I’m sure this has already been talked about but when I search I’m not finding what I am looking for.

Let’s say you have a hand that has a 36 percent chance to win by the river and you have 3 flops where this is the case and you go all the way to the river 3 times.

What are the odds that you will hit the winning hand by the river 1 of those 3 times?

Intuitively, I would think you have better than a 1/3 chance to hit your winning hand and you have 3 times to hit, if you don’t hit you are just running below EV.

However, is that incorrect? What is the percentage chance of winning 1 in 3 times when you have a 36 percent chance each time?

Assuming Independence, you can use the binomial formula:


0/3 times: 1 * 0.36^0 * 0.64^3 = 26.21%

1/3 times: = 3 * 0.36^1 * 0.64^2 = 44.23%

2/3 times: = 3 * 0.36^2 * 0.64^1 = 24.88%

3/3 times: 1 * 0.36^3 * 0.64^0 = 4.67%

About 44.23%

the chance of not hitting is 0.64*0.64*0.64 = 26%
the chance of hitting is 1-that = 84%

74% but yeah

There's a difference between hitting at least once (74%), and hitting exactly once (44%)

oops my bad

Given that your chance to hit it at least once is only 74 percent, does this formula take into account variance?

We know it’s not 100 percent chance since you can have all 3 chances and not win but I thought it was variance that would explain why you don’t have a guarantee to hit in those 3 chances.

Variance describes how "spread out" the outcomes are, as shown here. It's a measurement, not a force of nature.

If you always hit exactly 1/3 flops, that would be an outcome with no variance. But because poker has chance, the outcomes are spread out, resulting in variance.

You seem to think that a 1/3 chance repeated 3 times means you should always hit. But that's not how probabilities work.

Imagine a room filled with people comparing their birthdays. How many people must be in the room before two people share a birthday, on average? (Ignoring year)

I find it strange that you quote me saying we know it’s not 100 percent and your response to that you think I think 1/3 repeated 3 times is a guarantee.

What I was saying is, I thought that the reason you are not guaranteed to win is due to variance alone. It’s the same reason you can hit 3 times in a row when you don’t even have a 50 percent chance to win.

The question is does the 74 percent chance take into account the variance that happens?

I find it strange that you quote me saying we know it’s not 100 percent and your response to that you think I think 1/3 repeated 3 times is a guarantee.

What I was saying is, I thought that the reason you are not guaranteed to win is due to variance alone.

The question is does the formula account for variance or if there was no variance would you only hit 1/3 with 3 chances 74 percent of the time?

What do you think variance is and how it relates to the answer?

variance is not like a law of nature and science lol it just means "sometimes it happen, sometimes it dont' happen" so your question doesn't even make sense. If it was to always happen then it would always happen, but it only happens sometimes

When we say an event has a probability of 74%, that IS variance. Suppose if the event occurs, you win $1, and nothing happens if it does not occur. Your expected value is $0.74. Variance is just a number, one that can be fairly easily calculated once we know the possible outcomes and their probabilities.

In this case there are two outcomes - you win $1 with probability 0.74 or you win nothing with probability 0.26. In general, to calculate variance, for each possible outcome we take the value of that outcome, subtract the expected value and square the result. The variance is then just the sum of all of these numbers. In this case we get
0.74 x (1-0.74)^2 + 0.26 x (0 - 0.74)^2 = 0.2099.

This number, while mathematically valid, really means little. Technically, the reason is that the units of this number are dollars squared, a unit that really means nothing to us. To make better use of this measure of spread, it makes more sense to use the standard deviation, which is simply the square root of the variance (hence in this case it would have sensible units - dollars). Here the SD would be $0.458. So how do we make sense of this? Well in this case we really canÂ’t, but if we modify a bit we can. Suppose we perform our experiment a large number of times. I wonÂ’t go into detail, but if we take the average value of the outcome of this latrge number of trials, it will get close to 0.74. However it is very likely NOT to hit 0.74 exactly. How likely is any particular outcome to occur? We can use the SD to figure that out, remembering one key fact - the standard deviation of the average of N trials is equal to the standard deviation of one trial divided by the square root of N.

This is more clear in a specific example. Suppose we run 10000 trials. Our average should get very close to 0.74; the SD tells us how close. The SD of this average is 0.458/100 or 0.0046. As a general rule, about 2/3 of the time we will get an average that is within 1 SD of the expected value - in this case between 0.7354 and 0.7446. We will be within 2 SD 95% of the time (between 0.7308 and 0.7492). We will be within 3 SDs 99% of the time (0.7262 to 0.7538).

When we say an event has a probability of 74%, that IS variance. Suppose if the event occurs, you win $1, and nothing happens if it does not occur. Your expected value is $0.74. Variance is just a number, one that can be fairly easily calculated once we know the possible outcomes and their probabilities.

In this case there are two outcomes - you win $1 with probability 0.74 or you win nothing with probability 0.26. In general, to calculate variance, for each possible outcome we take the value of that outcome, subtract the expected value and square the result. The variance is then just the sum of all of these numbers. In this case we get
0.74 x (1-0.74)^2 + 0.26 x (0 - 0.74)^2 = 0.2099.

This number, while mathematically valid, really means little. Technically, the reason is that the units of this number are dollars squared, a unit that really means nothing to us. To make better use of this measure of spread, it makes more sense to use the standard deviation, which is simply the square root of the variance (hence in this case it would have sensible units - dollars). Here the SD would be $0.458. So how do we make sense of this? Well in this case we really canÂ’t, but if we modify a bit we can. Suppose we perform our experiment a large number of times. I wonÂ’t go into detail, but if we take the average value of the outcome of this latrge number of trials, it will get close to 0.74. However it is very likely NOT to hit 0.74 exactly. How likely is any particular outcome to occur? We can use the SD to figure that out, remembering one key fact - the standard deviation of the average of N trials is equal to the standard deviation of one trial divided by the square root of N.

This is more clear in a specific example. Suppose we run 10000 trials. Our average should get very close to 0.74; the SD tells us how close. The SD of this average is 0.458/100 or 0.0046. As a general rule, about 2/3 of the time we will get an average that is within 1 SD of the expected value - in this case between 0.7354 and 0.7446. We will be within 2 SD 95% of the time (between 0.7308 and 0.7492). We will be within 3 SDs 99% of the time (0.7262 to 0.7538).