Re: The Cut-Card Effect

From Stanford Wong's BJ21

Posted by MathProf on 31 Jan 1999, 6:35 pm, in response to The Cut-Card Effect, posted by Border Crosser on 31 Jan 1999, 5:36 pm

To understand the cut-card effect, let us first consider a game in which 3 rounds are dealt. Period. And then we reshuffle. If we are a BS player, not-counting, then all of our odds on the second and third round are just like the first. Griffin has a proof of this in his text. On the first round, the probability of getting a 10 is 4/13; it is the same on the second and third.

Now let us suppose instead we play a game in which we deal to a certain pre-scribed shuffle point. Most of the time, we get 3 rounds. But sometimes we get a 4th.

Now the probabilities on the first three rounds are the same as on the first one. But what of the fourth? If all 4th rounds were dealt, then they would have the same probabilities. But they are not all dealt. They are only dealt when the first 3 rounds have consumed relatively few cards, and hence were rich in 10s. Thus the 4th round is dealt from a poor deck. This is the "cut-card effect".

Note that this means that the probability that you are a dealt a 10 is somewhat less than 4/13. It is 4/13 for the first 3 rounds, but it is less than it on the last round.

Now you are right that I did pose the question that, if BS is defined as the optimal strategy for the non-Counting player, should that be changed with a cut-card effect. BS assumes that 16/52 cards will be tens; which is obviouslynot quite the case with a fixed-shuffle point. But is also true that this a largely academic question.

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