Probability of choosing the bakery with the best bread



Let's say this: You want to buy the best bread in town, and only two stores are offering what you want. Both stores far away from each other, and you don't have time to make the trip to one and the other. The thing is, for some bakery reasons, from time to time, ''Store A'' have the best bread, and some other time it's ''Store B''.


So, since you can't go to both stores to test them out, you have to make a choice. You can either play Heads or Tails and have a 0.5 of success, or you decide to call contacts to help you choose. You call 5 of them.


1st person has 40% of chances to give you the right store, the 2nd one has 35%, 3rd 36%, 4th 42% and the last one a mighty 43%.


After calling all of them, you are surprised to see that ALL of them told you to go to "Store B".


My question: Should I roll a dice, or should I choose "Store B" even if all the percentages are lower than 50%? If "Store B" it is, how to collectively get all those percentages to work in my favor?


Sorry for the bounty, it's 29.96$, not 30, hope you can forgive me for that missing 0.04 


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Kav10 Kav10
  • Hi, I have to say it's really a clean document, and I really appreciate the effort you've put in it. I'm simply not sure how come asking 5 people will diminish the chances down to 10%, because in this case not only I will choose to roll a dice indeed, but by asking only one person will be increased at least 20% more chances than the combined of them 5, which is countertuitive in my perspective.

    • Kav10 Kav10

      Hi. I am glad you liked the effort. Thanks for the coffee. The reason it might sound counterintuitive is that the chance of all 5 friends saying the same thing is low. Couple that with all of them saying the correct bakery with less than 50% chance. The definitions are really important here. What I have calculated for you is the chance of B being the best bakery, given that all friends say bakery B is the best. The conditional probability, dependent on all friends saying its B.

  • Kav10 Kav10

    If you see any flaws in the approach and/or any miscalculations or typos, let me know. We can discuss it more, as needed.

  • Kav10 Kav10

    Looking at this from another angle, if we were to calculate the probability that at least one of your friends is correct in saying that Bakery B is the best, I would expect that to be significantly higher than 50%. However, based on the wording of your question, the probability that bakery B is the best bakery given that ALL friends say the same thing is calculated.

  • I see, and you're right for the wording I've used, it was mostly to determine if it was possible to get more than 50% of chances even with those lower percentages by anyway possible way to add them all up. I might have worded this wrong as a matter of fact. So do you think there's a way to have more than 50% of chances to get the best bakery considering my available datas? Maybe we can forget in that case all of them choose bakery B, just keep the remaining percentages.

  • This is mostly what I want to know. Thank's.

  • Kav10 Kav10

    Sure. That would be a separate question. To calculate the probability that at least one of your friends gives you the right store, we can calculate (1 - the probability that none of your friends give you the right store), or in other words P’=1-P. The probability of the first friend not giving you the right store is 60% (100% - 40%). For your other friends this is 65%, 64%, 58%, and 57%.

    • Kav10 Kav10

      Therefore, the probability of at least one friend giving you the right store is: 1 - (0.6 x 0.65 x 0.64 x 0.58 x 0.57) = 0.917 or approximately 92%. So you have a high probability of getting the right store if you ask your friends for help.

  • Thx very much. Have a nice day!

    • Kav10 Kav10

      Of course! You too.

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