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

Quote: mustangsally

pacomartin mentioned that the percentage chance of winning is based on the number of trials. He is absolutely correct.

So, you're saying with the information available that game A wins 2 in a row on average every 6 trials, and game B wins 2 in a row on average every 8 trials, you can't determine the percentage chance of winning 2 in a row for either game A or game B? I can.

Game A has a 33% chance of winning 2 in a row, and Game B has a 25% chance of winning 2 in a row.

I took whoever's advice it was to hurry up and ask yahoo, I'm pretty sure there's enough information to determine the percentage chance of winning 2 in a row for both game A and game B.

So, do we still have a bet mustangsally :)

P.S. A quick second point. The math pacomartin did is the percentage chance of winning in the next n trials. It's not what the average percentage chance of winning 2 in a row. I'm going to go ahead and say pacomartin's answer is the incorrect answer for the "average" percentage chance of winning 2 in a row for game A.

Quote: JyBrd0403

So, you're saying with the information available that game A wins 2 in a row on average every 6 trials, and game B wins 2 in a row on average every 8 trials, you can't determine the percentage chance of winning 2 in a row for either game A or game B? I can.

Game A has a 33% chance of winning 2 in a row,

Really?
In how many trials are you talking about?
2. are you taking about just 2 trials?
3
4
5
6 maybe this one?

Quote: JyBrd0403

and Game B has a 25% chance of winning 2 in a row.

Really?
In how many trials?
2
3
4
5
6 maybe this one?

Infinite trials mustangsally Infinite:) For as long as you play the game the "average" percentage chance of winning 2 in a row will remain the same. I mean, unless someones cheating. That's kinda one way to determine whether they're cheating or not.

Quote: JyBrd0403

Infinite trials mustangsally Infinite:) For as long as you play the game the "average" percentage chance of winning 2 in a row will remain the same. I mean, unless someones cheating. That's kinda one way to determine whether they're cheating or not.

Over infinite trials, the percentage chance of winning 2 in a row is 100% assuming it's possible to win at all (e.g. p > 0). And why did you put "average" in quotes? What are you trying to average, anyway?

FYI, the chances of a player in Game A winning the next two games are 0.25, while the chances of a player in Game B winning the next two are 0.177692. Note the distinction between winning the next two in a row, that is, two wins in two trials, as opposed to winning two in a row anywhere over infinite trials.

Haven't a clue what you're talking about guy. This is for a 50/50 game of chance, so I don't now what p > 0 means. The percentage chance of winning 2 in a row once in infinity would be extremely close to 100%, I agree. So what? I've been asking what the "average" chance of winning 2 in a row would be (stated in percentage terms) for a 50/50 game of chance. And, before we go on to the how many trials thing again, it's infinite trials. On average it's once every 6 trials, or stated in percentage terms, 33% period.

Is it possible for anyone, besides me, to state in percentage terms the average of winning 2 in a row in 6 trials? Because, I'm still guessing that in percentage terms that would be 33%. Just state that fact in percentage terms.

On average you will win 2 in a row every 6 trials or stated in percenatage terms __%

Sorry for being short with you, but I'm drunk and have to get up early tomorrow :)

Quote: pacomartin

I taught high school math about 20 years ago, and it was the fairly smart kids who could do cubic equations in 9th grade. Sometimes if you can easily see one solution, (it's pretty obvious that x=1 makes " 8x^3 - 9x^2 + 1 " equal to zero, then you can divide by x-1).

I think a lot of the stuff we do will be lost to future generations as equation solvers and even symbolic calculus integrators will be built into telephones. It's difficult to understand what an impact Napier had in his day on computation by working out the logarithm tables.

Knowing how to factor sure helped James Martin when playing the British gameshow Countdown. If you don't know how to play the game the rules are the numbers: 25, 50, 75, & 100 plus two random numbers are selected. In this incident the random numbers were 3 and 6. A final target is chosen by computer (in this 952), and the objective of the game is to get as close as the target as possible using basic arithmetic functions in 30 seconds.

The loser came up with (6+3)*100 + 50 + 75/25 = 953 missing the target by 1. He used all six numbers, but that is not a requirement. The winner got exactly 952. If you can factor, then try it yourself before you watch the video. The host was gobsmacked.

