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Case 1: If we bet \$1 on team #1 and it wins then we will get \$2

Case 2: If we bet \$1 on team #2 and it wins then we will get \$4

Case 3: If we bet \$1 on team #3 and it wins then we will get \$6

Case 4: If we bet \$1 on team #4 and it wins then we will get \$16

Case 5: If we bet \$1 on team #5 and it wins then we will get \$21

We can bet any amount on all teams at same time. So find the correct combination (amount on each and every team) so that we have to get the profit or at least no loss regardless of which team may win.

Jean-Claude Arbaut
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rajashekar
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  • I guess $Rs$ means rupees (you may remove the money symbol, as it's useless, and is unclear in many countries). Is the gain proportional to the bet? I mean, if I bet $n$ rupees on team $5$ and I win, will I win $21n$ rupees? – Jean-Claude Arbaut May 27 '15 at 11:56
  • Let the total bet be $t$. If we bet $>\frac{t}{3}$, $>\frac{t}{5}$, $>\frac{t}{7}$, $>\frac{t}{17}$, $>\frac{t}{22}$ on the $1$st, $2$nd, $3$rd, $4$th, $5$th teams respectively, we don't lose. We can, since $1/3+1/5+1/7+1/17+1/22<1$. – Alexey Burdin May 27 '15 at 12:02
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    @AlexeyBurdin is it really 3,5,7,17,22 ? I mean, if you bet t/3 on the first, you will only have 2t/3 left if it wins. With a total bet of t, it's a loss. Or you mean you also recover your bet, but I thought it was the meaning of the first "Rs.1" in what you get. – Jean-Claude Arbaut May 27 '15 at 12:34
  • I think @jca's interpretation is correct; that is, the paying odds are even money, 3:1, 5:1, 15:1, and 20:1. However, we then have $1/2+1/4+1/6+1/16+1/21 > 1$. – Brian Tung May 27 '15 at 18:05

1 Answers1

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Let $T_n$ be the fraction of our total money betted on the $n^{th}$ team.

Both Alexey Burdin and Brian Tung offered an initial solution, where $T_1=\frac{1}{2},\ T_2=\frac{1}{4},\ T_3=\frac{1}{6},\ T_4=\frac{1}{16},\ T_5=\frac{1}{21} $

See that $T_n\times W_n=1\ \ \forall n$ where $W_n$ are winnings under if $n^{th}$ team wins. That means that whatever happens our entire betting is given back. But do the fractions really sum the total? $\sum T_n = \frac{115}{112}>1$

That means that it is impossible to ensure no loss. Since you'd need to bet more than the total amount betted, what's impossible.

Knowing the probability for each team to win, though, it's possible to offer a betting that maximizes the expected return.

Masclins
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