# Probability of Multiple Weighted Coin Flips

I would like to calculate the probabilities of the outcomes of three weighted coins being flipped. I believe what I am looking for is a Poisson Binomial Distribution.

I am having trouble verifying/interpreting the results that I am finding on an online calculator.

Edit: Order does not matter in this question - the below is a table of the sums of outcomes.

+---------+---------+-------------+
| Outcome | Heads P | Probability |
+---------+---------+-------------+
| 3 heads |     .75 | .421875??   |
| 2 heads |     .75 | ??          |
| 1 heads |     .75 | ??          |
| 0 heads |     .75 | ??          |
|         |         |  (Sum = 1)  |
+---------+---------+-------------+


The .42 is calcualted for X>=3, but since there are only 3 flips it cannot be any greater. An alternate calculator provides a much lower answer, .03, which seems too low.

1. Is a Poisson binomial distribution the correct calculation for this answer? (X=3 trials, .75 success)
2. How would I find the probability of 2 out of 3 heads, 1 out of 3 heads, and no heads?

Thank you for taking the time to explain what I might be missing here.

• If the coins all have the same bias, the distribution for the count of heads among three flips is Binomial. – Graham Kemp Jul 24 '19 at 4:05

In a problem like this, where there are only 8 possible outcomes, sometimes it is simpler just to list them all and calculate the probability of each.

The probability of HHH is 0.75 ^ 3 which is about 0.4219.
The probability of TTT is 0.25 ^ 3 which is about 0.0156.

The probability of HHT, HTH, and THH is 0.1406 each (.75^2 * .25) so 0.4218 total.

The probability of HTT, THT, and TTH is 0.0469 each (.25^2 * .75) so 0.1407 total.

All of these 8 possible outcomes sum up to probability 1 (discarding roundoff error).

Just as a "side note", mathematics has a powerful set of tools, but for simple problems, all you need are simple tools. Match the tool the the problem and you will get results.

• Thanks David. I had been thinking of each of the 4 outcomes as an aggregate of totals and not considering their order. There truly are 8 outcomes, and you are certainly correct that this is an easier way to consider. My final results where order does not matter and filling out my table above are: All heads=.4219, 2 heads=0.4219, 1 heads= 0.1406, 0 heads=0.156 (Totaling 1) Does this seem correct? – Daniel Sims Jul 24 '19 at 4:13
• Order only "doesn't matter" if you cannot tell any difference. For example, HHH. However, if you have one "tail" (HTH for example), you CAN tell the difference between that and say THH. It is easy to check if you made a mistake cuz u know the total probability of all 8 possible outcomes should be 1 (or very close to it if you have roundoff error). In math, keep things simple if you can. – David Jul 24 '19 at 4:19
• In this case, it can be assumed that the coins are being thrown at the same time, and looking at the landed coins yields a total of heads. Correction on previous final totals: All heads=0.4219, 2 heads=0.4219, 1 heads= 0.1406, 0 heads=0.0156. (Interesting that 2 and 3 heads has the same probability.) – Daniel Sims Jul 24 '19 at 4:22
• 2 heads and 3 heads has the same probability simply because 2 heads is .75^2 * .25 but there are 3 possible outcomes (so it is basically .75^ 2 * (3*.25) which is .75^3. HHH is also .75^3 so that is why they "match", – David Jul 24 '19 at 4:24
• Thanks for looking at this! Appreciate your insight! – Daniel Sims Jul 24 '19 at 4:25

If the coins all have the same bias (0.75 for showing heads), then the distribution for the count of heads among three flips is Binomial.

$$X\sim\mathcal{Binomial}(3,0.75)\quad\iff\quad\mathsf P(X{=}x) ~=~ \binom{3}{x}\,0.75^x\,0.25^{3-x}\,\mathbf 1_{x\in\{0,1,2,3\}}$$

(Poisson Binomial is when the coins each have a distinct bias).

• Thank you for the distinction! – Daniel Sims Jul 24 '19 at 4:28