# Odds of picking at least one winning horse out of x races **URL:** \ **Category:** Factual Questions\ **Created:** [August 30, 2025, 12:17am UTC](https://boards.straightdope.com/t/odds-of-picking-at-least-one-winning-horse-out-of-x-races/1022592 "2025-08-30T00:17:50Z")\ **Posts on this page:** 12\ **Page:** 1
**Author:** ![HeyHomie](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/heyhomie/32/207_2.png) [@HeyHomie](https://boards.straightdope.com/u/HeyHomie)\ **Post date:** [August 30, 2025, 12:17am UTC](https://boards.straightdope.com/t/odds-of-picking-at-least-one-winning-horse-out-of-x-races/1022592/1 "2025-08-30T00:17:50Z")
Imagine I’m going to the track. There are going to be ten races today. I’m going to bet on one horse, at random, to Win, for each race. Because the number of horses in each race will vary, I’m thinking my intuitive approach (explained in a second) may be too simplistic. Imagine that tonight’s card has ten races. I’m shall now list the number of horses each race, and the odds of choosing the winner in THAT race. Race 1: 8 horses, 12.5% Race 2: 10 horses, 10% Race 3: 12 horses, 8.3% Race 4: 6 horses, 16.67% Race 5: 7 horses, 14.29% Race 6: 9 horses, 11.11% Race 7: 10 horses, 10% Race 8: 8 horses, 12.5% Race 9: 12 horses, 8.3% Race 10: 10 horses, 10% Intuitively, it seems to me that the solution is to aggregate those percentages and divide by 10, giving me odds of 11.34% that I’ll leave having picked one winner. Or in other words, I’d have to bet on ± 100 races before, statistically speaking, I’d be likely to pick a winner. But I feel like I’m looking at this too simplistically. The odds may even be incalculable since each race is its own … event (for lack of a better choice of words). Any maths or statistics Dopers want to take a stab at this? ---
**Author:** ![markn\_1](https://avatars.discourse-cdn.com/v4/letter/m/f9ae1b/32.png) [@markn\_1](https://boards.straightdope.com/u/markn_1)\ **Post date:** [August 30, 2025, 12:32am UTC](https://boards.straightdope.com/t/odds-of-picking-at-least-one-winning-horse-out-of-x-races/1022592/2 "2025-08-30T00:32:36Z")
In problems like this, the easiest approach is to calculate the probability of LOSING ALL the races, then subtract that from 1, giving you the probability of winning at least one race. The probability of losing all the races is just the product of the probability of losing each race. So in this case, the probability of losing them all is \frac{7}{8} \times \frac{9}{10} \times \frac{11}{12} \times \frac{5}{6} \times \frac{6}{7} \times \frac{8}{9} \times \frac{9}{10} \times \frac{7}{8} \times \frac{11}{12} \times \frac{9}{10} = 0.2978 So the probability of winning at least one is 0.7022. ---
**Author:** ![Stranger\_On\_A\_Train](https://avatars.discourse-cdn.com/v4/letter/s/13edae/32.png) [@Stranger\_On\_A\_Train](https://boards.straightdope.com/u/Stranger_On_A_Train)\ **Post date:** [August 30, 2025, 12:42am UTC](https://boards.straightdope.com/t/odds-of-picking-at-least-one-winning-horse-out-of-x-races/1022592/3 "2025-08-30T00:42:40Z")
Your odds of getting at least one win by randomly selecting horses in each race is about 70%. (1.0-odds of picking losers in all races.) The odds of getting at least two winners are more complicated but are no worse than 5.6%. Whether you can make a profit on this depends on the payouts for each pick, but of course the odds are selected to ensure that the track statistically makes a profit, so random picking of a statistically significant number of races will ensure that you take a loss. Stranger ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [August 30, 2025, 1:44am UTC](https://boards.straightdope.com/t/odds-of-picking-at-least-one-winning-horse-out-of-x-races/1022592/4 "2025-08-30T01:44:05Z")
> [@HeyHomie](#): > > Intuitively, it seems to me that the solution is to aggregate those percentages and divide by 10, giving me odds of 11.34% that I’ll leave having picked one winner. A moment’s thought should reveal that this certainly isn’t correct, given that you’re more likely than that to win just in race #1 (or 4, 5, or 8), and you must be more likely to win at least one overall than you are to win any individual race. What you calculated there is a sort of average chance of winning any given race, not the chance of winning one, ever. ---
**Author:** ![Steven\_Maven](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/steven_maven/32/3272_2.png) [@Steven\_Maven](https://boards.straightdope.com/u/Steven_Maven)\ **Post date:** [August 30, 2025, 8:31pm UTC](https://boards.straightdope.com/t/odds-of-picking-at-least-one-winning-horse-out-of-x-races/1022592/5 "2025-08-30T20:31:36Z")
> [@markn\_1](#): > > In problems like this, the **easiest** approach is to calculate the probability of LOSING ALL the races, then subtract that from 1, giving you the probability of winning at least one race. Not just “easiest” but the only valid formula ---
