.
Secondly, I assume that for the purposes of this analysis, we only care about one row winning. I also assume that the chance of the center player having two outer players both match the HOST during their two spins, or .09339, is too small to bother including it in the calculations. If the 9% chance that the center player has 2 opportunities to get winning rows on the spins is factored in, it results in a less than 1% change in any of the numbers, which does not justify including it in the calculations.
Lastly, I'm sure I've missed something or made something unclear here. Feel free to PM with any questions (or reply). If anyone wants a statistical analysis for other games, feel free to ask me
?