Let’s say I have a lottery game or something similar. There are 20 numbers. Let n = 20. Let’s say you choose four numbers ®, and that four numbers will be drawn. What are the odds (OK, the probability) that you have chosen the four numbers that were drawn? (Numbers are used only once, and they do not have to be picked in order.)
Prob = n! / r!(n - r)!
So:
Prob = 20! / 4!(20 - 4)!
Prob = 2.43290E18 / 24 x 2.09228E13
Prob = 4,845
So given 20 numbers, of which four will be drawn, The probability that I picked those four numbers is one in 4,845. Right?
OK, so continuing with the ‘lottery game’ scenario, let’s say there’s a bonus number. You pick one number of ten, and one number is drawn. You have a one-in-ten chance of having picked the number.
So we have 1/4845 to pick the right four numbers in the first half of the game, and 1/10 to pick the bonus number. What are the odds of picking the correct four numbers and also the bonus number? Are the ratios simply multiplied? What is the equation to calculate the probability of winning all of a given number of games? (And here I’m assuming that picking the four numbers and picking the bonus number count as two games. Please correct me if I’m wrong.)
As long as they’re independent events, this is it. The probably of getting all four numbers plus the bonus number would be 1/48450.
If the games are independent, you simply multiply the ratios. If they’re not independent, you have to understand the relationships between them to work out the probabilities accurately.
Yes. Assuming the “subgames” are independent, then you just multiply together the probability of winning each subgame. This works for an arbitrary number of games.
I just looked up the Washington State Lottery Mega Millions game. Five numbers are chosen from a field of 56 and one number is chosen from a field of 46.
So the odds are 1/3,819,816 and 1/46. So the probability of winning the jackpot is 1/175,711,536, right?
(This is fun!)
Oh – I know there’s a difference between odds and probability, but I’m not sure what it is. Clarification?
If the outcome of interest has a probability p of occurring, the odds in favor of that outcome are p/(1-p). The odds against it is just the reciprocal of that.
So the odds of rolling a six on a standard die is (1/6)/(5/6) or 1:5. The odds against rolling a six are 5:1.
Odds help to equalize a wager. If I were playing a game where I win by rolling a six, my jackpot (the amount you pay me if I win) should be five times my penalty (the amount I pay you if I lose) to make the expected value zero. That is your jackpot-to-penalty ratio should be the same as your odds against a win.
So if I am betting you that an unlikely event will happen, and this odds against this particular unlikely event are, say, 100:1, then you should offer to pay me $100 if I win and I pay you $1 if I lose. This seems steep, but recall, I have only a (1/101), or approx. 0.0099, probability of winning.
Incidentally, the calculation you did for your first question, choosing four balls out of 20, is referred to as “20 choose 4”, and can be written as the first number over the second number (but without a line or anything between them), all encased in big parentheses. Something like
/20\
( )
\ 4/
Man, this would be so much easier if web browsers recognized LaTeX!
And since you said you had a new calculator… if it’s fancy enough (which isn’t too fancy), it may have a permutations feature, which is typically labeled nCr for “n choose r”, which in your case would indeed be “20 choose 4”.
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