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December 3

A question about odds

I’m a relatively simple man when it comes to mathematics. I can understand that if you flip a coin the odds of heads are 1 in 2. And I’ve a vague understanding of the odds of flipping a coin ten times in a row and the odds of all ten being heads and how it’s 50/50 each time but not 50/50 for all 10 flips. I think?

But the ten coin flips depends presumably on the fact that after flipping a tail, the flipper can continue to flip again up to ten flips.

If you transpose the calculation to a scenario whereby flipping “tails” ends the game, is there a way to calculate the odds or does it change?

my example would be Russian roulette with a two chamber pistol and one round. On the assumption that each chamber is equally likely to come up, I assume the first shot is 50/50 chance however is the odds across ten shots also affected by the 50/50 chance of each shot killing the participant and therefore the remaining shots not being completed? Is the odds of the 10th shot some multiple the original 50/50 odds, the odds of ten empty chambers, and the probability of surviving for each individual shot before? 2A02:C7E:2EC1:8D00:F005:18BD:4756:D126 (talk) 19:34, 3 December 2023 (UTC)[reply]

The odds for a player who is still in the game remain the same at each round. The coin (or pistol) does not "know" what happened before; it has no memory and cannot adjust its behaviour based on its earlier behaviour. The belief that such a random system somehow remembers the past is known as the gambler's fallacy.

The version in which failure to reach a specific outcome (like ten times "heads") ends the game has exactly the same probability of success as the version in which the failing player plods on to the end, even though the outcome is already fixed. Here is a way to see this, if it is not intuitively obvious. Imagine a gazillion of players are playing this synchronously. Each has a light bulb that is initially on but turns off and remains off on failure. After the first round, half the lights will go off. At the next round, half the remaining lights will go off, so one quarter is still on. Each time the number halves, so after 10 rounds only a fraction 1/2 × 1/2 × 1/2 × 1/2 × 1/2 × 1/2 × 1/2 × 1/2 × 1/2 × 1/2 = 1/1024 of the lights is still on. From this description, it should be obvious that it does not make a difference whether a player whose light bulb was turned off keeps playing or goes home. --Lambiam 21:54, 3 December 2023 (UTC)[reply]

so I suppose the odds of someone making ten blank fires in a row is the same as the odds of someone surviving ten rounds of Russian roulette given that blank fire and survival are the same outcome, so it’s immaterial if a player exits on one live round or flips a coin to tails if the objective is ten in a row of one outcome. The objective if failed if the undesired outcome is met regardless of whether it’s coin coming up tails or the coin going down their throat and choking them. Interesting thank you. — Preceding unsigned comment added by 2A02:C7E:2EC1:8D00:F005:18BD:4756:D126 (talk) 00:17, 4 December 2023 (UTC)[reply]