1
$\begingroup$

There's a similar question asking about 17 times in a row, but there was some confusion for me about the answer.

I put 3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3 into Google and it gave me a result of 7.6255975e+12. My questions are:

a) is this even the correct way to calculate the odds of what me and my friend did? And if it is,

b) what kind of number is 7.6255975e+12 in layman's terms?

  • 0
    You can use exponent notation and enter 3^27 into Google. That way you don't have to write so many threes. In the answer Google gave you, the "e+12" means "$\times 10^{12}$". The probability of tying in all 27 games is indeed $1/3^{27}$ (assuming you both make your moves randomly, and all moves have equal probability). – 2017-02-05
  • 2
    The probability of tying $27$ games is $\frac{1}{3^{27}}$ only if both players are playing *randomly* and *independently* of each other. This is very unlikely when two human players are squaring off, especially when it's players that know each well (maybe you have an inkling of each other's likely "strategy"). The "real life" probability of a draw $27$ times in a row is likely a lot higher, but is very difficult to estimate. – 2017-02-05

3 Answers 3

1

Yes, this is correct. You can also just Google for 3^27, as the caret is the symbol for exponentation. In layman's terms, that number is 7.6 trillion (or 7.6 billion, depending on where you live). It's 7.6 million times a million. You'd expect to draw 27 times in a row once in every 7.6 trillion times you play 27 RPS games (so that's 200 trillion RPS games in total).

1

You should talk about probability here, instead of odds. The probability of drawing a round of rock-paper-scissor is $\frac{1}{3}$, so you simply multiply this probability 27 times, which yields

$\big( \tfrac{1}{3} \big)^{27} \simeq 1.3114 \times 10^{-13}$.

which corresponds to the probability of drawing 27 times.

1

Since I don't want to argue in the comments I will write up a more comprehensive answer.

I will assume this is an actual question as opposed to a homework problem the OP was too lazy to work out.

For a) no this is not the correct way to calculate the odds.

There are two essential problems which are linked by one overarching problem and that is you are using a very poor model for what is going on.

The model you seem to be using assumes two different things.

  1. That the chances of your win, your friends win and a draw are equal (and thus $1/3$). This can come from assuming that both you and your friend choose between rock, paper and scissors independently of each other and with the same probability for each.
  2. That the $27$ games you played were independent on each other. That is that the way the preceding games were played has no effect on the way the next game is played.

Both of these assumptions are false. People in general are extremely bad at doing anything randomly especially if it is a voluntary act such as choosing a number or choosing rock/paper/scissors so the idea that you and your friend both choose rock/paper/scissors with a $1/3$ chance each and that moreover these chances are independent on each other (within one game) is very unlikely. So it's most probably not the case that $1/3$ games is a draw.

Secondly it is even more unlikely that the distinct games are independent of each other. You certainly know what the outcomes of the game so far where and as almost certainly influenced by them.

All together, while your answer is pretty much correct assuming two computers with decent random number generators play the game, you get a 1 in $3^{27}$ chance of drawing 27 times in that case, it most probably has little to do with how you and your friend played the game.

As to b) assuming the chance where 1 in $3^{27}$ it's roughly 10 million times less likely than being struck by lightning and roughly 10 thousand times less likely than winning the powerball lottery.

  • 0
    I did say this in my comment. I don't think this answer adds very much else. – 2017-02-05
  • 0
    @deepak It adds nothing else. But it's an answer and not a comment and thus can be upvoted and accepted. I wrote it strictly because both of the other Answers are answering a) in the affirmative without qualifying in any way. If you want to write this answer I'll be happy to delete mine. – 2017-02-05
  • 0
    It goes without saying that there I'm not taking human factors into account, i.e., I consider that the players choose their strategies independently and at random. – 2017-02-05