Gambling Logic Question

[albionmoonlight]

To keep it simple, let's make it a two horse race.

Horse A and Horse B are going to race.

I do my research and conclude that Horse A has a 2/3 chance of winning and Horse B has a 1/3 chance of winning.

The odds come out, and they are structured so that Horse A pays out as if it had a 3/4 chance of winning and Horse B has a 1/4 chance of winning.

(let's leave the self adjusting nature of parimutuel betting out of this and just assume fixed odds even though it is a horse race).

So, does it make more sense for me to

(1) bet on Horse A, who I think is going to win, or

(2) bet on Horse B, who I think is going to lose, but who gives me better odds than I think it should be giving me, or

(3) not bet at all because I should bet neither on the horse I expect to lose nor on the horse giving me worse odds than I expect?

[GoldenEagle]

I think betting on Horse B would have a negative EV. He wins 33% but you only get paid 25% of the wager. Is that right or am I just wrong here?

[CraigSca]

Anytime the payoff chance is less than what you think it is, you should bet (I think - ha).

In your example, the house only thinks horse B will win 1/4th of the time and they adjust their odds accordingly. If you think he'll win 1/3rd of the time, you should come out ahead in the long run, provided you're right about the 1/3rd odds.

[albionmoonlight]

I think you aren't quite getting my set-up, GE (which might be my fault).

A horse that is expected to win, say, 1 out of a hundred races will pay more for a win than a horse expected to win 1 out of 2 races.

I am trying to say that Horse B will hit big if it wins because it is only supposed to win 1 out of 4 races. I, however, think that it will win 1 out of 3 races, so I am getting a bigger payment than I would expect.

[Mustang]

Originally Posted by GoldenEagle

I think betting on Horse B would have a negative EV. He wins 33% but you only get paid 25% of the wager. Is that right or am I just wrong here?

I would think that horse B wins 33% of the time but, is paid 75% (should be 66%)

[CraigSca]

Were you directing your reply to me, albion? If so, that was exactly my point.

If you think the horse will win more often than the house (in this case your 1/3rd vs. the house's 1/4th), you should make that bet.

[albionmoonlight]

Sorry, CraigSca. I was directing it at GE.

The board is really slow for me, so it is hard for me to get in at the right place in these threads.

[CraigSca]

Oh, no problem. I was just wondering if you got that I got you .

[GoldenEagle]

Originally Posted by albionmoonlight

I think you aren't quite getting my set-up, GE (which might be my fault).

A horse that is expected to win, say, 1 out of a hundred races will pay more for a win than a horse expected to win 1 out of 2 races.

I am trying to say that Horse B will hit big if it wins because it is only supposed to win 1 out of 4 races. I, however, think that it will win 1 out of 3 races, so I am getting a bigger payment than I would expect.

I understand now. I also should have thought about in terms of odds instead of percentages. I find it easier to think in terms of odds.

[Pumpy Tudors]

Here's how I see it in my head (and someone correct me if I'm wrong):

Since we're using fixed odds for simplicity's sake here, does this mean that we can assume that 75% of the betting pool is on horse A and 25% is on horse B? Assume that there's $1000 in the pool, and you put $1 of that in. If you bet as follows:

Horse A:
67% of the time, you profit $0.33 = +$0.22
33% of the time, you lose $1 = -$0.33
EV = -$0.11

Horse B:
33% of the time, you profit $3 = +$1.00
67% of the time, you lose $1 = $0.67
EV = +$0.33

You would have to bet on Horse B here.

I came up with the profits by assuming that 750 betting units (dollars) were on Horse A, and 250 units were on Horse B. If Horse A wins, you have to split the wagers on Horse B among the winners. Split $250 into 750 pieces and you get $0.33 profit for each dollar wagered. On the other side, if Horse B wins, you have to split the wagers on Horse A ($750) into 250 pieces, so the people who bet on Horse B profit $3 each.

Anyway, by my calculations, Horse B would give you a positive expectation, and Horse A would give you a negative expectation, making the choice easy. It's overlay, baby!

[albionmoonlight]

Yeah.

It just (and I am kind of ashamed to admit this) never occured to me until last night that you should bet on the horse that you expect to lose. It makes perfect sense, but it seems counter-intuitive. I understood the concept of EV, but I just hadn't applied it to this particular situation.