# Thread: Check math and help w/ Expected Value

1. ## Check math and help w/ Expected Value

Assume we know that 2 horses in two different races each have exactly a 3.30/1 and 2.8/1 chance of winning their respective races. We are offered a bet that pays 25/1 if both horse win their respective races. How do we calculate the expected value of this bet.

I avoided every math class I ever got close to, so please excuse my ignorance.

This is the track I'm on, please confirm I am doing it correctly, and help me get to the final answer that eludes me.

Calculate the probability for each horse:

Horse #1 3.30/1 = 1/(1+3.30) = .2325581
Horse #2 2.80/1 = 1/(1+2.80) = .2597402

Probability of both events happening is .2325581 * .2597402 = .0604046

Please confirm I have that part correct.

If that is correct, how do I use it to calculate the EV of the offered bet?

In my head I would look at that and ball park estimate its roughly 1/4 * 1/4 = 1/16

which means in the context of the offered bet I would loose $1 15 times and gain$25 1 time. So I would take the bet.

Is it 25/15 = 1.66? which would mean I have a 66% advantage, or gain 66 cents every time I bet a dollar given the circumstances I described?

How can I use the .0604046 figure to calculate exactly what the EV is rather than just that quick ...in the head math estimation with the fractions above.

2. I never understood betting odd, like 25/1. I assume it means for every 1 dollar you bet, if you win you get 25 times that back. If so, then that is what you use for your expected value:

Probability Both Horses Win - 0.06
Probability Both Horses Don't Win - 0.94

To calculate expected value we multiply the probability of success, by our random variable's 1 and 25 respectively:

0.06*25+0.94*1=EXP for every one dollar you bet.

So what we are doing to adding the probabilities of the two possible outcomes:

When you make a bet at 25/1 it means you lose 1 dollar if you lose the bet, and get your 1 dollar wager back plus 24 dollars.

.06 or 6% of the time we get $25 .94 0r 94% of the time though, we lose the 1, we get 0 so considering what you have just so kindly taught me, isn't it (.06*25) + (.94*0) = 1.50 Do I need to consider that I am getting my own dollar back.... I'm getting 24 dollars plus the original wager. so maybe I should then: 1.50 - 1.00 = .50 and the EV on the bet is 50 cents for every dollar bet. I may be way off base here, I am totally winging it... but that number sounds about right to me. 4. Close, but not quite. Our Probability Mass Function looks something like this: X | P(X) 25 0.06 -1 0.94 Where X is the amount of money we can expect to win on this "Double Down" type bet. If the two horses win, you will win 25 bucks. If both horses don't win, you win -1 dollar (or in proper English, you lose a dollar - your initial bet). Expected value is calculated as: $\Sigma_{i=1}^{n}X_1\cdot P(X_1)$ Thus, to find out the Expected value of our probability mass function (which is just another fancy way of saying the MEAN), you need to take the sum of your individual random variables (in this case you have two) multiplied by their individual probabilities. You most definitely should get a negative number (otherwise these betting places would go out of business super quick). 5. Jus 6. ANDS! "You most definitely should get a negative number (otherwise these betting places would go out of business super quick)." In the US all horse racing is parimutuel. Think of it more as a marketplace that is almost always efficient, but sometimes inefficiencies arise like the above example, which are actual odds and the payoff was actually$25.90. The track takes a cut from each betting pool, and will never go out of business based on who wins or loses. The track has no stake in the betting.

In my innumerate mind, I look at that betting opportunity as roughly a 1 in 16 shot that is paying 25 to 1, so I would make that bet.

Perhaps there is something I don't know or am not factoring in. You second post is all Greek to me...

If we can find the probability of two independent events both happening by multiplying each probability, and....here's the assumption, but for the purposes of discussion, we know for certain that the correct odds on each horse in the example are as stated, why am I wrong is thinking that the payoff in this particular case is extraordinarily high, and a good betting prospect.

7. I actually had to look up a few of those terms - . I'm going to have a smoke and think about it. My last post was the fundamental idea of "expected value" assuming no other rules are in place: for example the lottery - if there are 100 tickets, and you buy one ticket for 5 bucks, you have 0.01 chance of making 95 bucks, and .99 chance of losing 5 bucks; your expected value would by 0.01*95+.99*(-5)=-4. It doesn't look like the track is that simple, but the rules should apply. I will think on this.

8. Many thanks for indulging me... You should have seen me struggling with poker math while reading Sklansky's The Theory of Poker, but I cracked it eventually, and actually enjoyed the process, which is more than I can say for any math class I fell victim to.

Instead of the first go of:

"0.06*25+0.94*1=EXP for every one dollar you bet."

which produces a number that just doesn't feel right.

or this, my second stab at it:

"(.06*25) + (.94*0) = 1.50"

we do this inspired from your last post:

(.06*24) + (.94*-1) = .50 cents

Which surprisingly is the same as:

Do I need to consider that I am getting my own dollar back.... I'm getting 24 dollars plus the original wager. so maybe I should then:

"1.50 - 1.00 = .50 and the EV on the bet is 50 cents for every dollar bet.

I may be way off base here, I am totally winging it... but that number sounds about right to me."

9. So after reading a little bit heres my take on the situation: I think the standard EV that I posted above is still in affect here, with one or two qualifiers: if these horses run multiple times, and you are able to make multiple bets, then yes you could expect to win approxiamately one time out of sixteen, and net 10 dollars. If however this is a on-off, and these horses never race again, then it would be how I originally set it up.

The reason you do not reduce the 25 to 24 is because you aren't saying "Well I can win 25, but I bet a 1" - that's already taken into account by the random variable "-1". The random variables were are calling WINNINGS: you win 25 dollars, or you lose 1 dollar on this bet. We do not adjust that based on what we have wagered. So in fact you would expect to win .56 cents for every 1 dollar that you bet.

The "parimutuel" way that the track system is set up, shouldn't actually affect what you are trying to calculate here, unless a person is mixing their bets - i.e. betting in different tiered systems (from what I gathered from the wikipedia article on Parimutuel Betting). This system that they have in place seems kind of like the "tables" at a casino for each individual tier - you're placing a bet on a specific outcome with the amount won dependent on how high, or low, stakes you are betting in - however the guy playing at the 1 dollar tables, has the same odds of winning as the guy at the 100 dollar table; it's just their EXPECTED winnings that will be different because of the different amounts wagered. The 25/1 helps us to establish what the winnings will be.

So as I see it, if someone tells you two horses have a .06 chance of winning together, and they are offering 25 to 1 odds on that, I would probably take it as well if it were a one-off bet, as you'd only be expected to lose less than half of your initial bet. Now if it were to go down into the .40 or below winnings back, then I'd have to evaluate what I'm prepared to lose again.

10. Yes...that makes sense. and feels right.

The first post uses a +1 and produces a result to weird to be right. But with a negative one.... I'll buy that.

It makes perfect sense when you state it as .06 times you win 25 and .94 times you win -1...

Thanks for your help in this. Many many thanks.