Chilli Problem

Aquafina · #1 $Old$ November 1st, 2009, 02:13 PM

Two people are playing a game which involves taking it in turns to eat chillies. There are 5 mild chillies and 1 hot chilli. Assuming the game is over when the hot chilli is eaten (and that I don't like hot chillies), is it a disadvantage to go first? What is the probability that I will eat the chilli if I go first? How about if there are 6 mild and 2 hot?

Isnt the probability with 1 chili 1/6 on the first go and 1/5 on the second.

With 2 chilli, 1/4 on the 1st go, 2/7 on the 2nd go.

So its better to go first both times?

awkward · #2 $Old$ November 1st, 2009, 07:26 PM

Quote:

Originally Posted by Aquafina $View Post$

Two people are playing a game which involves taking it in turns to eat chillies. There are 5 mild chillies and 1 hot chilli. Assuming the game is over when the hot chilli is eaten (and that I don't like hot chillies), is it a disadvantage to go first? What is the probability that I will eat the chilli if I go first? How about if there are 6 mild and 2 hot?

Isnt the probability with 1 chili 1/6 on the first go and 1/5 on the second.

With 2 chilli, 1/4 on the 1st go, 2/7 on the 2nd go.

So its better to go first both times?

Hi Aquafina,

If we think of the chilies as being numbered from 1 to 6 in order of eating, the hot one is equally likely to be 1, 2, 3, 4, 5, or 6. The first player gets it if if it is number 1, 3, or 5; the second player gets it if it is number 2, 4, or 6. So each player gets the hot chili with probability 1/2.

ANDS! · #3 $Old$ November 1st, 2009, 09:01 PM

There is only one hot chilli pepper in the first example. So it is better to go first than second.

Aquafina · #4 $Old$ November 1st, 2009, 11:34 PM

Quote:

Originally Posted by awkward $View Post$

Hi Aquafina,

If we think of the chilies as being numbered from 1 to 6 in order of eating, the hot one is equally likely to be 1, 2, 3, 4, 5, or 6. The first player gets it if if it is number 1, 3, or 5; the second player gets it if it is number 2, 4, or 6. So each player gets the hot chili with probability 1/2.

Hi, so are you saying they are numbered from before?

Quote:

Originally Posted by ANDS! $View Post$

There is only one hot chilli pepper in the first example. So it is better to go first than second.

I dont get the question. If I go first when there is 1 chilli, then obviously the probability will be 1/6, but the second time will be 1/5, as either the game would have ended, or 1 mild would have drawn, then the 3rd time would be 1/4 etc... So obviously its just the first time for each one on this reasoning?

Even with 2 chillis, theres less probability if we go first since there will be more mild (which haven't been drawn yet). Is this the reasoning you used?

Robb · #5 $Old$ November 2nd, 2009, 12:07 AM

Quote:

What is the probability that I will eat the chilli if I go first?

$Pr(\mbox{You eat first})=P(\mbox{eat hot on your first draw})+P(\mbox{eat hot on your second draw})$ $+P(\mbox{eat hot on your third draw})$ $=\frac{1}{6}+\frac{5}{6}\cdot \frac{4}{5}\cdot \frac{1}{4}+\frac{5}{6} \cdot \frac{4}{5}\cdot \frac{3}{4} \cdot \frac{2}{3}\cdot \frac{1}{2}=\frac{1}{2}$
And similiarly if you eat second, the probability of you eating it is 0.5, so no advantage/disadvantage if you go first or second as you are equally likely to eat the hot chilli

If there are 2 hot chillies, and 6 mild;
$Pr(\mbox{You go first and eat it})=\frac{2}{8}+\frac{6}{8} \cdot \frac{5}{7}\cdot \frac{2}{6}+\frac{6}{8} \cdot \frac{5}{7}\cdot \frac{4}{6}\cdot \frac{3}{5}\cdot \frac{2}{4}+\frac{6}{8} \cdot \frac{5}{7}\cdot \frac{4}{6}\cdot \frac{3}{5}\cdot \frac{2}{4}\cdot \frac{1}{3}\cdot 1=\frac{4}{7}$
So the probability of you eating the chili if you eat second will be $\frac{3}{7}$ so it is a disadvantage to go first when there are 2 hot and 6 mild

jt5271 · #6 $Old$ November 5th, 2009, 07:36 AM

Prob(eating hot chili on first go) = 1/6
Prob(eating hot chili on second go) = prob(eating mild first) AND prob(eating hot second) = 5/6 * 1/5 = 1/6

So no advantage in first problem to going second.

If there are 2 hot chilis and 6 mild;

Prob(eating hot first) = 1/4
Prob(eating hot second) = prob(eating mild first) AND prob(eating hot second) = 6/8 * 2/7 = 3/14

So there is an advantage to going second in second trial.

jt5271 · #7 $Old$ November 5th, 2009, 07:48 AM

Prob(eating hot chili on first go) = 1/6
Prob(eating hot chili on second go) = prob(eating mild first) AND prob(eating hot second) = 5/6 * 1/5 = 1/6

So no advantage in first problem to going second.

If there are 2 hot chilis and 6 mild;

Prob(eating hot first) = 1/4
Prob(eating hot second) = prob(eating mild first) AND prob(eating hot second) = 6/8 * 2/7 = 3/14

So there is an advantage to going second in second trial.