Say you're in a Vegas sportsbook and someone named Ralph gives you two options: you could have 2 winning bets out of 3 games or you could have 5 winning bets out of 8 games...which would you take?
Say someone named Mike gives you two options: you could have 2 winning $100 bets out of 3 games or you could have 5 winning $100 bets out of 8 games...which would you take?
These are not the same question. And my head hurts.
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To Ralph, I would rather have the winning bets. If I had $100 to wager on three games, you'd have 2 winning $33.33 tickets. After the vig, that's a profit of $60.66. With $100 to wager on eight games, you'd have 5 winning $12.50 tickets. After the vig, that's a profit of $56.88. Makes sense right? A 66% success rate is better than a 62.5% success rate.
But with Mike, it's not the same question. 2 winning $100 bets out of 3, nets you a profit of $82. 5 winning $100 bets out of 8, nets you a profit of $155. So you had a lower success rate, but you wagered $800 instead of $300.
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Our playoff picks had me thinking. Mark went 5-5 and lost 45 fictional dollars. Because all Vegas bets are marked -110, meaning you have to bet 110 to win 100. In other words they only payout 91%.
Which means that just to break even, you need to have a success rate of 55%. This is the house edge. Or in the words of my father, how Vegas pays the electric bill for all those lights. Again, if you win 54% of your bets, you are losing money.
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I wanted to break down, what's the ideal number of concurrent bets to have to turn a profit. So in each category I'll list the possible success rate outcomes. Rates that turn a profit (above 55%) will be in bold.
1 bet
0%
100%
2 bets
0% 50% 50%
100%
So this is pretty obvious stuff. On one bet you either have a win (complete success) or a loss (complete failure). On two bets, A and B, both could lose, both could win, or A could win and B lose, or B could win and A could lose. But what you're looking at is 3 out of 4 times you're not making money. This seems like the worst thing to do.
3 bets
0% 33% 33% 33%
66% 66% 66% 100%
With three concurrent bets, there are eight possible outcomes. In only one out of 8 do you lose all your money. In three out of eight, you lose a bit. In three out of eight you win a bit. And in one out of eight, you win it all. This is my favorite. Winning 2 out of 3 bets seems feasible.
4 bets
0 25 25 25 25 50 50 50 50 50 50 50
75 75 75 75 100
Now we're up to 16 outcomes. And in only 5 do you make money. Even numbered bets are the worst.
5 bets
0 20 20 20 20 20 40 40 40 40 40 40 40 40 40 40
60 60 60 60 60 60 60 60 60 60 80 80 80 80 80 100
32 outcomes. 16 turn a profit. So it's pretty clear to me, that either 3 or 5 is the perfect number to bet.
Let's do a quick comparison. Say you have $300 to bet.
If you divide it into 3 bets and win 2, that's a profit of $82.
If you divide it into 5 bets and win 3, that's a profit of $43.8.
For me, it seems about as easy to come out 1 game over .500 whether you bet 3 or 5, and it's almost twice as profitable to go 2-1 than 3-2.
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If you have $300 to bet and put it in 1 bet, and get it right you profit $273.
If you have $300 to be and put it in 3 bets, and go 3-0, you profit $273.
Clearly it's harder to go 3-0 than it is 1-0. So if you're trying to make the best profit, it's best to just put it all on one game. It has the most reward, but also the most risk. 0-1 is the easiest way to lose $300. It's much more likely to go 0-1 than it is 0-3.
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To conclude, don't ever bet on an even number of games. If you're trying for the best risk-reward, bet on 1 game. If you're trying to protect yourself and have the best chances of making money, bet on 3 games.
