On average how many trials will it take to lose 2 in a row on a 50/50 game of chance?

Quote: JyBrd0403

Hey, I got a question while we're waiting for infinity, what's the average number of trials it takes to win 2 in a row on a 50/50 game. Answer is guess what 33%.

33% is not "a number of trials". The answer is 6 trials. If you want to put that in percentage form, it's 600%. But that's not the same as 33%.

You're not suggesting that the "average number of trials it takes to win 2 in a row" is 1/3 of a trial, are you? That's what 33.33...% equals.

Quote: MathExtremist

33% is not "a number of trials". The answer is 6 trials.

No, you're correct it is 6 trials, or 33% of the time. Finally we agree on something.

I should have said while we're waiting for infinity, what was the average number of times a 2 in a row occurred, the answer would be 33% of the time.

-If you're playing a 50/50 game, the probability of winning 2 games in a row over a million games is very close to 100%.
If you're playing a 50/50 game, and you consider all possible number of games before you see the 2 wins in a row, the mean (average) number of games you will play before you see those 2 wins in a row is 6 games. Sometimes you win 2 games in a row right away, sometimes you don't win 2 in a row until the 100th and 101st games. The weighted average of all of those possibilities is 6 games. -

I apologize if this was in your original post. I might have stopped reading after seeing the 6 trials and 2 trials thing. Yes, in case you're wondering, I'm drunk again tonight LOL

Quote: JyBrd0403

I should have said while we're waiting for infinity, what was the average number of times a 2 in a row occurred, the answer would be 33% of the time.

That's a different question. You would, for example, first need to define whether 3 in a row counts as just one instance of 2 in a row, or whether it counts as two instances of 2 in a row, or whether it doesn't count at all. For example, how many times does 2 wins in a row occur in the following sequence:

W W L W W W L L L W W W W

1, 3, and 6 are all plausible answers, and which you pick will make a big difference.

I'd pick 4.

I suppose we should tie up this thread with a review of the recursion formulas.
p is a probability of event happening (normally it is the probability of player losing, but can be 50% for coin toss, or of a player winning)
P is the probability of a streak of length k (player losses by default) out of j trials
E is the expected number of trials, or an average number of trials for a streak of length k.


P(j,k)=0 if j<k
E(1)=1/p
P(j,k)=P(j-1,k)+(1-P(j-1-k,k) )*(1-p)*p^k for j=k....
E(k) = ( 1 + E(k-1) )/(1-p) for k=2....

Unfortunately, I'm drunk tonight, and we'll tie this thread up with, I'm still saying it's 4.

Quote: MathExtremist

That's a different question. You would, for example, first need to define whether 3 in a row counts as just one instance of 2 in a row, or whether it counts as two instances of 2 in a row, or whether it doesn't count at all. For example, how many times does 2 wins in a row occur in the following sequence:

W W L W W W L L L W W W W

1, 3, and 6 are all plausible answers, and which you pick will make a big difference.

Exactly.
It all depends on the counting method used and the type of statistical problem that one is trying to solve.
And the math for each is different. (most are concerned with N, G and E, but that does not mean one is better than the other.)

counting method/ frequency
N(n,k) 4
M(n,k) 6
G(n,k) 3
E(n,k) 1

E under counts the runs of length 2, at first glance if someone that parlayed one time after a win did win 3 parlays of length 2.
But E still says only 1.

It is also how one interprets the data.
Many trip up right there and are forever lost.

from:
K. S. Kotwal · R. L. Shinde
Joint distributions of runs in a sequence
of higher-order two-state Markov trials
Received: 8 June 2004 / Revised: 11 April 2005 / Published online: 21 July 2006
© The Institute of Statistical Mathematics, Tokyo 2006

The pdf can be found online

Quote: mustangsally

Exactly.
It all depends on the counting method used and the type of statistical problem that one is trying to solve.
And the math for each is different. (most are concerned with N, G and E, but that does not mean one is better than the other.)

counting method/ frequency
N(n,k) 4
M(n,k) 6
G(n,k) 3
E(n,k) 1

E under counts the runs of length 2, at first glance if someone that parlayed one time after a win did win 3 parlays of length 2.
But E still says only 1.

It is also how one interprets the data.
Many trip up right there and are forever lost.

I understand all this, believe it or not, but, I don't understand why it's difficult to simply state that on average you'll win 2 in a row 33% of the time.

If I had said, on average you win 2 in a row 25% of the time, everyone would have said that's absolutely correct. Right? We wouldn't be having this conversation of what do you mean by 2 in a row, and it depends how you count it etc.

But, for some reason it's real difficult for you to state that if you win 2 in a row once every 6 trials, that this means that on average you would win 2 in a row 33% of the time.

I guess what I'm saying is the normal question people ask is "What is the chance of winning 2 in a row". My answer would be 33%.

Quote: JyBrd0403

But, for some reason it's real difficult for you to state that if you win 2 in a row once every 6 trials, that this means that on average you would win 2 in a row 33% of the time.

You keep saying this. It's still wrong. For starters, 1 out of 6 is 16.67%, not 33%. Now consider that the chances of winning 2 in a row over 2 trials is 25%, so how could the chances be lower over more trials?

Hint: learn the difference between expected wait time and probability.