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A man bets on number 16 on a roulette wheel 14 times and losses each time. On the $15^{th}$ span he does a quick calculation and finds out that the number 12 had appeared twice in the 14 spans and is therefore, unable to decide whether to bet on 16 or 12 in the $15^{th}$ span. Which will give him the best chance and what are the odds of winning on the bet that he takes? (Roulette has numbers 1 to 36).




$12; 72 : 1$


$12; 7 : 1$


Either $; 35 : 1$

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Option(D) is correct

Each of the span is an independent event and the outcome of the $15^{th}$ span will not depend on the outcome of the earlier spans.

(5) Comment(s)


I think the odds should be 1:35 as it represents the odds in favour of winning while 35:1 are odds against winning


@poonam, your explanation is wrong.

1/36 is the probability of getting any number ... so odds are 35:1 .


35 : 1 ????

plz explain?

Poonam Pipaliya

there is 36 numbers. and on 16 number he lost 14 times then it is obvious that he will not bet on 16. so eleminating 1(16) number he has 35 numbers. now total events are $^{36}C_1=36$ and favorable event is $^{35}C_1=35$

So $\text{prabability} = \dfrac{\text{fav events}}{\text{total events}}$

$=\dfrac{35}{36}$ this is an answer.

But now they asked for odds

Now check out all answers

ans D says that odds are $35:1=\dfrac{35}{35+1}=\dfrac{35}{36}$

Sushma Saroj

please explain to me.

Tell me how to find out the odds

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