I forgot almost everything, 20 or more years back i would make like you sugested f(1) and divide by x-1.(probably, if there is not another even faster method)

But now i was able only to write on paper the longer solution, which i have put in the post. This is very basic math i know, unfortunately by not using it anymore i forgot. I was just contempt that i was able to remember the x^2-1 =(x-1)(x+1) thing and the formula for calculating the 2 solutions x2,x3= (-b -/+sqrtdelta)\2a.

PS: i don't like these types of gameshows, they somehow suggest that you must be able to do these calculations in 30 seconds. If you cannot, then you're garbage. Lets face it, to be able to do these calculations very quickly requires alot of exercise.

Quote: JyBrd0403

Is it possible for anyone, besides me, to state in percentage terms the average of winning 2 in a row in 6 trials? Because, I'm still guessing that in percentage terms that would be 33%. Just state that fact in percentage terms.

Average does not mean 100%. So in 6 trials the percentage chance to win 2 in a row is about 67%.

2/6 = 0.333(3) but has nothing to do with your original question.

Thank goodness for Yahoo. Below is the answer from yahoo answers

Probability of A winning 2 in a row on average every 6 trials = 2/6=1/3
33.33%
Probability of B winning 2 in a row on average every 8 trials = 2/8=1/4
25 %

It couldn't get any easier. Those are the correct answers in case you were wondering.

Quote: JyBrd0403

Haven't a clue what you're talking about guy. This is for a 50/50 game of chance, so I don't now what p > 0 means. The percentage chance of winning 2 in a row once in infinity would be extremely close to 100%, I agree. So what? I've been asking what the "average" chance of winning 2 in a row would be (stated in percentage terms) for a 50/50 game of chance. And, before we go on to the how many trials thing again, it's infinite trials. On average it's once every 6 trials, or stated in percentage terms, 33% period.

Is it possible for anyone, besides me, to state in percentage terms the average of winning 2 in a row in 6 trials? Because, I'm still guessing that in percentage terms that would be 33%. Just state that fact in percentage terms.

On average you will win 2 in a row every 6 trials or stated in percenatage terms __%

Sorry for being short with you, but I'm drunk and have to get up early tomorrow :)

Drunk or not, your question has no meaning.

If you're playing a 50/50 game, the probability of winning 2 games in a row over 6 games is 67.1875%.
If you're playing a 50/50 game, the probability of winning 2 games in a row over 2 games is 25%.
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.

But dividing 2 by 6 is meaningless here. It doesn't represent any relevant concept. It would be like saying "the average total of two dice is 7", which it is, but then rolling the dice, getting a 4, and saying "that's 4/7 or 57.14%". That's meaningless. It's just a 4.

In other words, stating things in percentage terms is improper for the question you're asking.

Sure it has meaning. It means on average you'll win 2 in a row 33% of the time.

I thought I clarified earlier that this is for infinite trials, and here we are back with these 6 trials 2 trials stuff again.

So, with your crazy math what is the chances of winning 2 in a row once in infinity. Answer 100%

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%. That's some pretty damn helpful information. Don't you think mustangsally?

As far as the dice thing, I think you might need a drink, I don't have a clue what you're talking about.

See as far as useless information is concerned, I would have to put the 6 trials thing and 2 trials thing pretty high up there, no offense intended to anyone. See the math for that is saying that in 6 trials 67.7% of the time a 2 in a row will occur. (I won't get into this with you guys, but that's wrong too) So, 67.7% of the time a 2 in a row occurs, that means 33.3% of the time it goes past 6 trials before you see a win of 2 in a row. That's very unhelpful for anyone who's let's say gambling or trying to use this information in any sort of productive way. Why, because in infinite trials, the answer is 100%. I dare say, after just a 1000 trials I can pretty much give you a 99.99% chance that you would have had at least 1 2 in a row by then. It's useless info. Well, I shouldn't say useless, but it's useless for my purposes. What's more important is the average number of trials.

If, I go 18 trials without winning 2 in a row then I win 2 in a row three times back to back and it averages once every 6 trials. That's a damn good bit of information to have. At least I think so.