5.3 MM's

From various sources I have found the fraction of chocolate M&Ms candies are red. The sources found are the following:

• Source A: 28% of M&Ms are red, 20% of M&Ms are orange.
• Source B: 20% of M&Ms are red, 10% of M&Ms are orange
• Source C: 13% of M&Ms are red, 21% of M&Ms are orange.

From actually counting of a bag of M&Ms I found the following data:

• 3 red M&Ms in 17 total ($R=3, N=17$)

The question is, which source can we trust the most? Here we follow Bayes' recipe,

• Specify the prior probabilities for the models being considered

  $$\begin{eqnarray*} P(A)=P(B)=P(C)=1/3 \end{eqnarray*}$$

• Write the top of Bayes' Rule (i.e. likelihood $\times$ prior) for all models being considered

  $$\begin{eqnarray*} P(A|R=3, N=17) &\sim& \binom{17}{3} 0.28^{3}(1-0.28)^{17-3}\times \frac{1}{3}\\ P(B|R=3, N=17) &\sim& \binom{17}{3} 0.20^{3}(1-0.20)^{17-3}\times \frac{1}{3}\\ P(C|R=3, N=17) &\sim& \binom{17}{3} 0.13^{3}(1-0.13)^{17-3}\times \frac{1}{3} \end{eqnarray*}$$

• Add these values for all models, to get $K$

  $$\begin{eqnarray*} P(A|R=3, N=17) &\sim& 0.05006\\ &&+\\ P(B|R=3, N=17) &\sim& 0.07975\\ &&+\\ P(C|R=3, N=17) &\sim& 0.07087 \\ \cline{1-3} K&=& 0.20068 \end{eqnarray*}$$

• Divide each of the values by this sum, $K$, to get the final probabilities

  $$\begin{eqnarray*} P(A|R=3, N=17) &=& 0.05006/0.20068 = 0.250\\ P(B|R=3, N=17) &=& 0.07975/0.20068 = 0.397\\ P(C|R=3, N=17) &=& 0.07087/0.20068 = 0.353 \end{eqnarray*}$$

So we are most confident in Source B, although none of them really changed by a lot - there is no clear winner.

Updating with other data

• 5 orange M&Ms in 16 total ($G=5, N=16$)

Again, we follow the same recipe, starting with out posterior probabilities from above as our starting prior probabilities - they are prior to the new data.

• Specify the prior probabilities for the models being considered

  $$\begin{eqnarray*} P(A|{\rm old\ data}) &=& 0.250\\ P(B|{\rm old\ data}) &=& 0.07975/0.20068 = 0.397\\ P(C|{\rm old\ data}) &=& 0.07087/0.20068 = 0.353 \end{eqnarray*}$$

• Write the top of Bayes' Rule (i.e. likelihood $\times$ prior) for all models being considered

  $$\begin{eqnarray*} P(A|G=5, N=16 \text{ \textbf{and} old data}) &\sim& \binom{16}{5} 0.20^{5}(1-0.20)^{16-5}\times 0.250\\ P(B|G=5, N=16 \text{ \textbf{and} old data}) &\sim& \binom{16}{5} 0.10^{5}(1-0.10)^{16-5}\times 0.397\\ P(C|G=5, N=16 \text{ \textbf{and} old data}) &\sim& \binom{16}{5} 0.21^{5}(1-0.21)^{16-5}\times 0.353 \end{eqnarray*}$$

• Add these values for all models, to get $K$

  $$\begin{eqnarray*} P(A|{\rm data}) &\sim& 0.0300\\ &&+\\ P(B|{\rm data}) &\sim& 0.00544\\ &&+\\ P(C|{\rm data}) &\sim& 0.0471 \\ \cline{1-3} K&=& 0.08254 \end{eqnarray*}$$

• Divide each of the values by this sum, $K$, to get the final probabilities

  $$\begin{eqnarray*} P(A|{\rm data}) &=& 0.0300/0.08254 = 0.363\\ P(B|{\rm data}) &\sim& 0.00544/0.08254 = 0.0659\\ P(C|{\rm data}) &\sim& 0.0471/0.08254 = 0.5706 \end{eqnarray*}$$

Given this new data, we update our state of knowledge, and we're much more confident that Source C is the best one. It is clear that Source B is unlikely, with a probability of only about 6.5%. We could extend this example with more data, and more models if we'd like.

Adapted from Statistical Inference for Everyone, by Brian Blais (Bryant University), licensed under CC BY-SA 4.0 (dual-licensed under the GNU FDL 1.2 or later; this adaptation uses the CC BY-SA grant). Changes were made; this adaptation is distributed under the same license. License: CC-BY-SA-4.0.