When I retired, I decided that among other things, I wanted to learn to code. I’d had a bunch of data scientists reporting to me over the years, but other than some surface-level SAS and R, I hadn’t run anything but canned models, and so I’ve been doing all sorts of things since then - maps, models, simulations, etc.
Recently a question occurred to me, and I decided to model it out via simulation: suppose I roll two dice over and over, adding them together and keeping track of the totals, and when one of the numbers has been rolled 20 times, I declare it the winner - woo hoo! - and then start over again. We know 7 will be the most common roll (1 in 6 rolls in the long run), but how frequently will it win the races?
I made a guess and then wrote the simulations, and the result wasn’t what I expected. Spoiler box now: I assumed that the 7 would win a significant % percent of the time, because it only needs to outrun the other numbers individually, not collectively (insert “I only have to outrun you, not the bear” joke); still, at a count of 20-to-win, the simulations had the 7 winning over 50% of the time, and I’d guessed something like 1/3 of the time. So, my intuition may be off, but it also might be my coding.
Anyway, interested in any thoughts on all that, or some verification/refutation.
In a single roll of the two dice, (i.e., if we were doing “first-to-1” instead of “first-to-20”), 7 would have an edge, but you’d still get a reasonable chance of all of the other numbers.
The more you increase the number needed for a “win”, though, the more that advantage is amplified. For example,
Let’s suppose, for simplicity, that 2 is only competing against 7. If we get snakeyes or craps, we add it to the tally, and if we roll anything else, we discard it. There are six ways to roll a 7, and only one way to roll a 2, so on each individual roll, there’s a 1 in 7 chance that it’ll be a 2, and a 6 in 7 chance that it’ll be a 7.
But “first to 20” is equivalent to “best out of 39”. In other words, if I make that roll 39 times, what are the odds that I’ll hit the underdog over half of the time? That’s quite a longshot, probably significantly less than 1 in 100,000. And it’s actually even worse than that, because 2 isn’t just competing against 7, but also against all the other numbers.
I play backgammon which uses two dice. Knowing the probability of rolling any total is useful.
Dice Roll Probability
The chance of rolling a total of 2 is 2.78 percent
The chance of rolling a total of 3 is 5.56 percent
The chance of rolling a total of 4 is 8.33 percent
The chance of rolling a total of 5 is 11.11 percent
The chance of rolling a total of 6 is 13.89 percent
The chance of rolling a total of 7 is 16.67 percent
The chance of rolling a total of 8 is 13.89 percent
The chance of rolling a total of 9 is 11.11 percent
The chance of rolling a total of 10 is 8.33 percent
The chance of rolling a total of 11 is 5.56 percent
The chance of rolling a total of 12 is 2.78 percent
Simulating 1 million runs even though it doesn’t offer any real benefit: chance of 7 winning was 52.66%. I didn’t get a single 2 or 12 winner. 36/39 for 3/11.
My intuition was that 7 would have had a higher winning percentage.
Upping the number for 30 to win, after 100k trials, only bumps up 7’s chances to 60%. At 100 to win 7’s chances are 83%