Florida Fantasy 5 sums probability

Reply #9

Wavepack — Fri, 03 Feb 2023 18:13:47 GMT

^ I replied to the wrong post, so the quote is the wrong one.

Reply #8

Wavepack — Fri, 03 Feb 2023 18:08:51 GMT

Good point about picking without replacement complicating the answer. However, I contend that the sum X1+ ... + X5 can be computed analytically via conditional probabilities and Bayes rule (along with convolution of the pdfs of the terms in the sum X1 + X2|X1 + X3|X2,X1 + X4|X3,X2,X1 + X5|X4,X3,X2,X1. Way more than five convolutions needed.

Reply #7

cottoneyedjoe — Fri, 03 Feb 2023 16:31:30 GMT

For comparison to California's 5/39 game (also called Fantasy 5), the probability that 5 quickpicks all have different sums is about 88.8%. And the probability that 11 lines of quickpicks all have different sums is about 51.5

Reply #6

cottoneyedjoe — Fri, 03 Feb 2023 16:27:47 GMT

It's not as simple as finding the convolution/sum of a distribution function. In this case, the distribution of the sum of the numbers on a random FL Fantasy 5 ticket is already known, call it P(X). The question is finding the probability that five i.i.d random variables X1, X2, X3, X4, and X5 whose underlying distribtution is P(X) are all distinct .

P(X) itself is not a convolution of five discrete uniform distributions because a FL Fantasy 5 ticket must have five distinct numbers.

Yo... [ More ]

Reply #5

Wavepack — Fri, 03 Feb 2023 14:05:42 GMT

You can derive the answer exactly without resorting to simulation with a PRNG.

The pdf of the sum of random variables is all of the pdfs convolved together. (In the limit, you get a Guassian pdf, which is the Central Limit Theorem). For example, if your RVs have uniform pdfs, as they are supposed to have in a lottery, then the sum of two of those RVs has a triangular pdf.

To answer your question, you'd convolve five uniform pdfs together (a few lines of Matlab code). Keep in mind the su... [ More ]

Reply #4

db101 — Fri, 03 Feb 2023 03:35:57 GMT

Well it worked to something like this:

10 lines, about 55.3%

11 lines, about 48.5%

12 lines, about 41.8%

My intuition was a little off on this one.

Reply #3

db101 — Wed, 01 Feb 2023 19:32:09 GMT

I suspect you're right. I'm going to work out how many lines of quick picks give a 50% chance of having all sums distinct. My instinct says between 15 and 20. We shall see.

Reply #2

cottoneyedjoe — Tue, 31 Jan 2023 18:49:09 GMT

It would be a total slog to get the exact probability because the denominator of the fraction would be on the order of 10^27. I see no shortcut to working out exactly because the 156 possible sums are not distributed uniformly as you probably realized. The distr is bell-shaped with tails at min=15, max=170, and a peak at median=92.5.

I also ran 1,000,000 simulations of 5 quick picks, and the sums were all distinct 87.9083% of the time, very close to your estimate. I'd stick with the estimate... [ More ]

Reply #1

db101 — Tue, 31 Jan 2023 16:31:55 GMT

With random simulations its about 87.88% of the time. Is an exact probability feasible with a shortcut or is it too much work to bother

Florida Fantasy 5 sums probability

db101 — Tue, 31 Jan 2023 16:21:55 GMT

I think if I put my mind to it I could work this one out long hand, but to save time I just ran a million simulations and estimated the answer.

The question is suppose you buy 5 quick picks of Florida Fantasy 5, which is a 5/36 game. What is the probability that each of the five lines has a different sum