Tournament Sports Betting

[BMAD]

64 teams are going to play in a tournament. You desire to bet your $192 savings on these teams but you know absolutely nothing about these teams so you will use a fair coin to select the team that you think will win the contest. Every correct bet doubles your money so a $1 bet awards you back your dollar plus an additional dollar and you cannot bet on two teams if they are playing each other in the same round.

The tournament is a single elimination contest, hence there will be 6 rounds. Your goal from betting is to attempt to make the most money at the end of the tournament.

You have two friends who proposed betting strategies:

Jon says that you should bet $32 for each round evenly. So in the first round, you put $1 on each team you think will win, then in the second round $2, culminating the final round with $32 on the winning team.

Eric, on the other hand, says you should bet $3 on each team in the first round, $3 on each team in the second round, and so on.

If you bet on the same teams regardless on the strategy chosen, which strategy will produce the most money?

[dgreening]

Some thoughts - unless I am missing something

in round 1, there are 64 teams in 32 games. If you choose randomly, you should win about half the games [16 in round 1] for each win you will get your bet back + that same amount [2*$bet].

Using Jon's strategy [doubling the bets every round]

you bet $1 on 32 games [$32 outlay], ------------ win half the bets [16] and get twice your bet [$2] for every win or 16*$2 = $32 ...... break even

Round 2: bet $2 on 16 games [$32 outlay],------ win half the bets [8] and get twice your bet [$4] for every win or 8*$4 = $32 .......... break even

Round 3: bet $4 on 8 games [$32 outlay],------ win half the bets [4] and get twice your bet [$8] for every win or 4*$8 = $32 .......... break even

Round 4: bet $8 on 4 games [$32 outlay],------ win half the bets [2] and get twice your bet [$16] for every win or 2*$16 = $32 .......... break even

Round 5: bet $16 on 2 games [$32 outlay],------ win half the bets [1] and get twice your bet [$32] for every win or 1*$32 = $32 .......... break even

Round 5: bet $32 on 1 games [$32 outlay],------ win half the bets [0.5] and get twice your bet [$64] for every win or 0.5*$64 = $32 .......... break even

By my calculations, you walk away [on average] with the same amount of money you started with using Jon's strategy

Eric's strategy [$3 on every game]

Round 1: bet $3 on 32 games [$96 outlay], ------------ win half the bets [16] and get twice your bet [$6] for every win or 16*$6 = $96 ...... break even

Round 2: bet $3 on 16 games [$48 outlay],------ win half the bets [8] and get twice your bet [$6] for every win or 8*$6 = $48 .......... break even

Round 3: bet $3 on 8 games [$24 outlay],------ win half the bets [4] and get twice your bet [$6] for every win or 4*$6 = $24 .......... break even

Round 4: bet $3 on 4 games [$12 outlay],------ win half the bets [2] and get twice your bet [$ 6] for every win or 2*$6 = $12 .......... break even

Round 5: bet $3 on 2 games [$6 outlay],------ win half the bets [1] and get twice your bet [$6] for every win or 1*$6 = $6 .......... break even

Round 6: bet $3 on 1 games [$3 outlay],------ win half the bets [0.5] and get twice your bet [$6] for every win or 0.5*$6 = $3 .......... break even

Thus Eric's strategy results in an average gain of $0

Bottom Line: neither approach will on average net you any money - both are equally good [bad].