Viewing a response to: @themarkymark/would-you-switch
I would say it would always be a 50/50 chance of picking the right door. In first try it would be 50/50 between winner pick or non-winner, even if one options has two doors left. Once you pick the wrong door, it is still 50/50 chance to pick the winner between remaining doors. Opened door becomes irrelevant at that point.
| author | geekgirl |
|---|---|
| permlink | qt9nod |
| category | hive-167922 |
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| created | 2021-05-17 19:33:48 |
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How could it be a 50/50 chance when you are picking one out of three on the first try? You are a Python dev, program it out, you will see it isn't 50/50 in either situation. Posted Using [LeoFinance <sup>Beta</sup>](https://leofinance.io/@themarkymark/re-geekgirl-54ucci)
| author | themarkymark |
|---|---|
| permlink | re-geekgirl-54ucci |
| category | hive-167922 |
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| created | 2021-05-17 19:37:03 |
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Because ultimately everything comes down to 0 vs 1, True or False. Especially in this case, since I can open only one door at a time, my options are: a. Door 1 vs (Door 2 + Door 3) b. Door 2 vs (Door 1 + Door 3) c. Door 3 vs (Door 2 + Door 3) Whichever I choose, I face with 50/50 chance. The only difference is, with the first try I have 50% chance to get the correct one, and 50% to get another opportunity to try. With the second try, I have 50% chance to get the correct one, and 50% to lose. But also, coins only have two sides. I would toss a coin to decide for me. And for that reason I need to make it 50/50. P.S. I misunderstood the puzzle at first. But I would still use the coin toss, and treat each step as 50/50 chance. So can't tell if I would switch or stay until I flip the coin.
| author | geekgirl |
|---|---|
| permlink | qt9qf0 |
| category | hive-167922 |
| json_metadata | {"app":"hiveblog/0.1"} |
| created | 2021-05-17 20:33:00 |
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| stemgeeks | 0 | 20,685,612,596 | 20% | ||
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