Calculating odds for rolling dice.?

Question

If I have 10 dies and each of them have 10 sides. What are the odds that the numbers I rolled, added up, will equal 92 or higher?

Also, can anyone give me a formula for calculating the odds when rolling dice? Dice of any amount of sides and in any amount of number.

Answer 1

10 dies - 10 sides each...

the lowest you could get is 10



10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10

a total of 100

so if rolling a 92 or higher is 8/90 because you can't get lower then 10. 8/90 reduces to 4/45 ~ about 8.9%





ex: if i wanted a 56 or LOWER on a roll with 10 die with 10 sides each

then i take the number and put it over 90 = 56/90

reduces to 28/45 ~ 62%



or



if i want a number or HIGHER...ex: 22, i then take the number subtract it FROM 90, for 90 - 22 = 68

leaving me with a 78/90 - reduces to 39/45 ~ 86%

Answer 2

Gianlino's "generating function" approach is cool. What it's based on is what's called linear recurrence...but grinding out the derivatives may be harder than setting up a spreadsheet to do the work. The idea is to find p(n) as the probability of hitting n at some point. p(0) = 1, since every game starts at 0. If you define p(n)=0 for negative n (can't get a negative score) then for n>1 you have: p(n) = (1/6)p(n-1) + (1/6) p(n-2) + ... + (1/6) p(n-6) This is because in order to get a score of n, you have to roll the number k (1 <= k <= 6) when the previous score was (n-k). The expression above take into account all 6 ways this can happen. Each probability turns out to be the average of the previous 6 probabilities. This is very easy to put into a cell formula for OpenOffice Calc or MS Excel. Or you can grind out the exact fractions, but the denominators will get large. Here are the first few: p(1) = (1) / 6 = 1/6 p(2) = (1/6 + 1)/6 = 7/36 p(3) = (7/36 + 1/6 + 1) / 6 = 49/216 p(4) = ( 49/216 + 7/36 + 1/6 + 1) / 6 = 343/1296 Note that this is 7^(n-1) / 6^n, and will be for n up to 6. Except for the first term, you have a geometric series, and you can work this out by induction, if you like. Starting with p(7), though, not all of the previous terms will be summed, so the formula stops working. The "average of the previous 6" recurrence rule still works, though, and that's what you should use. I used OpenOffice Calc, and got the following relevant probabilities: p(11) = 106442161/362797056= 0.2933931222 p(12) = 633074071/2176782336= 0.2908302133 p(13) = 3647371105/13060694016= 0.2792631923 I claim the rational fractions to be exact, and the decimal fractions to be pretty good approximates. To not bust, you need one of three mutually exclusive things to happen: a) get to 11, with probability P(11) = p(11) ~= 0.2934 b) get to 12 without getting to 11 first: P(12) = p(12) - p(11)/6 ~= 0.2419 c) get to 13 without getting to 11 or 12 first: P(13) = p(13) - p(11)/6 - p(12)/6 ~= 0.1819 The capital P(n) is the probability of n being your final score. You will not bust P(11)+P(12)+P(13) of the time, or about 0.7172 probability of not busting. So, you lose outright 0.2828 of the time. You will finish with 11 with p(11) ~= 0.2934, and the dealer will finish with 11 or better (no bust) the same 0.7172 of the time that you would have, so you lose in this case with probability 0.2934 * 0.7172. That's another 0.2104 probability to lose. You will finish at 12 with probability P(12) ~= 0.2419, but lose when the computer finished with 12 or 13 for a combined probability of P(12)*[P(12) + P(13)] ~= 0.1025 You will finish at 13, but lose to a dealer 13 with probabilty P(13)^2 ~= 0.0331 Your total probability of losing is then 0.2828 + 0.2104 + 0.1024 + 0.0331 ~= 0.6287 In odds terms, that's a little worse than 8-to-5 against the player. If the house paid $8 on a $5 bet, the expected return to the house would be about 17.3 cents per game. Edit: Thanks to gianlino for pointing out my error: The final term should be P(13)*p(13) ~= 0.0508 instead of 0.0331, giving 0.6464 as the probability to lose, and more like 9-to-5 odds. At $9 payoff on a $5 bet, the house wins about a nickel per game.

Answer 3

There are 10^10 total number of ways the dies could roll,

starting from (1,1,1,1,1,1,1,1,1,1) and ending in (10,10,10,10,10,10,10,10,10,10).

Of these 10^10 , we must count how many procude a sum of 92.

One example this be (10,10,10,10,10,10,10,10,10,8) =98 an example.

If they are all 9, it would not produce a sum of 92 or higher.

So, 9 of them must be 10's and the other could be 2 through 10.

x x x x x x x x x x

10 spots for 2 and the other spots for 10s.

10 spots for 3 and the other spots for 10s.

.

.

10 spots for 9 and the other spots for 10s.

90 ways

All are 10 = 1 way.

91 / 10^10 is the probability.

=91 / 10,000,000,000