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simple math question

Discussion in 'Non-Vegas Chat' started by chef, Feb 17, 2016.

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  1. chef

    chef Resident Buffetologist

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    What is the probability of heads coming up three times in three coins flips? And, is there a simple formula for this?
     
  2. ngrund

    ngrund Low-Roller

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    the probability is.125 or 1/8

    the formula is .5 X .5 X .5 or 1/2 X 1/2 X 1/2
     
  3. ngrund

    ngrund Low-Roller

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  4. chef

    chef Resident Buffetologist

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    thanks for the quick response ngrund.
     
  5. thefish2010

    thefish2010 Low-Roller

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    For any question like this, you just take the probability of the outcome you are looking for, and then raise it to the power of the number of trials.

    So on a coin flip, the probability of heads is 1/2, and you're talking about 3 in a row. The answer is therefore 1/2^3, which is 1/8.

    If you wanted to figure out the odds of a roulette number hitting multiple times, you'd follow the same formula. On a double zero roulette game, the probability of any single number hitting on a given spin is 1/38. So for two in a row it would be 1/38^2, or 1/1,444. 3 in a row would be 1/38^3, which is 1/54,872. 4 times would be 1/38^4, or 1/2,085,136
     
  6. chef

    chef Resident Buffetologist

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    After reviewing the results, I don't think the question was phrased properly. Sorry.
    I was using the coin-flip analogy for the probability of a sports team winning three times against three equal opponents in an elimination tournament.

    Game 1 - Team A beats Team B, with both having a 50 percent chance of winning.
    Game 2 - Team A then beats Team C, with both having a 50 percent chance of winning.
    Game 3 - Team A then beats Team D, with both having a 50 percent chance of winning.

    What then is the probability of Team A winning all three games? I know it is 50 percent for the first game, but after that, it gets fuzzy for me.
     
  7. journaljim

    journaljim Low-Roller

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    Hi Chef,
    There are far smarter people out here than me, but if the teams are truly equal opponents, it's still 1/8 or 12.5% over all. You are right, it would be 50% for the first game, and for each game actually, but to string 3 in a row would only happen 1 out 8 times on average. Of course, to achieve that you've got to look at the long run. If you schedule 24 games (8 sets of 3), than Team A would have a probability of sweeping a 3 game series once out of the eight sets, or 12.5%. Of course, we are talking hypotheticals so it could take hundreds of sets for that to play out or they could sweep the first three.
     
  8. chef

    chef Resident Buffetologist

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    This does make some sense to me - because for Team A to win three games (or be heads three times in a row), an elimination tournament must have eight teams (12.5%).
    But, where it gets confusing is that not all eight teams are equal. Team A, B, C and D are equals and each could beat the other on any given night, realizing that in a 50% scenario, those odds decrease from game 1 to game 2 to game 3.
    Maybe there isn't a formula that can be used and if I haven't explained myself properly, more apologies. My objective was to provide a probability factor for one of the local teams I cover for the newspaper.
     
  9. Breeze147

    Breeze147 Button Man

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    This doesn't answer your question, but I once rolled a 3, 3 consecutive times at The Fremont and a woman just to my left was on the three and pressed the full amount twice. In other words, she bet on it 3 times.

    Figure those odds!!!

    Sorry, OP, I just had to tell that story.
     
  10. AndyAkeko

    AndyAkeko Time magazine's 2006 Person of the Year

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    It's impossible to quantify the exact odds without some bias regarding the other seven teams. But in your scenario, let's say that the #1 seed is far superior to the #8 seed, a little better than the 4/5 seed (who they would play in the semis) and nearly even with the 2 and 3 seeds (who would be on the other side of the bracket). In that case, they have a better than 12.5% to win the whole tournament because it's a near certainty they will win their first game. Assuming they'd beat the 8th seed 90% of the time, the odds now become .9x.5x.5 for 22.5% from the beginning of the tournament. If we say the odds they'd beat the 4 or 5 seed is closer, say, 6 out of ten times, then the odds are .9x.6x.5 or 27%.

    Now, that's still a little simplistic because we're assuming the higher seeds will win, and a true calculation would involve the odds of the 2,3,6 or 7 seed coming out of the other side of the bracket for the final. But in almost all cases, a team's chances to win the tournament improve as the tournament progresses, simply because there are fewer teams. (There are scenarios where the odds could decrease because they get a highly more unfavourable match-up in a later round than expected, like if the team can easily handle the 2-3-7 teams but would get wiped out by the 6th seed... but that's more of a theoretical situation than a practical one).
     
  11. thefish2010

    thefish2010 Low-Roller

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    You're making this harder than it is. The probability that they will win 3 games in a row is one out of eight. Each time the odds of the team winning are 1/2. To do it three times in a row, the probability is 1/2^3, which is 1/8 (aka 12.5%). Twice in a row would be 1/2^2, which is 1/4. Once is 1/2^1, or 1/2. As I said earlier, you just take the probability of the outcome you are looking for, and then raise it to the power of the number of trials.

    If the odds were different for each game, you would just multiply the odds together. Let's say the odds of team A were 50% for game 1, 30% for game 2, and 10% for game 3. 0.5*0.3*0.1=0.015, which means there is a 1.5% chance that they would win all three in that scenario.
     
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