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The odds of picking of 1st, 2nd and 3rd

Written by Sherwin, published on March 18, 2011

numbers in sequence

I was recently watching a television show, The Mentalist, in which the protagonist, Patrick Jane, pulled a slick trick of picking the first, second and third place horses at a race track. He told a guy he would do it and the bet was $500.

He won the bet. The trick was that he placed a bazillion fifty cent bets covering all of the permutations. Of course he didnt’ tell the mark that this was what he was going to do. And didn’t do it for money. He discloses later that he spent 5000 dollars doing it.

But I was interested instantly in how this might work. So here it is.

How does it work?

Let’s assume, for simplicity, that he can exclude some of the horses from hitting the top three. So let’s say that there are ten horses left, of which any three could win first, second and third. Setting aside the order of the wins for a moment, how many groups of three are there in a pool of ten horses? Well, forgetting some math, I did this instead to try to enumerate all of the possible collections of “three horses.”

Each digit represents a horse. Three digits together represent three winning horses, in no particular order.


134 234
135 235
136 236
137 237
138 238
139 239
130 230

145 245 345
146 246 346
147 247 347
148 248 348
149 249 349
140 240 340

156 256 356 456
157 257 357 457
158 258 358 458
159 259 359 459
150 250 350 450

167 267 367 467 567
168 268 368 468 568
169 269 369 469 569
160 260 360 460 560

178 278 378 478 578 678
179 279 379 479 579 679
170 270 370 470 570 670

189 289 389 489 589 689 789
180 280 380 480 580 680 780

190 290 390 490 590 690 790 890

Now I get that this is not the most elegant demonstration of how to enumerate all of the groupings of three (winning horses), from a collection of ten. And it’s not exactly a proof. But as a demonstration, it works. The number of groupings is:

# = (8 x 1) + (2 x 7) + (3 x 6) + (4 x 5) + (5 x 4) + (6 x 3) + (7 x 2) + (8 x 1)

# = 8 + 14 + 18 + 20 + 20 + 18 + 14 + 8 = 40 + 36 + 28 + 16 = 120

But the order of the three horses that win matters. Patrick Jane said he would pick the first, second and third place horses. And of any three horses that win, there are six permutations of their order. How to prove this is left as an exercise for the reader. But 120 groupings multiplied by a factor of 6 possible orders gives us 720 bets that Jane had to make. At fifty cents each, he therefore spent 360 dollars.

But this is assuming that he could narrow the field to ten horses. Every extra horse over ten is an exponential growth in the number of bets he would have to place and the money he would have to spend.

Now he said he spent five thousand dollars. So the question is, given that he bet on every permutation, how many horses were in the field?

Seriously, though, I’m interested in how someone might work this out.

I also might not be right about how I did this.

Interesting test of this method

Given that this demonstration is not really a proof, I think there is a fun way to test the method a little, by way of convincing ourselves that I’m right.

Imagine for a moment what the odds are of a single horse being in an particular grouping of horses. Let’s imagine for example, that the three winning horses are a single stone in a bag full of stones where each stone represents a unique grouping of horses. Not prejudging how many stones are in the bag, assuming there are ten horses in the field, we can still safely assume that if you reach into the bag, that every horse has an equal chance of being in the winning grouping (stone). Put differently, if the race is perfectly random, the horses have an equal chance of being in the winning trio.

This means that we can look at my enumeration above and test it to see if every horse has an equal number of occurrences in the winning trio. If we find that there are a different number of digits, in my method above, that would prove that the enumeration had failed.

As it turns out, each digit (horse) appears in 36 groupings. It’s actually really fun to pick a digit and then count the occurrences. Neat patterns emerge. 36 doesn’t appear to be an accident does it? We might use this as a clue to create a formula for solving this kind of problem for N horses.


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Archived comments

  1. Sherwin says:

    Yes! It’s the same idea isn’t it? If you can be right even just three times in a row, and not count the wrongs, then you can probably secure a lot of cash from folks that are too greedy to see the deception.

  2. sarah says:

    This makes me think of spam stock recommendations. If you’re going to tell half your suckers that a stock will go up tomorrow and half it will go down, then repeat the next day with whoever received the correct prediction, how many suckers do you have to start with to have some left after a few rounds of your dazzling investment insights?

    (You like math. I like cons.)