**Author:** ![markn\_1](https://avatars.discourse-cdn.com/v4/letter/m/f9ae1b/32.png) [@markn\_1](https://boards.straightdope.com/u/markn_1)\ **Post date:** [August 30, 2025, 8:59pm UTC](https://boards.straightdope.com/t/odds-of-picking-at-least-one-winning-horse-out-of-x-races/1022592/6 "2025-08-30T20:59:22Z")
Well, you could approach it in any of several more complicated ways. For example, it is the probability of getting exactly one win, OR getting exactly two wins, OR … OR getting exactly ten wins. The answer will come out the same of course, but the math will be more complicated. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [August 31, 2025, 3:12pm UTC](https://boards.straightdope.com/t/odds-of-picking-at-least-one-winning-horse-out-of-x-races/1022592/7 "2025-08-31T15:12:48Z")
Yeah, if it were the only valid formula, then it’d be impossible to ask questions like “What are the odds of picking at least three winners”. Those questions are usually more difficult to answer, but they can still be answered. ---
**Author:** ![Stranger\_On\_A\_Train](https://avatars.discourse-cdn.com/v4/letter/s/13edae/32.png) [@Stranger\_On\_A\_Train](https://boards.straightdope.com/u/Stranger_On_A_Train)\ **Post date:** [August 31, 2025, 8:05pm UTC](https://boards.straightdope.com/t/odds-of-picking-at-least-one-winning-horse-out-of-x-races/1022592/8 "2025-08-31T20:05:11Z")
> [@markn\_1](#): > > Well, you could approach it in any of several more complicated ways. For example, it is the probability of getting exactly one win, OR getting exactly two wins, OR … OR getting exactly ten wins. The specific and composite odds for getting at least a specified number of wins are: | Wins [N] | Probability of exactly N wins | Probability of at least N wins | | --- | --- | --- | | 1 | 0.3849 | 0.7022 | | 2 | 0.2223 | 0.3173 | | 3 | 0.0755 | 0.0951 | | 4 | 0.0167 | 0.0195 | | 5 | 2.53e-03 | 2.81e-03 | | 6 | 2.63e-04 | 2.82e-04 | | 7 | 1.87e-05 | 1.95e-05 | | 8 | 8.63e-07 | 8.87e-07 | | 9 | 2.35e-08 | 2.38e-08 | | 10 | 2.87e-10 | 2.87e-10 | The better is actually more likely to get at least two wins from a purely random draw than they are of losing all races (again, assuming uniform odds for each race). Of course, most horses that are likely to win aren’t going to pay anything close to 5:1 so the likelihood of the random draw strategy to net a profit is pretty low unless you engage in some parimutuel fuckery (see _The Grifters_ for an example) to bias the payouts in favor of your picks. Stranger ---
**Author:** ![Maserschmidt](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/maserschmidt/32/18829_2.png) [@Maserschmidt](https://boards.straightdope.com/u/Maserschmidt)\ **Post date:** [August 31, 2025, 8:32pm UTC](https://boards.straightdope.com/t/odds-of-picking-at-least-one-winning-horse-out-of-x-races/1022592/9 "2025-08-31T20:32:35Z")
> [@Stranger\_On\_A\_Train](#): > > the likelihood of the random draw strategy to net a profit is pretty low I simulated 1000 bettors, 10 races of 10 horses each, a range of probabilities/odds (with two strong favorites at 30% and 20%, etc.), everyone including random-me betting $10/race, and a track take of 15%. This is the kind of thing I do while watching football. Expected net return per 10-race session: $ -15.5 [only did 100,000 simulations, this would —\> -$15 with more] Standard deviation of net return: $ 122.96 Probability of making a profit: 0.2759 Pretty low, and a lot of times it pays out it’s nominal. ---
**Author:** ![Maserschmidt](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/maserschmidt/32/18829_2.png) [@Maserschmidt](https://boards.straightdope.com/u/Maserschmidt)\ **Post date:** [August 31, 2025, 8:33pm UTC](https://boards.straightdope.com/t/odds-of-picking-at-least-one-winning-horse-out-of-x-races/1022592/10 "2025-08-31T20:33:43Z")
Oops, somehow my bracketed -$15 triggered some Discoursery. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [September 1, 2025, 1:22am UTC](https://boards.straightdope.com/t/odds-of-picking-at-least-one-winning-horse-out-of-x-races/1022592/11 "2025-09-01T01:22:10Z")
I fixed it. Just like Discourse uses asterisks to mark off _italics_, it uses dollar signs to mark off \LaTeX (often used for equations and such). You can prevent that from happening by putting a \backslash in front of one or more of the symbols. ---
**Author:** ![Maserschmidt](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/maserschmidt/32/18829_2.png) [@Maserschmidt](https://boards.straightdope.com/u/Maserschmidt)\ **Post date:** [September 1, 2025, 1:35am UTC](https://boards.straightdope.com/t/odds-of-picking-at-least-one-winning-horse-out-of-x-races/1022592/12 "2025-09-01T01:35:47Z")
Got it, thank you!