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A casino gave you a $1 chip which is non redeemable for cash, you can only use it to play the roulette. If you play it either you win and get cash redeemable chips in return or you lose the chip, but since the chip is not redeemable for cash you are not losing cash, only the chance to play. Assume that this roulette has only 36 slots numbered 1 to 36.

Payoffs for $1 bet are as follows:

a) red or black       $1 
b) any single number $35

The expected value for the bets would be:

a) 0.5 * $1 = $0.5
b) 1/36 * $35 = $0.97222

Therefore betting on a single number has a higher expected value. Is this reasoning correct?

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    $\begingroup$ Yes - if you are only allowed one bet $\endgroup$
    – Henry
    Oct 29, 2017 at 13:38
  • $\begingroup$ @Henry you can bet again if you win. Otherwise you lose the chip. $\endgroup$
    – Roland
    Oct 29, 2017 at 13:47

1 Answer 1

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Yes, it is correct. You should have $2$ or $36$ dollars if you win, but there is a "tax" of $1$. That is a bigger fraction of $2$ than $36$.

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  • $\begingroup$ Ok. I was confused because if you are betting cash the expected value is the same for both bets: math.stackexchange.com/q/1475920/144957 $\endgroup$
    – Roland
    Oct 29, 2017 at 14:02
  • $\begingroup$ That is correct. Both bets are fair with your initial stake returned. Losing the return of the stake hurts more on the bet you win more. $\endgroup$ Oct 29, 2017 at 14:32
  • $\begingroup$ Suppose you won. You can still use the non redeemable chip for another bet. Does this change anything? $\endgroup$
    – Roland
    Oct 29, 2017 at 14:37
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    $\begingroup$ That is right. Now it doesn't matter which you bet. You can also bet in any pattern you like and the sum will be $1$ if you keep going until you lose the first time. $\endgroup$ Oct 30, 2017 at 13:16
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    $\begingroup$ Yes, you can. You can bet on red, which has expected value of $\frac 12$ and a chance of $\frac 12$ to continue. You can (if you win the first bet) bet on a number, which has expected value $\frac 12 \cdot \frac {35}{36}$ and a $\frac 1{36}$ chance to go on from here. You now only have $\frac 1{72}$ chance of getting a third bet. If you sum the expected values you will again get $1$ $\endgroup$ Oct 30, 2017 at 15:02

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